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17. Vector Calculus with Applications

17.1 INTRODUCTION

In vector calculus, we deal with two types of functions: Scalar Functions (or Scalar Field) and

Vector Functions (or Vector Field).

Scalar Point Function

A scalar function 𝐹(𝑥, 𝑦, 𝑧)defined over some region R of space is a function which associates, to

each point 𝑃(𝑥, 𝑦, 𝑧)in R, a scalar value 𝐹 𝑃 = 𝐹(𝑥, 𝑦, 𝑧). And the set of all scalars 𝐹(𝑃)for all

values of P in R is called the scalar field over R.

Precisely, we can say that scalar function defines a scalar field in a region or on a space or a curve.

Examples are the temperature field in a body, pressure field in the air in earth’s atmosphere.

Moreover, if the position vector of the point P is 𝑟 , then we may also write the scalar field as

𝐹 𝑃 = 𝐹(𝑟 ). This notation emphasizes the fact that the scalar value 𝐹(𝑟 ) is associated with the

position vector 𝑟 in the region R.

E.g. 1) The distance 𝐹 𝑃 of any point 𝑃(𝑥, 𝑦, 𝑧) from a fixed point 𝑃′(𝑥 ′, 𝑦′, 𝑧′) in the space is a

scalar function whose domain of definition is the whole space and is given by

𝐹 𝑃 = (𝑥 − 𝑥 ′)2 + (𝑦 − 𝑦 ′)2 + (𝑧 − 𝑧 ′)2. Also 𝐹 𝑃 defines a scalar field in space.

E.g. 2) The function 𝐹 𝑥, 𝑦, 𝑧 = 𝑥𝑦2 + 𝑦𝑧 + 𝑥2 for the point (𝑥, 𝑦, 𝑧) inside the unit sphere

𝑥2 + 𝑦2 + 𝑧2 = 1 is a scalar function and also is defines a scalar field throughout the sphere.

Note: In the physical problems, the scalar function F depends on time variable t in addition to the

point P and then we write it as 𝐹 𝑃, 𝑡 = 𝐹 𝑟 , 𝑡 = 𝐹(𝑥, 𝑦, 𝑧, 𝑡) . The example of such a time

dependent scalar function is the temperature distribution throughout a block of metal heated in such

a way that its temperature varies with time.

Vector Point Function

A vector functions 𝐹 𝑥, 𝑦, 𝑧 defined over some region R of space is a function which associates, to

each point 𝑃(𝑥, 𝑦, 𝑧) in R, a vector value 𝐹 𝑃 = 𝐹 (𝑥, 𝑦, 𝑧) and the set of all vectors 𝐹 𝑃 for all

points P in R is called the vector field over R

Moreover, if the position vector of the point P is 𝑟 , then we may write the vector field as

𝐹 𝑃 = 𝐹 (𝑟 ). This notation emphasizes the fact that the vector value 𝐹 (𝑟 ) is associated with the

position vector 𝑟 in the region R. Also the general form (component form) of the vector function is

𝐹 𝑟 = 𝐹1 𝑟 𝑖 + 𝐹2 𝑟 𝑗 + 𝐹3 𝑟 𝑘 , where the components 𝐹1 𝑟 , 𝐹2 𝑟 and 𝐹3 𝑟 are the scalar

functions.

E.g. 1) The function 𝐹 𝑥, 𝑦, 𝑧 = 2𝑥𝑦𝑖 + sin 𝑥 𝑗 + 3𝑧2𝑘 for point P( x, y, z ) inside an ellipsoid

𝑥2

9+

𝑦2

16+

𝑧2

4 = 1 is a vector function and defines a vector field throughout the ellipsoid.

E.g. 2) The force field given by 𝐹 𝑥, 𝑦, 𝑧 = 𝑥𝑖 + 2𝑦𝑗 + 𝑧2𝑘 is a vector field.

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Note: Like the time dependent scalar field, time dependent vector field also exists. Such a field depends on

time variable t in addition to the point in the region R and may be expressed as

𝐹 𝑟 , 𝑡 = 𝐹1 𝑟 , 𝑡 𝑖 + 𝐹2 𝑟 , 𝑡 𝑗 + 𝐹3 𝑟 , 𝑡 𝑘 , where 𝐹1 , 𝐹2 and 𝐹3 are scalar functions. An example of time

dependent vector field is the fluid velocity vector in the unsteady flow of water around a bridge support

column, because this velocity depends on the position vector 𝑟 in the water and the time variable t and is

given as 𝑉 (𝑟 , 𝑡).

Vector Function of Single Variable

A vector function 𝐹 of single variable t is a function which assigns a vector value 𝐹 (𝑡) to each

scalar value t in interval 𝑎 ≤ 𝑡 ≤ 𝑏 . In the component form, it may be written as

𝐹 𝑡 = 𝐹1 𝑡 𝑖 + 𝐹2 𝑡 𝑗 + 𝐹3 𝑡 𝑘 where 𝐹1, 𝐹2 and 𝐹3 are called components and are scalar functions

of the same single variable t.

For example, the functions given by𝐹 𝑡 = 𝑡 𝑖 + sin(𝑡 − 2) 𝑗 + cos 3𝑡 𝑘 and 𝐺 𝑡 = 𝑡2𝑖 + 𝑒𝑡𝑗 + log 𝑡 𝑘

are vector functions of a single variable t.

Limit of a Vector Function of Single Variable

A vector function 𝐹 𝑡 = 𝐹1 𝑡 𝑖 + 𝐹2 𝑡 𝑗 + 𝐹3(𝑡)𝑘 of single variable t is said to have a limit

𝐿 = 𝐿1𝑖 +𝐿2𝑗 +𝐿3𝑘 as 𝑡 → 𝑡0 , if 𝐹 𝑡 is defined in the neighborhood of t0 and Lt𝑡→𝑡0 𝐹 𝑡 − 𝐿 = 0

or Lt𝑡→𝑡0

𝐹1 𝑡 − 𝐿1 = Lt𝑡→𝑡0 𝐹2 𝑡 − 𝐿2 = Lt𝑡→𝑡0

𝐹3 𝑡 − 𝐿3 = 0, then we write it as Lt𝑡→𝑡0𝐹 (𝑡) = 𝐿 .

Continuity of a Vector Function of Single Variable

A vector function 𝐹 𝑡 = 𝐹1 𝑡 𝑖 + 𝐹2 𝑡 𝑗 + 𝐹3(𝑡)𝑘 of a single variable t is said to be continuous at

t = t0, if it is defined in some neighborhood of t0 and Lt𝑡→𝑡0𝐹 𝑡 = 𝐹 𝑡0 .

Moreover, 𝐹 𝑡 is said to be continuous at 𝑡 = 𝑡0 if and only if its three components F1, F2 and F3

are continuous as 𝑡 = 𝑡0.

DIFFERENTIAL VECTOR CALCULUS

17.2 DIFFERENTIATION OF VECTORES

Differentiability of a Vector Function of Single Variable

A vector function 𝐹 𝑡 = 𝐹1 𝑡 𝑖 + 𝐹2 𝑡 𝑗 + 𝐹3(𝑡)𝑘 of a single variable t defined over the interval

𝑎 ≤ 𝑡 ≤ 𝑏 is said to be differentiable at t = t0 if the following limit exists.

Lt𝑡→𝑡0

𝐹 𝑡 −𝐹 (𝑡0)

𝑡−𝑡0 = 𝐹 ′(𝑡0)

And 𝐹 ′(𝑡0) is called the derivative of 𝐹 𝑡 at t = t0.

Also 𝐹 𝑡 is said to be differentiable over the interval 𝑎 ≤ 𝑡 ≤ 𝑏, if it is differentiable at each of the

points of the interval. In component form, 𝐹 𝑡 is said to be differentiable at t = t0 if and only if its

three components are differentiable at t = t0. In general, the derivative of 𝐹 (𝑡) is given by

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𝐹 ′ 𝑡 = Lt𝑡→𝑡0

𝐹 𝑡+∆𝑡 −𝐹 (𝑡)

∆𝑡, provided the limit exists and in terms of components

𝐹 ′ 𝑡 = 𝐹1′ 𝑡 𝑖 + 𝐹2

′ 𝑡 𝑗 + 𝐹3′ (𝑡)𝑘 or

𝑑𝐹

𝑑𝑡=

𝑑𝐹1

𝑑𝑡𝑖 +

𝑑𝐹2

𝑑𝑡𝑗 +

𝑑𝐹3

𝑑𝑡𝑘 .

In the similar manner, 𝑑2𝐹

𝑑𝑡 2=

𝑑

𝑑𝑡 𝑑𝐹

𝑑𝑡 ,

𝑑3𝐹

𝑑𝑡 3=

𝑑

𝑑𝑡

𝑑2𝐹

𝑑𝑡 2 =𝑑2

𝑑𝑡 2 𝑑𝐹

𝑑𝑡 .

Rules for Differentiation of Vector Functions

If 𝐹 𝑡 , 𝐺 𝑡 & 𝐻 𝑡 are the vector functions and 𝑓(𝑡) is a scalar function of single variable 𝑡 defined

over the interval 𝑎 ≤ 𝑡 ≤ 𝑏, then

1. 𝑑𝐶

𝑑𝑡 = 0 , where 𝑐 is a constant vector.

2. 𝑑 𝐶 𝐹 𝑡

𝑑𝑡 = 𝐶

𝑑𝐹

𝑑𝑡, where C is a constant.

3. 𝑑 𝐹 𝑡 ±𝐺 (𝑡)

𝑑𝑡=

𝑑𝐹

𝑑𝑡±

𝑑𝐺

𝑑𝑡

4. 𝑑 𝑓 𝑡 𝐹 𝑡

𝑑𝑡 = 𝑓 𝑡

𝑑𝐹

𝑑𝑡+

𝑑𝑓

𝑑𝑡 𝐹 (𝑡)

5. 𝑑 𝐹 𝑡 ∙ 𝐺 (𝑡)

𝑑𝑡 =

𝑑𝐹

𝑑𝑡 ∙ 𝐺 (𝑡) + 𝐹 (𝑡) ∙

𝑑𝐺

𝑑𝑡

6. 𝑑 𝐹 𝑡 ×𝐺 (𝑡)

𝑑𝑡 =

𝑑𝐹

𝑑𝑡 × 𝐺 (𝑡) + 𝐹 (𝑡) ×

𝑑𝐺

𝑑𝑡

7. 𝑑

𝑑𝑡 𝐹 𝑡 , 𝐺 𝑡 , 𝐻 (𝑡) =

𝑑𝐹

𝑑𝑡, 𝐺 𝑡 , 𝐻 (𝑡) + 𝐹 𝑡 ,

𝑑𝐺

𝑑𝑡, 𝐻 (𝑡) + 𝐹 𝑡 , 𝐺 𝑡 ,

𝑑𝐻

𝑑𝑡

8. 𝑑

𝑑𝑡 𝐹 𝑡 × 𝐺 𝑡 × 𝐻 𝑡 =

𝑑𝐹

𝑑𝑡× 𝐺 𝑡 × 𝐻 (𝑡) + 𝐹 𝑡 ×

𝑑𝐺

𝑑𝑡× 𝐻 (𝑡) + 𝐹 𝑡 × 𝐺 𝑡 ×

𝑑𝐻

𝑑𝑡

9. If 𝐹 𝑡 is differentiable function of 𝑡 and 𝑡 = 𝑡(𝑠) is differentiable function then

𝑑𝐹

𝑑𝑠=

𝑑𝐹

𝑑𝑡 𝑑𝑡

𝑑𝑠 .

Observations:

(i) If 𝐹 (𝑡) has a constant magnitude, then 𝐹 ∙ 𝑑𝐹

𝑑𝑡 = 0. For 𝐹 𝑡 ∙ 𝐹 𝑡 = 𝐹 (𝑡)

2= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,

implying 𝐹 ∙ 𝑑𝐹

𝑑𝑡 = 0 or 𝐹 ⊥

𝑑𝐹

𝑑𝑡 .

(ii) If 𝐹 (𝑡) has a constant (fixed) direction, then 𝐹 × 𝑑𝐹

𝑑𝑡 = 0 .

Let 𝐹 𝑡 = 𝑓(𝑡)𝐺 (𝑡), where 𝐺 (𝑡) is a unit vector in the direction of 𝐹 (𝑡).

∴ dF

dt=

𝑑 𝑓 𝑡 𝐺 𝑡

𝑑𝑡 = 𝑓 𝑡

𝑑𝐺

𝑑𝑡+

𝑑𝑓

𝑑𝑡 𝐺 𝑡 =

𝑑𝑓

𝑑𝑡 𝐺 𝑡 𝑠𝑖𝑛𝑐𝑒, 𝐺 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑠𝑜

𝑑𝐺

𝑑𝑡= 0

and 𝐹 × 𝑑𝐹

𝑑𝑡= 𝑓(𝑡)𝐺 (𝑡) ×

𝑑𝑓

𝑑𝑡 𝐺 𝑡 = 𝑓 𝑡

𝑑𝑓

𝑑𝑡 𝐺 𝑡 × 𝐺 𝑡 = 0 𝑠𝑖𝑛𝑐𝑒, 𝐺 × 𝐺 = 0

Theorem 1: Derivative of a constant vector is a zero vector. A vector is said to be constant if

both its magnitude and direction are constant (fixed).

Proof: Let 𝑟 = 𝑐 be a constant vector, then 𝑟 + 𝛿𝑟 = 𝑐 .

On subtraction, 𝛿𝑟 = 0 .Which further implies that 𝛿𝑟

𝛿𝑡 = 0 .

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Implying,𝐿𝑡

𝛿𝑡 → 0 𝛿𝑟

𝛿𝑡 = 0 i.e.

𝑑𝑟

𝑑𝑡 = 0 .

Theorem 2: The necessary and sufficient condition for the vector function 𝑭 of a single

variable t to have constant magnitude is 𝑭 ∙ 𝒅𝑭

𝒅𝒕 = 𝟎.

Proof:

Necessary condition: Suppose 𝐹 has constant magnitude, so 𝐹 𝑡 ∙ 𝐹 𝑡 = 𝐹 (𝑡) 2

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

=> 𝑑

𝑑𝑡 𝐹 ∙ 𝐹 = 0 i.e. 𝐹 ∙

𝑑𝐹

𝑑𝑡 +

𝑑𝐹

𝑑𝑡 ∙ 𝐹 = 0

=> 2𝐹 ∙ 𝑑𝐹

𝑑𝑡 = 0 i.e. 𝐹 ∙

𝑑𝐹

𝑑𝑡 = 0.

Sufficient condition: Suppose 𝐹 ∙ 𝑑𝐹

𝑑𝑡 = 0 => 2𝐹 ∙

𝑑𝐹

𝑑𝑡 = 0

=> 𝐹 ∙ 𝑑𝐹

𝑑𝑡 +

𝑑𝐹

𝑑𝑡 ∙ 𝐹 = 0 =>

𝑑

𝑑𝑡 𝐹 ∙ 𝐹 = 0

=> 𝐹 ∙ 𝐹 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 => 𝐹 2

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Therefore 𝐹 has a constant magnitude.

Theorem 3: The necessary and sufficient condition for the vector function 𝑭 of a single

variable t to have a constant direction is 𝑭 × 𝒅𝑭

𝒅𝒕 = 𝟎 .

Proof: Suppose that 𝑓 is a unit vector in the direction of 𝐹 and 𝐹 = 𝐹 , then 𝑓 = 𝐹

𝐹 i.e.

𝐹 = 𝐹𝑓 … (1)

And 𝑑𝐹

𝑑𝑡 = 𝐹

𝑑𝑓

𝑑𝑡+

𝑑𝐹

𝑑𝑡 𝑓 … (2)

Thus 𝐹 × 𝑑𝐹

𝑑𝑡 = 𝐹𝑓 × (𝐹

𝑑𝑓

𝑑𝑡+

𝑑𝐹

𝑑𝑡 𝑓 ) (using (1) and (2))

= 𝐹2𝑓 × 𝑑𝑓

𝑑𝑡 +𝐹

𝑑𝐹

𝑑𝑡 (𝑓 × 𝑓 )

= 𝐹2𝑓 × 𝑑𝑓

𝑑𝑡 (since 𝑓 × 𝑓 = 0 ) … (3)

Necessary condition: Suppose 𝐹 has a constant direction, then 𝑓 has a constant direction and

constant magnitude. So 𝑑𝑓

𝑑𝑡 = 0 . Thus from (3), 𝐹 ×

𝑑𝐹

𝑑𝑡 = 0 .

Sufficient condition : Suppose that 𝐹 × 𝑑𝐹

𝑑𝑡 = 0 .

Then by (3), 𝐹2𝑓 × 𝑑𝑓

𝑑𝑡 = 0 i.e. 𝑓 ×

𝑑𝑓

𝑑𝑡 = 0 … (4)

Since 𝑓 has a constant magnitude, so, by theorem 2, 𝑓 ∙ 𝑑𝑓

𝑑𝑡 = 0 … (5)

Form (4) and (5), 𝑑𝑓

𝑑𝑡 = 0.

Which implies 𝑓 is a constant vector i.e. 𝑓 has a constant direction. Hence 𝐹 has a constant

direction.

Example 1: Show that if 𝒓 = 𝒂 𝐬𝐢𝐧𝝎𝒕 + 𝒃 𝐜𝐨𝐬 𝝎𝒕 where 𝒂 , 𝒃 and 𝝎 are constants, then

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𝒅𝟐𝒓

𝒅𝒕𝟐 = −𝝎𝟐𝒓 and 𝒓 ×

𝒅𝒓

𝒅𝒕 = −𝝎(𝒂 × 𝒃) .

Solution: Given 𝑟 = 𝑎 sin 𝜔𝑡 + 𝑏 cos 𝜔𝑡

Differentiating w. r. to t, 𝑑𝑟

𝑑𝑡 = 𝑎 𝜔 cos 𝜔𝑡 − 𝑏 𝜔 sin 𝜔𝑡

Again differentiating w. r. to t, 𝑑2𝑟

𝑑𝑡 2 = −𝑎 𝜔2 sin 𝜔𝑡 − 𝑏 𝜔2 cos 𝜔𝑡

= − 𝜔2 𝑎 sin 𝜔𝑡 + 𝑏 cos 𝜔𝑡 = −𝜔2𝑟

Also 𝑟 × 𝑑𝑟

𝑑𝑡 = 𝑎 sin 𝜔𝑡 + 𝑏 cos 𝜔𝑡 × (𝑎 𝜔 cos 𝜔𝑡 − 𝑏 𝜔 sin 𝜔𝑡)

= 𝑎 × 𝑎 𝜔 sin 𝜔𝑡 cos 𝜔𝑡 + 𝑏 × 𝑎 𝜔 cos2 𝜔𝑡 − 𝑎 × 𝑏 𝜔 sin2 𝜔𝑡 − 𝑏 × 𝑏 𝜔 sin 𝜔𝑡 cos 𝜔𝑡

= − 𝑎 × 𝑏 𝜔(cos2 𝜔𝑡 + sin2 𝜔𝑡) (since 𝑎 × 𝑎 = 𝑏 × 𝑏 = 0 )

= − 𝑎 × 𝑏 𝜔 = −𝜔(𝑎 × 𝑏 ).

Example 2: If 𝒂 = 𝒙𝟐𝒚𝒛 𝒊 − 𝟐𝒙𝒛𝟑 𝒋 + 𝒙𝒛𝟐 𝒌 and 𝒃 = 𝟐𝒛 𝒊 + 𝒚 𝒋 − 𝒙𝟐 𝒌 , find 𝝏𝟐

𝝏𝒙𝝏𝒚 𝒂 × 𝒃 at

(1, 0, -2).

Solution: Here 𝑎 × 𝑏 = 𝑥2𝑦𝑧 𝑖 − 2𝑥𝑧3 𝑗 + 𝑥𝑧2 𝑘 × 2𝑧 𝑖 + 𝑦 𝑗 − 𝑥2 𝑘

= 𝑥2𝑦2𝑧 𝑘 + 𝑥4𝑦𝑧 𝑗 + 4𝑥𝑧4 𝑘 + 2𝑥3𝑧3 𝑖 + 2𝑥𝑧3 𝑗 − 𝑥𝑦𝑧2 𝑖

(∵ 𝑖 × 𝑖 = 𝑗 × 𝑗 = 𝑘 × 𝑘 = 0 𝑎𝑛𝑑 𝑖 × 𝑗 = 𝑘 , 𝑗 × 𝑖 = −𝑘 𝑒𝑡𝑐.)

= 2𝑥3𝑧3 − 𝑥𝑦𝑧2 𝑖 + 𝑥4𝑦𝑧 + 2𝑥𝑧3 𝑗 + 𝑥2𝑦2𝑧 + 4𝑥𝑧4 𝑘

Now 𝜕2

𝜕𝑥𝜕𝑦 𝑎 × 𝑏 =

𝜕2

𝜕𝑥𝜕𝑦 2𝑥3𝑧3 − 𝑥𝑦𝑧2 𝑖 + 𝑥4𝑦𝑧 + 2𝑥𝑧3 𝑗 + 𝑥2𝑦2𝑧 + 4𝑥𝑧4 𝑘

= 𝜕

𝜕𝑥

𝜕

𝜕𝑦 2𝑥3𝑧3 − 𝑥𝑦𝑧2 𝑖 + 𝑥4𝑦𝑧 + 2𝑥𝑧3 𝑗 + 𝑥2𝑦2𝑧 + 4𝑥𝑧4 𝑘

= 𝜕

𝜕𝑥 −𝑥𝑧2 𝑖 + 𝑥4𝑧 𝑗 + 2𝑥2𝑦𝑧 𝑘 = −𝑧2 𝑖 + 4𝑥3𝑧 𝑗 + 4𝑥𝑦𝑧 𝑘

At the point (1, 0, -2) 𝜕2

𝜕𝑥𝜕𝑦 𝑎 × 𝑏 = −4𝑖 − 8𝑗

Example 3: If 𝑷 = 𝟓𝒕𝟐𝒊 + 𝒕𝟑𝒋 − 𝒕𝒌 and 𝑸 = 𝟐 𝐬𝐢𝐧 𝒕 𝒊 − 𝐜𝐨𝐬 𝒕 𝒋 + 𝟓𝒕𝒌 , then find (a) 𝒅

𝒅𝒕 𝑷 ∙ 𝑸

(b) 𝒅

𝒅𝒕 𝑷 × 𝑸 .

Solution: Consider 𝑃 = 5𝑡2𝑖 + 𝑡3𝑗 − 𝑡𝑘 and 𝑄 = 2 sin 𝑡 𝑖 − cos 𝑡 𝑗 + 5𝑡𝑘

So 𝒅𝑃

𝒅𝒕 = 10𝑡 𝑖 + 3𝑡2𝑗 − 𝑘 and

𝒅𝑄

𝒅𝒕 = 2 cos 𝑡 𝑖 + sin 𝑡 𝑗 + 5𝑘

a) 𝑑

𝑑𝑡 𝑃 ∙ 𝑄 =

𝒅𝑃

𝒅𝒕 ∙ 𝑄 + 𝑃 ∙

𝒅𝑄

𝒅𝒕

= 10𝑡 𝑖 + 3𝑡2𝑗 − 𝑘 ∙ 2 sin 𝑡 𝑖 − cos 𝑡 𝑗 + 5𝑡𝑘

+ 5𝑡2𝑖 + 𝑡3𝑗 − 𝑡𝑘 ∙ 2 cos 𝑡 𝑖 + sin 𝑡 𝑗 + 5𝑘

= 20 𝑡 sin 𝑡 − 3𝑡2 cos 𝑡 − 5𝑡 + 10 𝑡2 cos 𝑡 + 𝑡3 sin 𝑡 − 5𝑡

= 𝑡3 sin 𝑡 + 7𝑡2 cos 𝑡 + 20 𝑡 sin 𝑡 − 10 𝑡

6

b) 𝑑

𝑑𝑡 𝑃 × 𝑄 =

𝒅𝑃

𝒅𝒕 × 𝑄 + 𝑃 ×

𝒅𝑄

𝒅𝒕

= 10𝑡 𝑖 + 3𝑡2𝑗 − 𝑘 × 2 sin 𝑡 𝑖 − cos 𝑡 𝑗 + 5𝑡𝑘

+ 5𝑡2𝑖 + 𝑡3𝑗 − 𝑡𝑘 × 2 cos 𝑡 𝑖 + sin 𝑡 𝑗 + 5𝑘

= 𝑖 𝑗 𝑘

10 𝑡 3𝑡2 −12 sin 𝑡 − cos 𝑡 5𝑡

+ 𝑖 𝑗 𝑘

5𝑡2 𝑡3 −𝑡2 cos 𝑡 sin 𝑡 5

= 𝑖 15𝑡3 − cos 𝑡 + 𝑗 −2 sin 𝑡 − 50 𝑡2 + 𝑘 −10𝑡 cos 𝑡 − 6𝑡2 sin 𝑡

+ 𝑖 5𝑡3 + 𝑡 sin 𝑡 + 𝑗 −2𝑡 cos 𝑡 − 25 𝑡2 + 𝑘 5𝑡2 sin 𝑡 − 2𝑡3 cos 𝑡

= 𝑖 20𝑡3 + 𝑡 sin 𝑡 − cos 𝑡 − 𝑗 2𝑡 cos 𝑡 + 2 sin 𝑡 + 75 𝑡2

−𝑘 2𝑡3 cos 𝑡 + 10𝑡 cos 𝑡 + 𝑡2 sin 𝑡

Example 4: If 𝒅𝑼

𝒅𝒕 = 𝑾 × 𝑼 and

𝒅𝑽

𝒅𝒕 = 𝑾 × 𝑽 , then prove that

𝒅

𝒅𝒕 𝑼 × 𝑽 = 𝑾 × (𝑼 × 𝑽 ).

Solution: Given 𝑑𝑈

𝑑𝑡 = 𝑊 × 𝑈 and

𝑑𝑉

𝑑𝑡 = 𝑊 × 𝑉 … (1)

Consider 𝑑

𝑑𝑡 𝑈 × 𝑉 =

𝑑𝑈

𝑑𝑡 × 𝑉 + 𝑈 ×

𝑑𝑉

𝑑𝑡 = 𝑊 × 𝑈 × 𝑉 + 𝑈 × 𝑊 × 𝑉

= 𝑊 ∙ 𝑉 𝑈 − 𝑈 ∙ 𝑉 𝑊 + 𝑈 ∙ 𝑉 𝑊 − 𝑈 ∙ 𝑊 𝑉 = 𝑊 ∙ 𝑉 𝑈 − 𝑈 ∙ 𝑊 𝑉

= 𝑊 × (𝑈 × 𝑉 )

𝑢𝑠𝑖𝑛𝑔 𝑎 × 𝑏 × 𝑐 = 𝑎 ∙ 𝑐 𝑏 − 𝑏 ∙ 𝑐 𝑎 𝑎𝑛𝑑 𝑎 × 𝑏 × 𝑐 = 𝑎 ∙ 𝑐 𝑏 − 𝑎 ∙ 𝑏 𝑐

7.3 CURVES IN SPACE

1. Tangent Vector:

Let 𝑟 𝑡 = 𝑥 𝑡 𝑖 + 𝑦 𝑡 𝑗 + 𝑧(𝑡)𝑘 be the position vector

of a point P. Then for different values of the scalar

parameter t, point P traces the curve in space (Fig.

17.1). For neighboring point Q with position vector

𝑟 (𝑡 + 𝛿𝑡) , 𝛿𝑟 = 𝑟 𝑡 + 𝛿𝑡 − 𝑟 𝑡 implying 𝛿𝑟

𝛿𝑡=

𝑟 𝑡+𝛿𝑡 −𝑟 (𝑡)

𝛿𝑡 is directed along the chord PQ.

As 𝛿𝑡 → 0, 𝛿𝑟

𝛿𝑡 becomes the tangent to the space curve

at P provided there exists a non zero limit.

Thus a vector 𝑑𝑟

𝑑𝑡= 𝑟 ′ is a tangent to the space curve 𝑟 = 𝐹 (𝑡).

Let P0 be a fixed point on the space curve corresponding to t=t0, and the arc length 𝑃0𝑃 = 𝑠 , then

𝛿𝑠

𝛿𝑡=

𝛿𝑠

𝛿𝑟

𝛿𝑟

𝛿𝑡=

𝑎𝑟𝑐 𝑃𝑄

𝑐𝑕𝑜𝑟𝑑 𝑃𝑄 𝛿𝑟

𝛿𝑡 . … (1)

As 𝑄 → 𝑃 along the curve QP, i.e. 𝛿𝑡 → 0, then the 𝑎𝑟𝑐 𝑃𝑄

𝑐𝑕𝑜𝑟𝑑 𝑃𝑄 → 1 and

𝑑𝑠

𝑑𝑡=

𝑑𝑟

𝑑𝑡 = 𝑟 ′(𝑡) … (2)

7

If 𝑑𝑟

𝑑𝑡 is continuous, then 𝑠 =

𝑑𝑟

𝑑𝑡 𝑑𝑡

𝑡

𝑡0 = 𝑥′ 2+ 𝑦′ 2+ 𝑧′ 2𝑑𝑡

𝑡

𝑡0 … (3)

Further, if we take s as the parameter in place of t, then the magnitude of the tangent vector i.e.

𝑑𝑟

𝑑𝑠 = 1. Thus denoting the unit tangent vector by 𝑇 , we have 𝑇 =

𝑑𝑟

𝑑𝑠 …(4)

Example 5: Find the unit tangent vector at any point on the curve 𝒙 = 𝒕𝟐 + 𝟐, 𝒚 = 𝟒𝒕 − 𝟓,

𝒛 = 𝟐𝒕𝟐 − 𝟔𝒕 where t is variable. Also determine the unit tangent vector at t = 2.

Solution: Let 𝑟 be the position vector of any point 𝑥, 𝑦, 𝑧 on the given curve,

then 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘

∴ 𝑟 = 𝑡2 + 2 𝑖 + 4𝑡 − 5 𝑗 + 2𝑡2 − 6𝑡 𝑘

The vector tangent to the curve at any point 𝑥, 𝑦, 𝑧 is 𝑑𝑟

𝑑𝑡 = 2𝑡 𝑖 + 4 𝑗 + 4𝑡 − 6 𝑘

Now 𝑑𝑟

𝑑𝑡 = 2𝑡 2 + 4 2 + 4𝑡 − 6 2 = 2 5𝑡2 − 12𝑡 + 13

Therefore unit tangent vector at 𝑥, 𝑦, 𝑧 =𝑑𝑟

𝑑𝑡 𝑑𝑟

𝑑𝑡 =

2𝑡 𝑖 + 4 𝑗 + 4𝑡−6 𝑘

2 5𝑡2−12𝑡+13

And the unit tangent vector at t = 2 is 𝑑𝑟

𝑑𝑡 𝑑𝑟

𝑑𝑡

𝑡=2 =

2𝑡 𝑖 + 4 𝑗 + 4𝑡−6 𝑘

2 5𝑡2−12𝑡+13 𝑡 = 2

=2𝑖 +2𝑗 +𝑘

3.

2. Principal Normal: Since 𝑇 is a unit vector, so 𝑑𝑇

𝑑𝑠∙ 𝑇 = 0 i.e. either

𝑑𝑇

𝑑𝑠 is perpendicular to 𝑇 or

𝑑𝑇

𝑑𝑠 = 0, in which case 𝑇 is a constant vector w.r.t. the arc length s and so has a fixed direction i.e.

the curve is a straight line. Now, if we denote the unit normal vector to the curve at P by 𝑁 , then 𝑑𝑇

𝑑𝑠

is in the direction of 𝑁 which is known as the principal normal to the curve at P. The plane of 𝑇

and 𝑁 is called osculating plane of the curve at P.

3. Binormal: A unit vector 𝐵 = 𝑇 × 𝑁 is called the binormal at P. As 𝑇 and 𝑁 both are unit vectors,

so 𝐵 is also a unit vector normal to both 𝑇 and 𝑁

i.e. to the osculating plane of 𝑇 and 𝑁 .

Thus at each point P on the curve C, there are

three mutually perpendicular unit vectors 𝑇 , 𝑁

and 𝐵 , which form a moving trihedral such that

𝐵 = 𝑇 × 𝑁 , 𝑁 = 𝐵 × 𝑇 , 𝑇 = 𝑁 × 𝐵 . … (1)

This moving trihedral determines three

fundamental planes at each point of the curve C.

(i) The osculating plane of 𝑇 and 𝑁 .

(ii) The normal plane of 𝑁 and 𝐵 .

(iii) The rectifying plane of 𝐵 and 𝑇 .

4. Curvture: The arc rate of turning of the tangent viz. 𝑑𝑇

𝑑𝑠 is called the curvature of the curve

and is denoted by . As 𝑑𝑇

𝑑𝑠 is in the direction of the principal normal 𝑁 , therefore

𝑑𝑇

𝑑𝑠 = 𝑁 … (2)

8

5. Torsion: As the binormal 𝐵 is a unit vector, so 𝑑𝐵

𝑑𝑠 ∙ 𝐵 = 0. Also 𝐵 ∙ 𝑇 = 0, therefore

𝑑𝐵

𝑑𝑠 ∙ 𝑇 + 𝐵 ∙

𝑑𝑇

𝑑𝑠 = 0 or

𝑑𝐵

𝑑𝑠 ∙ 𝑇 + 𝐵 ∙ 𝑁 = 0 or

𝑑𝐵

𝑑𝑠 ∙ 𝑇 = 0. Hence,

𝑑𝐵

𝑑𝑠 is perpendicular to both 𝐵 and 𝑇 and

is, therefore, parallel to 𝑁 . The arc rate of turning of the binormal viz. 𝑑𝐵

𝑑𝑠 is called torsion of the

curve and is denoted by 𝜏. So, we can write it as

𝑑𝐵

𝑑𝑠 = −𝜏𝑁 (the negative sign indicates that for 𝜏 > 0,

𝑑𝐵

𝑑𝑠 has a direction of −𝑁 ) …(3)

Further, we know that 𝑁 = 𝐵 × 𝑇 , which on differentiation gives ,

𝑑𝑁

𝑑𝑠 =

𝑑𝐵

𝑑𝑠 × 𝑇 + 𝐵 ×

𝑑𝑇

𝑑𝑠 = −𝜏𝑁 × 𝑇 + 𝐵 × 𝑁

𝑑𝑁

𝑑𝑠 = 𝜏𝐵 − 𝑇 (using (1)) … (4)

The relations in (1), (2) and (3) constitutes the well known Frenet Formulas for the curve C.

Observations:

(i) 𝜌 =1

is called the radius of curvature.

(ii) 𝜍 =1

𝜏 is called the radius of torsion.

(iii) 𝜏 = 0 for a plane curve.

Example 6: Find 𝑵 (𝒕) and 𝑵 (𝟏) for the curve represented by 𝒓 𝒕 = 𝟑𝒕 𝒊 + 𝟐𝒕𝟐𝒋 .

Solution: For given 𝑟 𝑡 , we have 𝑟 ′ 𝑡 = 3𝑖 + 4𝑡𝑗 and 𝑟 ′ 𝑡 = 9 + 16𝑡2

Which implies that the unit tangent vector 𝑇 𝑡 = 𝑟 ′ 𝑡

𝑟 ′ 𝑡 =

3𝑖 +4𝑡𝑗

9+16𝑡2 … (1)

Differentiating 𝑇 𝑡 w. r. to t, 𝑇 ′ 𝑡 = 1

9+16𝑡2 4𝑗 −

16𝑡

9+16𝑡2 32

(3𝑖 + 4𝑡𝑗 ) = 12(−4𝑡𝑖 +3𝑗 )

9+16𝑡2 3/2 …(2)

And 𝑇 ′ 𝑡 = 12 9+16𝑡2

9+16𝑡2 3=

12

9+16𝑡2 … (3)

Therefore, the principal unit normal vector is 𝑁 𝑡 = 𝑇 ′(𝑡)

𝑇 ′(𝑡) =

−4𝑡𝑖 +3𝑗

9+16𝑡2 … (4)

But at 𝑡 = 1, the principal unit normal vector is 𝑁 1 = 1

5 (−4𝑖 + 3𝑗 ).

Example 7: Find the angle between the tangents to the curve 𝒓 = 𝒕𝟐 𝒊 + 𝟐𝒕 𝒋 − 𝒕𝟑 𝒌 at the point

t = ± 1.

Solution: Differentiating the given curve w. r. to t, we get

𝑑𝑟

𝑑𝑡 = 2𝑡 𝑖 + 2 𝑗 − 3𝑡2 𝑘 which is the tangent vector to the curve at any point t.

Let 𝑇1 & 𝑇2

are the tangent vectors to the curve at t = 1 and t = -1 respectively, then

𝑇1 = 2 𝑖 + 2 𝑗 − 3 𝑘 and 𝑇2

= −2 𝑖 + 2 𝑗 − 3 𝑘

Let 𝜃 be the angle between the tangents 𝑇1 & 𝑇2

, then

𝐶𝑜𝑠 𝜃 = 𝑇1 ∙ 𝑇2

𝑇1 𝑇2 =

2 𝑖 +2 𝑗 −3 𝑘 ∙ −2 𝑖 +2 𝑗 −3 𝑘

2 𝑖 +2 𝑗 −3 𝑘 −2 𝑖 +2 𝑗 −3 𝑘 =

−4 + 4 + 9

17 17=

9

17

9

∴ 𝜃 = 𝑐𝑜𝑠−1 9

17

Example 8: Find the curvature and torsion of the curve 𝒙 = 𝒂 𝒄𝒐𝒔 𝒕, 𝒚 = 𝒂 𝒔𝒊𝒏 𝒕, 𝒛 = 𝒃𝒕. (This curve is drawn on a circular cylinder cutting its generators at a constant angle and is known as a circular helix)

Solution: Equation of the given curve in vector form is

𝑟 = 𝑎 cos 𝑡 𝑖 + 𝑎 sin 𝑡 𝑗 + 𝑏𝑡 𝑘

Differentiating w. r. to t,

𝑑𝑟

𝑑𝑡 = −𝑎 sin 𝑡 𝑖 + 𝑎 cos 𝑡 𝑗 + 𝑏 𝑘

Now, the arc length of the curve from P0 (t = 0) to any

point P (t) is given by

𝑠 = 𝑑𝑟

𝑑𝑡 𝑑𝑡

𝑡

0 = 𝑎2 + 𝑏2 𝑡

∴ 𝑑𝑠

𝑑𝑡 = 𝑎2 + 𝑏2

Now, the unit tangent vector,

𝑇 =𝑑𝑟

𝑑𝑠=

𝑑𝑟 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

−𝑎 sin 𝑡 𝑖 + 𝑎 cos 𝑡 𝑗 + 𝑏 𝑘

𝑎2+𝑏2

So 𝑑𝑇

𝑑𝑠=

𝑑𝑇 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

−𝑎 cos 𝑡 𝑖 − 𝑎 sin 𝑡 𝑗

𝑎2+𝑏2

∴ = 𝑑𝑇

𝑑𝑠 =

𝑎

𝑎2+𝑏2 is the curvature of the given curve.

Also, the unit normal vector is 𝑁 = −(cos 𝑡 𝑖 + sin 𝑡 𝑗 ) and 𝐵 = 𝑇 × 𝑁 = (b sin 𝑡 𝑖 −𝑏 cos 𝑡𝑗 +𝑎𝑘 )

𝑎2+𝑏2

So 𝑑𝐵

𝑑𝑠=

𝑑𝐵 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

𝑏(cos 𝑡 𝑖 +sin 𝑡 𝑗)

𝑎2+𝑏2 = −𝜏𝑁 = 𝜏(cos 𝑡 𝑖 + sin 𝑡 𝑗 )

Hence 𝜏 = 𝑏

𝑎2+𝑏2 .

Example 9: A circular helix is given by the equation 𝒓 = 𝟐 𝐜𝐨𝐬 𝒕 𝒊 + 𝟐 𝐬𝐢𝐧 𝒕 𝒋 + 𝒌 . Find the

curvature and torsion of the curve at any point and show that they are constant.

Solution: Equation of the given curve in vector form is

𝑟 = 2 cos 𝑡 𝑖 + 2 sin 𝑡 𝑗 + 𝑘

Differentiating w. r. to t, 𝑑𝑟

𝑑𝑡 = −2 sin 𝑡 𝑖 + 2 cos 𝑡 𝑗 + 0 𝑘

Now, the arc length of the curve from P0 (t = 0) to any point P (t) is given by

𝑠 = 𝑑𝑟

𝑑𝑡 𝑑𝑡

𝑡

0 = 2 𝑡 implying

𝑑𝑠

𝑑𝑡 = 2

Now, the unit tangent vector, 𝑇 =𝑑𝑟

𝑑𝑠=

𝑑𝑟 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

−2 sin 𝑡 𝑖 + 2 cos 𝑡 𝑗 + 0 𝑘

2

So 𝑑𝑇

𝑑𝑠=

𝑑𝑇 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

− cos 𝑡 𝑖 − sin 𝑡 𝑗

2

∴ = 𝑑𝑇

𝑑𝑠 =

1

2 is the curvature of the given curve and is a constant.

Also, the unit normal vector is 𝑁 = −(cos 𝑡 𝑖 + sin 𝑡 𝑗 ) and

𝐵 = 𝑇 × 𝑁 = − sin 𝑡 𝑖 + cos 𝑡 𝑗 × − cos 𝑡 𝑖 − sin 𝑡 𝑗 = 𝑘

So 𝑑𝐵

𝑑𝑠=

𝑑𝐵 𝑑𝑡

𝑑𝑠 𝑑𝑡 = 0 = −𝜏𝑁 = 𝜏(cos 𝑡 𝑖 + sin 𝑡 𝑗 )

10

Hence 𝜏 = 0 is the torsion of the given curve and is constant.

Example 10: Show that for the curve 𝒓 = 𝒂 𝟑𝒕 − 𝒕𝟑 𝒊 + 𝟑𝒂𝒕𝟐 𝒋 + 𝒂(𝟑𝒕 + 𝒕𝟑)𝒌 , the curvature

equals torsion.

Solution: Given curve is 𝑟 = 𝑎 3𝑡 − 𝑡3 𝑖 + 3𝑎𝑡2 𝑗 + 𝑎(3𝑡 + 𝑡3)𝑘

Differentiating w. r. to t, 𝑑𝑟

𝑑𝑡= 𝑎 3 − 3𝑡2 𝑖 + 6𝑎𝑡 𝑗 + 𝑎 3 + 3𝑡2 𝑘

Now, the arc length of the curve P0 (t = 0) to any point P (t) is given by

𝑠 = 𝑑𝑟

𝑑𝑡 𝑑𝑡

𝑡

0= 𝑎 3 − 3𝑡2

2 + 6𝑎𝑡 2 + 𝑎 3 + 3𝑡2

2 𝑑𝑡

𝑡

0

= 3𝑎 2 𝑡2 + 1 𝑑𝑡𝑡

0= 3𝑎 2

𝑡3

3+ 𝑡

∴ 𝑑𝑠

𝑑𝑡 = 3𝑎 2 𝑡2 + 1

Now, the unit tangent vector,

𝑇 = 𝑑𝑟

𝑑𝑠=

𝑑𝑟 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

𝑎 3−3𝑡2 𝑖 +6𝑎𝑡 𝑗 +𝑎 3+3𝑡2 𝑘

3𝑎 2 𝑡2+1 =

1−𝑡2 𝑖 +2𝑡 𝑗 + 1+𝑡2 𝑘

2 𝑡2+1

So 𝑑𝑇

𝑑𝑠=

𝑑𝑇 𝑑𝑡

𝑑𝑠 𝑑𝑡 =

−2𝑡 𝑖 + 1−𝑡2 𝑗

3𝑎 1+𝑡2 3

∴ = 𝑑𝑇

𝑑𝑠 =

1

3𝑎 1+𝑡2 2 is the curvature of the given curve.

Also, the unit normal vector is 𝑁 =−2𝑡 𝑖 + 1−𝑡2

𝑗

1+𝑡2

and

𝐵 = 𝑇 × 𝑁 = 1−𝑡2 𝑖 +2𝑡 𝑗 + 1+𝑡2 𝑘

2 1+𝑡2

So 𝑑𝐵

𝑑𝑠=

𝑑𝐵 𝑑𝑡

𝑑𝑠 𝑑𝑡 = −

−2𝑡 𝑖 + 1−𝑡2 𝑗

3𝑎 1+𝑡2 3 = −

1

3𝑎 1+𝑡2 2 .

−2𝑡 𝑖 + 1−𝑡2 𝑗

1+𝑡2 = −𝜏𝑁

∴ 𝜏 =1

3𝑎 1+𝑡2

2 is the torsion of the given curve.

Hence curvature equals torsion for the given curve.

ASSIGNMENT 1

1. If 𝑎 = 𝑥2𝑦𝑧 𝑖 − 2𝑥𝑧3𝑗 + 𝑥𝑧2𝑘 and 𝑏 = 2𝑧 𝑖 + 𝑦 𝑗 − 𝑥2𝑘 , find 𝜕2

𝜕𝑥𝜕𝑦 𝑎 × 𝑏 at (1, 0, -2).

2. Given 𝑟 = 𝑡𝑚 𝐴 + 𝑡𝑛 𝐵 , where 𝐴 and 𝐵 are constant vectors, show that, if 𝑟 and 𝑑2𝑟

𝑑𝑡2 are parallel

vectors, then 𝑚 + 𝑛 = 1, unless 𝑚 = 𝑛.

3. Find the equation of tangent line to the curve 𝑥 = 𝑎 cos 𝜃 , 𝑦 = 𝑎 sin 𝜃 , 𝑧 = 𝑎𝜃 tan 𝛼 at = 𝜋

4.

11

4. Find the unit tangent vector at any point on the curve 𝑥 = 𝑡2 + 2, 𝑦 = 4𝑡 − 5, 𝑧 = 2𝑡2 − 6𝑡, where t

is any variable. Also determine the unit tangent vector at the point 𝑡 = 2.

5. If 𝑟 = 𝑎 cos 𝑡 𝑖 + 𝑎 sin 𝑡 𝑗 + 𝑎𝑡 tan 𝛼 𝑘 , find the value of (a) 𝑑𝑟

𝑑𝑡×

𝑑2𝑟

𝑑𝑡2 (b)

𝑑𝑟

𝑑𝑡

𝑑2𝑟

𝑑𝑡2

𝑑3𝑟

𝑑𝑡3 .

Also find the unit tangent vector at any point t on the curve.

6. Find the equation of the osculating plane and binormal to the curve

(a) 𝑥 = 𝑒𝜃 cos 𝜃 , 𝑦 = 𝑒𝜃 sin 𝜃 , 𝑧 = 𝑒𝜃 at 𝜃 = 0 (b) 𝑥 = 2 cosh𝜃

2, 𝑦 = 2 sinh

𝜃

2, 𝑧 = 2𝜃 at 𝜃 = 0

7. Find the curvature of the (a) ellipse 𝑟 = 𝑎 cos 𝑡 𝑖 + 𝑏 sin 𝑡 𝑗 (b) parabola 𝑟 = 2𝑡 𝑖 + 𝑡2 𝑗 at point

𝑡 = 1.

17.4 VELOCITY AND ACCELERATION

1. Velocity: Let the position of particle P at a time t on the curve C is 𝑟 𝑡 and it comes to point Q

at time 𝑡 + 𝛿𝑡 having position 𝑟 𝑡 + 𝛿𝑡 , then 𝛿𝑟 = 𝑟 𝑡 + 𝛿𝑡 − 𝑟 𝑡 i.e. 𝛿𝑟

𝛿𝑡 is directed along PQ.

As 𝑄 → 𝑃 along C, the line PQ becomes tangent at P to the curve C.

So 𝑣 𝑡 = 𝑑𝑟

𝑑𝑡 = Lt𝛿𝑡→0

𝛿𝑟

𝛿𝑡 is the tangent vector to C at point P which is the velocity vector 𝑣 (𝑡) of

the motion and its magnitude gives the speed 𝑣 = 𝑑𝑠

𝑑𝑡, where s is the arc length of P from a fixed

point P0 (s=0) on C.

2. Acceleration: Acceleration vector 𝑎 𝑡 of a particle is the derivative of the velocity vector 𝑣 (𝑡),

and it is given by 𝑎 𝑡 = 𝑑𝑣

𝑑𝑡=

𝑑2𝑟

𝑑𝑡2 . It is an interesting fact that the magnitude of acceleration is not

always the rate of change of 𝑣 = 𝑣 , as 𝑎 𝑡 is not always tangential to the curve C. There are two

components of acceleration, which are given as: (i) Tangential Acceleration (ii) Normal

Acceleration

Observation: The acceleration is the time rate of change of 𝑣 𝑡 = 𝑑𝑠

𝑑𝑡, if and only if the normal acceleration

is zero, for then 𝑎 = 𝑑2𝑠

𝑑𝑡2 𝑑𝑟

𝑑𝑠 =

𝑑2𝑠

𝑑𝑡2 .

3. Relative Velocity and Acceleration: Let two particles P and

Q moving along the curves C1 and C2 have position vectors 𝑟1 (𝑡)

and 𝑟2 (𝑡) at time t, so that 𝑟 𝑡 = 𝑃𝑄 = 𝑟2 𝑡 − 𝑟1 (𝑡)

Differentiating w. r. to t, 𝑑𝑟

𝑑𝑡=

𝑑𝑟 2

𝑑𝑡−

𝑑𝑟 1

𝑑𝑡 … (1)

This defines the relative velocity of Q w. r. t. P and states that the

velocity of Q relative to P = Velocity vector of Q – Velocity

vector of P.

Again differentiating (1) w. r. to t., we have

𝑑2𝑟

𝑑𝑡2=

𝑑2𝑟 2

𝑑𝑡2−

𝑑2𝑟 1

𝑑𝑡 2

This defines the relative acceleration of Q w. r. t. and states that Acceleration of Q relative to P =

Acceleration of Q – Acceleration of P.

12

Example 11: Find the tangential and normal acceleration of a particle moving in a plane

curve in Cartesian coordinates.

Solution: Let 𝑟 be the position vector the point P, a function of a scalar t. In particular, if the scalar

variable t is taken as an arc length s along the curve C measured from some fixed point, that is,

𝑥 = 𝑥 𝑠 , 𝑦 = 𝑦 𝑠 , 𝑧 = 𝑧(𝑠) then 𝑟 = 𝑥 𝑠 𝑖 + 𝑦 𝑠 𝑗 + 𝑧(𝑠)𝑘

So that 𝑑𝑟

𝑑𝑠=

𝑑𝑥

𝑑𝑠𝑖 +

𝑑𝑦

𝑑𝑠𝑗 +

𝑑𝑧

𝑑𝑠𝑘 … (1)

And 𝑑𝑟

𝑑𝑠

2

= 𝑑𝑥

𝑑𝑠

2+

𝑑𝑦

𝑑𝑠

2+

𝑑𝑧

𝑑𝑠

2 … (2)

For two dimension curves we have in calculus

𝑑𝑠 2 = 𝑑𝑥 2 + 𝑑𝑦 2

which when extended to the space, becomes

𝑑𝑠 2 = 𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2

Or 𝑑𝑥

𝑑𝑠

2+

𝑑𝑦

𝑑𝑠

2+

𝑑𝑧

𝑑𝑠

2= 1

Therefore (2) gives 𝑑𝑟

𝑑𝑠

2

= 1

That means, 𝑑𝑟

𝑑𝑠 is a unit vector along the tangent and (1) represents a unit tangent vector along the

curve C in space.

Therefore, Velocity 𝑣 of the particle at any point of the curve is given by

𝑣 =𝑑𝑟

𝑑𝑡= 𝑑𝑟

𝑑𝑠𝑑𝑠𝑑𝑡

= 𝑣 𝑇 … (3)

where 𝑣 =𝑑𝑠

𝑑𝑡 and 𝑇 =

𝑑𝑟

𝑑𝑠 is the unit vector along the tangent.

Thus 𝑣 =𝑑𝑠

𝑑𝑡 is the tangential component of the velocity and the normal component of the velocity

is zero.

Next, acceleration 𝑎 =𝑑𝑣

𝑑𝑡= 𝑑(𝑣𝑇)

𝑑𝑡= 𝑑𝑣

𝑑𝑡 𝑇 + 𝑣

𝑑𝑇

𝑑𝑡

or 𝑎 =𝑑2𝑠

𝑑𝑡2 𝑇 + 𝑑𝑠

𝑑𝑡

𝑑𝑇

𝑑𝑠

𝑑𝑠

𝑑𝑡=

𝑑𝑣

𝑑𝑡 𝑇 + 𝑣2

𝑑𝑇

𝑑𝜓

𝑑𝜓

𝑑𝑠

=𝑑𝑣

𝑑𝑡 𝑇 +

𝑣2

𝜌 𝑑𝑇

𝑑𝜓 (since radius of curvature, 𝜌 =

𝑑𝜓

𝑑𝑠 ) … (4)

From the adjoining figure, 𝑇 = 𝑃𝑄 is along the tangent at P to the curve C and 𝑁 is the unit vector

along the normal to P.

∴ 𝑇 = cos 𝜓 𝑖 + sin 𝜓 𝑗 and 𝑁 = cos 𝜋

2+ 𝜓 𝑖 + sin

𝜋

2+ 𝜓 𝑗 = −sin 𝜓 𝑖 + cos 𝜓 𝑗

Now 𝑑𝑇

𝑑𝜓 = −sin 𝜓 𝑖 + cos 𝜓 𝑗 = 𝑁

Therefore, equation (4) becomes 𝑎 =𝑑𝑣

𝑑𝑡 𝑇 +

𝑣2

𝜌𝑁

Which shows that tangential and normal components of acceleration at the point P are 𝑑𝑣

𝑑𝑡 and

𝑣2

𝜌 .

Since 𝑑𝑣

𝑑𝑡=

𝑑𝑣

𝑑𝑠

𝑑𝑠

𝑑𝑡 = v

𝑑𝑣

𝑑𝑠, so the tangential component of acceleration is also written as v

𝑑𝑣

𝑑𝑠.

13

Example 12: Find the radial and transverse acceleration of a particle moving in a plane curve

in Polar coordinates.

Solution: Let the position vector of a moving particle 𝑃(𝑟, 𝜃) be 𝑟 so that

𝑟 = 𝑟 𝑟 = 𝑟 (cos 𝜃 𝑖 + sin 𝜃 𝑗 ) at any time t.

Then the velocity of the particle is 𝑣 = 𝑑𝑟

𝑑𝑡=

𝑑𝑟

𝑑𝑡 𝑟 + 𝑟

𝑑𝑟

𝑑𝑡

As 𝑟 = (cos 𝜃 𝑖 + sin 𝜃 𝑗 ) so 𝑑𝑟

𝑑𝑡= (−sin𝜃 𝑖 + cos 𝜃 𝑗 )

𝑑𝜃

𝑑𝑡

Therefore, 𝑑𝑟

𝑑𝑡 is perpendicular to 𝑟 and

𝑑𝑟

𝑑𝑡 =

𝑑𝜃

𝑑𝑡 i.e. if 𝑢

is a unit vector perpendicular to 𝑟 , then 𝑑𝑟

𝑑𝑡=

𝑑𝜃

𝑑𝑡 𝑢

And thus, 𝑣 = 𝑑𝑟

𝑑𝑡=

𝑑𝑟

𝑑𝑡 𝑟 + 𝑟

𝑑𝜃

𝑑𝑡 𝑢

So the radial and transverse components of the velocity are 𝑑𝑟

𝑑𝑡 and

𝑑𝜃

𝑑𝑡.

Also 𝑎 = 𝑑𝑣

𝑑𝑡=

𝑑2𝑟

𝑑𝑡 2 𝑟 +

𝑑𝑟

𝑑𝑡

𝑑𝑟

𝑑𝑡 +

𝑑𝑟

𝑑𝑡

𝑑𝜃

𝑑𝑡 𝑢 +𝑟

𝑑2𝜃

𝑑𝑡 2 𝑢 + 𝑟

𝑑𝜃

𝑑𝑡

𝑑𝑢

𝑑𝑡

= 𝑑2𝑟

𝑑𝑡 2− 𝑟

𝑑𝜃

𝑑𝑡

2

𝑟 + 2𝑑𝑟

𝑑𝑡

𝑑𝜃

𝑑𝑡+ 𝑟

𝑑2𝜃

𝑑𝑡2 𝑢

𝑠𝑖𝑛𝑐𝑒 𝑢 = − sin 𝜃 𝑖 + cos 𝜃 𝑗 𝑔𝑖𝑣𝑒𝑠 𝑑𝑢

𝑑𝑡= −

𝑑𝜃

𝑑𝑡𝑟

Thus radial and transverse components of the acceleration are 𝑑2𝑟

𝑑𝑡 2− 𝑟

𝑑𝜃

𝑑𝑡

2

and

2𝑑𝑟

𝑑𝑡

𝑑𝜃

𝑑𝑡+ 𝑟

𝑑2𝜃

𝑑𝑡 2 .

Example 13: A particle moves along the curve 𝒓 = 𝐭𝟑 − 𝟒𝐭 𝒊 + 𝐭𝟐 + 𝟒𝐭 𝒋 + 𝟖𝒕𝟐 − 𝟑𝒕𝟑 𝒌

where t denotes the time. Find the magnitudes of acceleration along the tangent and normal at

time t = 2.

Solution: The velocity of the particle is 𝑣 = 𝑑𝑟

𝑑𝑡 = 3𝑡2 − 4 𝑖 + 2𝑡 + 4 𝑗 + 16𝑡 − 9𝑡2 𝑘

And the acceleration is 𝑎 = 𝑑2𝑟

𝑑𝑡 2 = 6𝑡 𝑖 + 2 𝑗 + 16 − 18𝑡 𝑘

At t = 2, 𝑣 = 8 𝑖 + 8 𝑗 − 4 𝑘 and 𝑎 = 12 𝑖 + 2 𝑗 − 20 𝑘

Since the velocity vector is also the tangent vector to the curve, so the magnitude of acceleration

along the tangent at t = 2 is = 𝑎 ∙ 𝑣

𝑣 = 12 𝑖 + 2 𝑗 − 20 𝑘 ∙

8 𝑖 +8 𝑗 −4 𝑘

64+64+16

= 12 8 + 2 8 + −20 (−4)

12 = 16

And, the magnitude of acceleration along the normal at t = 2 is

𝑎 − 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑎 along the tangent at 𝑡 = 2 = 12 𝑖 + 2 𝑗 − 20 𝑘 − 16 8 𝑖 +8 𝑗 −4 𝑘

12

= 4 𝑖 −26 𝑗 −44 𝑘

3 = 2 73

Example 14: A person going east wards with a velocity of 4 km per hour, finds that the wind

appears to blow directly from the north. He doubles his speed and the wind seems to come

from north-east. Find the actual velocity of the wind.

14

Solution: Let the actual velocity of the wind is 𝑣 = 𝑥𝑖 + 𝑦𝑗 , where 𝑖 and 𝑗 represent velocities of 1

km per hour towards the east and north respectively. As the person is going eastwards with a

velocity of 4 km per hour, his actual velocity is 4𝑖 .

Then the velocity of the wind relative to the man is 𝑥𝑖 + 𝑦𝑗 − 4𝑖 , which is parallel to −𝑗 , as it

appears to blow from the north. Hence x = 4.

When the velocity of the person becomes 8𝑖 , the velocity of the wind relative to a man is 𝑥𝑖 + 𝑦𝑗 −

8𝑖 . But this is parallel to − 𝑖 + 𝑗 .

∴ 𝑥−8

𝑦 = 1 which gives 𝑦 = −4 (using (1))

Hence the actual velocity of the wind is 4𝑖 − 4𝑗 i.e. 4 2 km per hour towards south east.

Example 15: A particle moves along the curve 𝒙 = 𝒕𝟑 + 𝟏, 𝒚 = 𝒕𝟐, 𝒛 = 𝟐𝒕 + 𝟑 where t is the

time. Find the components of velocity and acceleration at t = 1in the direction of the vector

𝒊 + 𝒋 + 𝟑𝒌 .

Solution: Let 𝑟 be the position vector of the particle at any time t,

then 𝑟 = 𝑡3 + 1 𝑖 + 𝑡2 𝑗 + 2𝑡 + 3 𝑘

So the velocity is 𝑣 = 𝑑𝑟

𝑑𝑡 = 3𝑡2 𝑖 + 2𝑡 𝑗 + 2 𝑘

And the acceleration is 𝑎 = 𝑑2𝑟

𝑑𝑡 2 = 6𝑡 𝑖 + 2 𝑗 + 0 𝑘

At t =1, 𝑣 = 3 𝑖 + 2 𝑗 + 2 𝑘 and 𝑎 = 6 𝑖 + 2 𝑗 + 0 𝑘

Also the unit vector in the direction of the given vector 𝑖 + 𝑗 + 3𝑘 is = 𝑖 +𝑗 +3𝑘

𝑖 +𝑗 +3𝑘 =

𝑖 +𝑗 +3𝑘

11

Now the component of velocity at t = 1, in the direction of the vector 𝑖 + 𝑗 + 3𝑘 =

3 𝑖 +2 𝑗 +2 𝑘 ∙ 𝑖 +𝑗 +3𝑘

11=

3+2+6

11 = 11

And the component of acceleration at t = 1, in the direction of the vector 𝑖 + 𝑗 + 3𝑘 =

6 𝑖 +2 𝑗 +0 𝑘 ∙ 𝑖 +𝑗 +3𝑘

11=

6+2+0

11=

8

11

ASSIGNMENT 2

1. The particle moves along a curve 𝑥 = 𝑒−𝑡 , 𝑦 = 2 cos 3𝑡 , 𝑧 = 2 sin 3𝑡, where t is the time variable.

Determine its velocity and acceleration vectors and also the magnitudes of velocity and

acceleration at 𝑡 = 0.

2. A particle (with position vector 𝑟 ) is moving in a circle with constant angular velocity 𝜔. Show

by vector methods, that the acceleration is equal to −𝜔2𝑟 .

3. A particle moves on the curve 𝑥 = 2𝑡2 , 𝑦 = 𝑡2 − 4𝑡, 𝑧 = 3𝑡 − 5, where t is the time. Find the

components of velocity and acceleration at time t = 1 in the direction 𝑖 − 3𝑗 + 2𝑘 .

4. The position vector of a particle at time t is 𝑟 = cos(𝑡 − 1) 𝑖 + sinh 𝑡 − 1 𝑗 + 𝑎𝑡3𝑘 . Find the

condition imposed on a by requiring that at time t = 1, the acceleration is normal to the position

vector.

15

5. A particle moves so that its position vector is given by 𝑟 = cos 𝜔𝑡 𝑖 + sin 𝜔𝑡 𝑗 . Show that the

velocity 𝑣 of the particle is perpendicular to 𝑟 and 𝑟 × 𝑣 is a constant vector.

6. A particle moves along a catenary 𝑠 = 𝑐 tan 𝜓. The direction of acceleration at any point makes

equal angles with the tangent and normal to the path at that point. If the speed at vertex (𝜓 = 0)

be 𝑣0, show that the magnitude of velocity and acceleration at any point are given by 𝑣0 𝑒𝜓 and

2

𝑐𝑣0

2𝑒2𝜓 cos2 𝜓 respectively.

7. The position vector of a moving particle at a time t is 𝑟 = 𝑡2 𝑖 − 𝑡3 𝑗 + 𝑡4 𝑘 . Find the tangential

and normal components of acceleration at t = 1.

8. A vessel A is a sailing with a velocity of 11 knots per hour in the direction south-east and a

second vessel B is sailing with a velocity of 13 knots per hour in a direction 300 of north. Find

the velocity of A relative to B.

9. The velocity of a boat relative to water is represented by 3𝑖 + 4𝑗 and that of water relative to

earth is 𝑖 − 3𝑗 . What is the velocity of the boat relative to earth if 𝑖 and 𝑗 represent one KM an

hour east and north respectively?

10. A person travelling towards the north-east with a velocity of 6 KM per hour finds that the wind

appears to blow from the north, but when he doubles his speed it seems to come from a

direction inclined at an angle tan−1 2 to the north of east. Show that the actual velocity of the

wind is 3 2 KM per hour towards the east.

17.5 DEL APPLIED TO SCALAR POINT FUNCTIONS: GRADIENT [KUK 2009]

Del Operator: Del operator is a vector differential operator and is written as

𝛁 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧

Gradient of a Scalar Function: Let ∅(𝑥, 𝑦, 𝑧) be a scalar function of three variables defined over a

region R of space. Then gradient of is a vector function defined as

∅ = 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧,

wherever the partial derivatives exists. It may also be denoted as 𝑔𝑟𝑎𝑑(∅). The del operator is also

called the gradient operator.

Level Surface: Let ∅(𝑥, 𝑦, 𝑧) be a scalar valued function and C is a constant. The surface given by

∅ 𝑥, 𝑦, 𝑧 = 𝐶 through a point 𝑃(𝑟 ) is such that at each point on it the function has same value, is

called the level surface of ∅ 𝑥, 𝑦, 𝑧 through P, e.g. equi-potential or isothermal surfaces.

In other words, locus of the point 𝑃(𝑟 ) satisfying ∅ 𝑟 = 𝐶 form a surface through P. This surface is

called the level surface through P.

Gradient as Normal or Geometrical Interpretation of Gradient

16

Let ∅ 𝑟 = ∅ 𝑥, 𝑦, 𝑧 is a scalar function where 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 . And ∅ 𝑥, 𝑦, 𝑧 = 𝐶 is the level

surface of ∅ through 𝑃(𝑟 ). Let 𝑄(𝑟 + 𝛿𝑟 ) be a point on neighboring level surface ∅ + 𝛿∅, then

𝛁∅ ∙ δr = 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 ∙ 𝛿𝑥𝑖 + 𝛿𝑦𝑗 + 𝛿𝑧𝑘

= 𝜕∅

𝜕𝑥 𝛿𝑥 +

𝜕∅

𝜕𝑦 𝛿𝑦 +

𝜕∅

𝜕𝑧 𝛿𝑧 = 𝛿∅ … (1)

Now if P and Q lie on same level surface i.e. ∅ & ∅ + 𝛿∅ are

same, then 𝛿∅ = 0.

Implies 𝛁∅ ∙ δr = 0 (using (1))

Therefore, 𝛁∅ is perpendicular (normal) to every δr lying on

this surface.

Hence, 𝛁∅ is normal to the surface ∅ 𝑥, 𝑦, 𝑧 = 𝐶 and we can write 𝛁∅ = 𝛁∅ n , where 𝑛 is unit

normal vector to the surface.

See the Fig. 17.7, if the perpendicular distance PM between the surfaces through P and Q be 𝛿𝑛, then

rate of change of ∅ along the normal to the surface through P is

𝜕∅

𝜕𝑛= 𝐿𝑡𝛿𝑛→0

𝛿∅

𝛿𝑛 = 𝐿𝑡𝛿𝑛→0

𝛁∅∙δr

𝛿𝑛 (using (1))

= Lt𝛿𝑛→0 𝛁∅ n ∙δr

𝛿𝑛 = 𝛁∅ Lt𝛿𝑛→0

n ∙δr

𝛿𝑛 … (2)

= 𝛁∅ Lt𝛿𝑛→0δn

𝛿𝑛 = 𝛁∅ (𝐴𝑠 𝑛 ∙ 𝛿𝑟 = 𝛿𝑟 cos 𝜃 = 𝛿𝑛)

Hence the magnitude of 𝛁∅ i.e. 𝛁∅ = 𝜕∅

𝜕𝑛 … (3)

Thus 𝑔𝑟𝑎𝑑 ∅ is normal vector to the level surface ∅ 𝑥, 𝑦, 𝑧 = 𝐶 and its magnitude represents the rate

of change of ∅ along this normal.

Directional Derivative: If 𝛿𝑟 denotes the length PQ and 𝒖 be the unit vector in the direction of PQ,

the limiting value of 𝛿𝑓

𝛿𝑟 as 𝛿𝑟 → 0 (i.e.

𝜕𝑓

𝜕𝑟) is known as the directional derivative of f along the

direction PQ.

Since 𝛿𝑟 =𝛿𝑛

cos 𝛼=

𝛿𝑛

𝑛 ∙𝑢 , therefore

𝜕𝑓

𝜕𝑟= Lt𝛿𝑟→0 𝑛 ∙ 𝑢

𝛿𝑓

𝛿𝑛 = 𝑢 ∙

𝛿𝑓

𝛿𝑛 𝑛 = 𝑢 ∙ 𝛁𝑓

Thus the directional derivative of f in the direction of 𝑢 is the resolved part of 𝛁𝑓 in the direction of 𝑢 .

Since 𝛁𝑓 ∙ 𝑢 = 𝛁𝑓 cos 𝛼 ≤ 𝛁𝑓

It follows that 𝛁𝑓 gives the maximum rate of change of f.

Properties of Gradient Operator: Let ∅(𝑥, 𝑦, 𝑧) & ψ(𝑥, 𝑦, 𝑧) are two differentiable scalar functions

defined over some region R. then the gradient operator has following properties:

17

(i) Gradient of a constant multiple of scalar function ∅

𝑔𝑟𝑎𝑑 𝐶∅ = 𝐶 𝑔𝑟𝑎𝑑(∅) or ∇ 𝐶∅ = 𝐶 ∇∅

Proof: Consider 𝑔𝑟𝑎𝑑 𝐶∅ = ∇ 𝐶∅

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝐶∅

= 𝑖 𝜕

𝜕𝑥(𝐶∅) + 𝑗

𝜕

𝜕𝑦(𝐶∅) + 𝑘

𝜕

𝜕𝑧(𝐶∅)

= 𝐶 𝑖 𝜕∅

𝜕𝑥+ 𝐶 𝑗

𝜕∅

𝜕𝑦+ 𝐶 𝑘

𝜕∅

𝜕𝑧

= 𝐶 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧

= 𝐶 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∅

= 𝐶 ∇∅ = 𝐶 𝑔𝑟𝑎𝑑 ∅

Hence 𝑔𝑟𝑎𝑑 𝐶∅ = 𝐶 𝑔𝑟𝑎𝑑(∅).

(ii) Gradient of sum or difference of two scalar functions

𝑔𝑟𝑎𝑑 ∅ ± 𝜓 = 𝑔𝑟𝑎𝑑 ∅ ± 𝑔𝑟𝑎𝑑(𝜓) or ∇ ∅ ± 𝜓 = ∇∅ ± ∇𝜓

Proof: Consider 𝑔𝑟𝑎𝑑 ∅ ± 𝜓 = ∇ ∅ ± 𝜓

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∅ ± 𝜓

= 𝑖 𝜕

𝜕𝑥 ∅ ± 𝜓 + 𝑗

𝜕

𝜕𝑦 ∅ ± 𝜓 + 𝑘

𝜕

𝜕𝑧 ∅ ± 𝜓

= 𝑖 𝜕∅

𝜕𝑥±

𝜕𝜓

𝜕𝑥 + 𝑗

𝜕∅

𝜕𝑦±

𝜕𝜓

𝜕𝑦 + 𝑘

𝜕∅

𝜕𝑧±

𝜕𝜓

𝜕𝑧

= 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 ± 𝑖

𝜕𝜓

𝜕𝑥+ 𝑗

𝜕𝜓

𝜕𝑦+ 𝑘

𝜕𝜓

𝜕𝑧

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∅ ± 𝑖

𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝜓

= ∇∅ ± ∇𝜓 = 𝑔𝑟𝑎𝑑 ∅ ± 𝑔𝑟𝑎𝑑(𝜓).

Hence 𝑔𝑟𝑎𝑑 ∅ ± 𝜓 = 𝑔𝑟𝑎𝑑 ∅ ± 𝑔𝑟𝑎𝑑(𝜓).

(iii) Gradient of product of two scalar functions

𝑔𝑟𝑎𝑑 ∅𝜓 = ∅ 𝑔𝑟𝑎𝑑 𝜓 + 𝜓 𝑔𝑟𝑎𝑑(∅) or ∇ ∅ 𝜓 = ∅ ∇𝜓 + 𝜓 ∇∅

Proof: Consider 𝑔𝑟𝑎𝑑 ∅𝜓 = ∇ ∅ 𝜓

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∅ 𝜓

= 𝑖 𝜕

𝜕𝑥 ∅ 𝜓 + 𝑗

𝜕

𝜕𝑦 ∅ 𝜓 + 𝑘

𝜕

𝜕𝑧 ∅ 𝜓

18

= 𝑖 ∅ 𝜕𝜓

𝜕𝑥+ 𝜓

𝜕∅

𝜕𝑥 + 𝑗 ∅

𝜕𝜓

𝜕𝑦+ 𝜓

𝜕∅

𝜕𝑦 + 𝑘 ∅

𝜕𝜓

𝜕𝑧+ 𝜓

𝜕∅

𝜕𝑧

= ∅ 𝑖 𝜕𝜓

𝜕𝑥+ 𝑗

𝜕𝜓

𝜕𝑦+ 𝑘

𝜕𝜓

𝜕𝑧 + 𝜓 𝑖

𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧

= ∅ ∇𝜓 + 𝜓 ∇∅ = ∅ 𝑔𝑟𝑎𝑑 𝜓 ± 𝜓 𝑔𝑟𝑎𝑑(∅)

Hence 𝑔𝑟𝑎𝑑 ∅𝜓 = ∅ 𝑔𝑟𝑎𝑑 𝜓 ± 𝜓 𝑔𝑟𝑎𝑑(∅).

(iv) Gradient of quotient of two scalar functions

𝑔𝑟𝑎𝑑 ∅

𝜓 =

𝜓 𝑔𝑟𝑎𝑑 ∅ −∅ 𝑔𝑟𝑎𝑑 (𝜓)

𝜓 2 or ∇

∅

𝜓 =

𝜓 ∇ ∅ −∅ ∇(𝜓)

𝜓 2, provided 𝜓 ≠ 0.

Proof: Consider 𝑔𝑟𝑎𝑑 ∅

𝜓 = ∇

∅

𝜓

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧

∅

𝜓

= 𝑖 𝜕

𝜕𝑥

∅

𝜓 + 𝑗

𝜕

𝜕𝑦

∅

𝜓 + 𝑘

𝜕

𝜕𝑧

∅

𝜓

= 𝑖 𝜓

𝜕∅

𝜕𝑥 − ∅

𝜕𝜓

𝜕𝑥

𝜓2 + 𝑗 𝜓

𝜕∅

𝜕𝑦 − ∅

𝜕𝜓

𝜕𝑦

𝜓2 + 𝑘 𝜓

𝜕∅

𝜕𝑧 − ∅

𝜕𝜓

𝜕𝑧

𝜓2

=1

𝜓2 𝜓 𝑖

𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 − ∅ 𝑖

𝜕𝜓

𝜕𝑥+ 𝑗

𝜕𝜓

𝜕𝑦+ 𝑘

𝜕𝜓

𝜕𝑧

=1

𝜓2 𝜓 𝑔𝑟𝑎𝑑(∅) − ∅ 𝑔𝑟𝑎𝑑(𝜓) =

𝜓 𝑔𝑟𝑎𝑑 ∅ −∅ 𝑔𝑟𝑎𝑑 (𝜓)

𝜓 2

=𝜓 ∇ ∅ −∅ ∇(𝜓)

𝜓 2

Hence 𝑔𝑟𝑎𝑑 ∅

𝜓 =

𝜓 𝑔𝑟𝑎𝑑 ∅ −∅ 𝑔𝑟𝑎𝑑 (𝜓)

𝜓 2 .

Example 16: If ∅ = 𝟑𝒙𝟐𝒚 − 𝒚𝟑𝒛𝟐 , then find 𝒈𝒓𝒂𝒅 ∅ at (1, -2, -1).

Solution: 𝑔𝑟𝑎𝑑 ∅ = ∇ 3𝑥2𝑦 − 𝑦3𝑧2

= 𝑖 𝜕

𝜕𝑥 3𝑥2𝑦 − 𝑦3𝑧2 + 𝑗

𝜕

𝜕𝑦 3𝑥2𝑦 − 𝑦3𝑧2 + 𝑘

𝜕

𝜕𝑧 3𝑥2𝑦 − 𝑦3𝑧2

= 𝑖 6𝑥𝑦 + 𝑗 3𝑥2 − 3𝑦2𝑧2 + 𝑘 −2𝑦3𝑧

At (1, -2, -1), 𝑔𝑟𝑎𝑑 ∅ = 𝑖 6 1 −2 + 𝑗 3 1 2 − 3 −2 2 −1 2 + 𝑘 −2 −2 3 −1

= −12𝑖 − 9𝑗 − 16𝑘

Example 17: Prove that 𝛁 𝒓𝒏 = 𝒏 𝒓𝒏−𝟐 𝒓 , where 𝒓 = 𝒙𝒊 + 𝒚𝒋 + 𝒛𝒌 .

Solution: Here 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 and 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2

So differentiating partially w. r. t. x, 𝜕𝑟

𝜕𝑥=

𝑥

𝑟, Similarly,

𝜕𝑟

𝜕𝑦=

𝑦

𝑟 and

𝜕𝑟

𝜕𝑧=

𝑧

𝑟 … (1)

19

Consider ∇ 𝑟𝑛 = 𝑖 𝜕

𝜕𝑥 𝑟𝑛 + 𝑗

𝜕

𝜕𝑦 𝑟𝑛 + 𝑘

𝜕

𝜕𝑧 𝑟𝑛

= 𝑖 𝑛 𝑟𝑛−1 𝜕𝑟

𝜕𝑥 + 𝑗 𝑛 𝑟𝑛−1 𝜕𝑟

𝜕𝑦 + 𝑘 𝑛 𝑟𝑛−1 𝜕𝑟

𝜕𝑧

= 𝑖 𝑛 𝑟𝑛−1 𝑥

𝑟 + 𝑗 𝑛 𝑟𝑛−1 𝑦

𝑟 + 𝑘 𝑛 𝑟𝑛−1 𝑧

𝑟

= 𝑖 𝑛 𝑟𝑛−2 𝑥 + 𝑗 𝑛 𝑟𝑛−2 𝑦 + 𝑘 𝑛 𝑟𝑛−2 𝑧

= 𝑛 𝑟𝑛−2 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = 𝑛 𝑟𝑛−2 𝑟

Hence ∇ 𝑟𝑛 = 𝑛 𝑟𝑛−2 𝑟

Example 18: Find the unit vector normal to the surface 𝒙𝟑 + 𝒚𝟑 + 𝟑𝒙𝒚𝒛 = 𝟑 at

(i) the point (1, 2, -1) * (ii) the point (1, 3, -1)** [KUK *2006, **2011]

Solution: We know that a vector normal to a surface is given by its gradient, so if 𝑛 is the vector

normal to the given surface then

𝑛 = ∇ 𝑥3 + 𝑦3 + 3𝑥𝑦𝑧 − 3 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝑥3 + 𝑦3 + 3𝑥𝑦𝑧 − 3

= 3𝑥2 + 3𝑦𝑧 𝑖 + 3𝑦2 + 3𝑥𝑧 𝑗 + 3𝑥𝑦 𝑘

(i) At point (1, 2, -1), 𝑛 = 3 1 2 + 3 2 −1 𝑖 + 3 2 2 + 3 1 −1 𝑗 + 3 1 2 𝑘

= −𝟑𝑖 + 9𝑗 + 6𝑘

Also 𝑛 = −3 2 + 9 2 + 6 2 = 126

Therefore the unit normal vector to the given surface at a point (1, 2, -1) is

𝑛 =1

126 −𝟑𝑖 + 9𝑗 + 6𝑘 .

(ii) At point (1, 3, -1), 𝑛 = 3 1 2 + 3 3 −1 𝑖 + 3 3 2 + 3 1 −1 𝑗 + 3 1 3 𝑘

= −6𝑖 + 24𝑗 + 9𝑘

Also 𝑛 = −6 2 + 24 2 + 9 2 = 693

Therefore the unit normal vector to the given surface at a point (1, 3, -1) is

𝑛 =1

693 – 6 𝑖 + 24 𝑗 + 9 𝑘 .

Example 19: Show that 𝒈𝒓𝒂𝒅 𝒆(𝒙𝟐+𝒚𝟐+𝒛𝟐) = 𝟐𝒆𝒓𝟐, 𝐰𝐡𝐞𝐫𝐞 𝒓𝟐 = 𝒓 𝟐 = 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐.

Solution: Here 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2

So differentiating w. r. t. x, 𝜕𝑟

𝜕𝑥=

𝑥

𝑟, Similarly,

𝜕𝑟

𝜕𝑦=

𝑦

𝑟 and

𝜕𝑟

𝜕𝑧=

𝑧

𝑟.

Now 𝑔𝑟𝑎𝑑 𝑒 𝑥2+𝑦2+𝑧2 = ∇ 𝑒𝑟2= 𝑖

𝜕

𝜕𝑥 𝑒𝑟2

+ 𝑗 𝜕

𝜕𝑦 𝑒𝑟2

+ 𝑘 𝜕

𝜕𝑧 𝑒𝑟2

= 𝑖 𝑒𝑟2 2𝑟

𝜕𝑟

𝜕𝑥 + 𝑗 𝑒𝑟2

2𝑟𝜕𝑟

𝜕𝑦 + 𝑘 𝑒𝑟2

2𝑟𝜕𝑟

𝜕𝑧

20

= 𝑒𝑟2 2𝑟 𝑖

𝜕𝑟

𝜕𝑥+ 𝑗

𝜕𝑟

𝜕𝑦+ 𝑘

𝜕𝑟

𝜕𝑧 = 𝑒𝑟2

2𝑟 𝑖 𝑥

𝑟+ 𝑗

𝑦

𝑟+ 𝑘

𝑧

𝑟

= 2𝑒𝑟2𝑟

Example 20: If 𝒖 = 𝒙 + 𝒚 + 𝒛, 𝒗 = 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 𝐚𝐧𝐝 𝒘 = 𝒚𝒛 + 𝒛𝒙 + 𝒙𝒚 , then show that

𝒈𝒓𝒂𝒅 𝒖, 𝒈𝒓𝒂𝒅 𝒗 𝐚𝐧𝐝 𝒈𝒓𝒂𝒅 𝒘 are coplanar.

Solution: Consider 𝑔𝑟𝑎𝑑 𝑢 = 𝑖 𝜕𝑢

𝜕𝑥+ 𝑗

𝜕𝑢

𝜕𝑦+ 𝑘

𝜕𝑢

𝜕𝑧 = 𝑖 1 + 𝑗 1 + 𝑘 1 = 𝑖 + 𝑗 + 𝑘

𝑔𝑟𝑎𝑑 𝑣 = 𝑖 𝜕𝑣

𝜕𝑥+ 𝑗

𝜕𝑣

𝜕𝑦+ 𝑘

𝜕𝑣

𝜕𝑧= 𝑖 2𝑥 + 𝑗 2𝑦 + 𝑘 2𝑧 = 2𝑥 𝑖 + 2𝑦 𝑗 + 2𝑧 𝑘

𝑔𝑟𝑎𝑑 𝑤 = 𝑖 𝜕𝑤

𝜕𝑥+ 𝑗

𝜕𝑤

𝜕𝑦+ 𝑘

𝜕𝑤

𝜕𝑧= 𝑦 + 𝑧 𝑖 + 𝑧 + 𝑥 𝑗 + 𝑥 + 𝑦 𝑘

We know that three vectors 𝑎 , 𝑏 𝑎𝑛𝑑 𝑐 are coplanar if their scalar triple product is zero i.e.

𝑎 𝑏 𝑐 = 0

Consider 𝑔𝑟𝑎𝑑 𝑢 𝑔𝑟𝑎𝑑 𝑣 𝑔𝑟𝑎𝑑 𝑤 = 1 1 1

2𝑥 2𝑦 2𝑧𝑦 + 𝑧 𝑧 + 𝑥 𝑥 + 𝑦

= 2 1 1 1𝑥 𝑦 𝑧

𝑦 + 𝑧 𝑧 + 𝑥 𝑥 + 𝑦 (taking common 2 from R2)

= 2 1 1 1

𝑥 + 𝑦 + 𝑧 𝑥 + 𝑦 + 𝑧 𝑥 + 𝑦 + 𝑧𝑦 + 𝑧 𝑧 + 𝑥 𝑥 + 𝑦

(adding R2 and R3)

= 2(𝑥 + 𝑦 + 𝑧) 1 1 11 1 1

𝑦 + 𝑧 𝑧 + 𝑥 𝑥 + 𝑦

(taking common (x + y + z) from R2)

= 2 𝑥 + 𝑦 + 𝑧 0 = 0

Hence 𝑔𝑟𝑎𝑑 𝑢, 𝑔𝑟𝑎𝑑 𝑣 and 𝑔𝑟𝑎𝑑 𝑤 are coplanar.

Example 21: Show that 𝒈𝒓𝒂𝒅 𝒇 𝒓 × 𝒓 = 𝟎 .

Solution: Here 𝑔𝑟𝑎𝑑 𝑓 𝑟 = ∇ 𝑓(𝑟) = 𝑖 𝜕

𝜕𝑥 𝑓(𝑟) + 𝑗

𝜕

𝜕𝑦 𝑓(𝑟) + 𝑘

𝜕

𝜕𝑧 𝑓(𝑟)

= 𝑖 𝑓 ′(𝑟) 𝜕𝑟

𝜕𝑥 + 𝑗 𝑓 ′(𝑟)

𝜕𝑟

𝜕𝑦 + 𝑘 𝑓 ′(𝑟)

𝜕𝑟

𝜕𝑧

= 𝑓 ′(𝑟) 𝑖 𝜕𝑟

𝜕𝑥+ 𝑗

𝜕𝑟

𝜕𝑦+ 𝑘

𝜕𝑟

𝜕𝑧 = 𝑓 ′(𝑟) 𝑖

𝑥

𝑟+ 𝑗

𝑦

𝑟+ 𝑘

𝑧

𝑟

= 𝑓 ′(𝑟) 𝑟

𝑟

21

Now 𝑔𝑟𝑎𝑑 𝑓 𝑟 × 𝑟 = 𝑓 ′(𝑟) 𝑟

𝑟× 𝑟 = 𝑓 ′(𝑟)

1

𝑟 𝑟 × 𝑟 = 0

𝑠𝑖𝑛𝑐𝑒 𝑟 × 𝑟 = 0

Example 22: Find the directional derivative of 𝒇 𝒙, 𝒚, 𝒛 = 𝒙𝟐𝒚𝟐𝒛𝟐 at the point (1, 1, -1) in the

direction of the tangent to the curve 𝒙 = 𝒆𝒕, 𝒚 = 𝟐 𝒔𝒊𝒏 𝒕 + 𝟏, 𝒛 = 𝒕 − 𝒄𝒐𝒔 𝒕 at t = 0.

Solution: Consider ∇ 𝑓 𝑥, 𝑦, 𝑧 = ∇ 𝑥2𝑦2𝑧2 = 2𝑥𝑦2𝑧2 𝑖 + 2𝑦𝑥2𝑧2 𝑗 + 2𝑧𝑥2𝑦2 𝑘

At (1, 1, -1), ∇ 𝑓 𝑥, 𝑦, 𝑧 = 2 𝑖 + 2 𝑗 − 2 𝑘

Now 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = 𝑒𝑡 𝑖 + 2 sin 𝑡 + 1 𝑗 + (𝑡 − cos 𝑡)𝑘

So tangent to the curve is 𝑑𝑟

𝑑𝑡 = 𝑒𝑡 𝑖 + 2 cos 𝑡 𝑗 + (1 + sin 𝑡)𝑘

At t = 0, 𝑑𝑟

𝑑𝑡 = 𝑖 + 2𝑗 + 𝑘

And the unit tangent vector is 𝑑𝑟

𝑑𝑡/

𝑑𝑟

𝑑𝑡 =

𝑖 +2𝑗 +𝑘

6

So the required directional derivative in the direction of the tangent is ∇𝑓 𝑥, 𝑦, 𝑧 ∙ 𝑑𝑟

𝑑𝑡/

𝑑𝑟

𝑑𝑡

= 2 𝑖 + 2 𝑗 − 2 𝑘 ∙ 𝑖 + 2𝑗 + 𝑘 / 6 =4

6=

2 3

3

Example 23: If the directional derivative ∅ = 𝒂 𝒙𝟐𝒚 + 𝒃 𝒚𝟐𝒛 + 𝒄 𝒛𝟐𝒙 at the point (1, 1, 1) has

maximum magnitude 15 in the direction parallel to the line 𝒙−𝟏

𝟐=

𝒚−𝟑

−𝟐=

𝒛

𝟏 , find the values of

a, b and c. [Madrass 2004]

Solution: Consider ∇ ∅ = ∇ (𝑎 𝑥2𝑦 + 𝑏 𝑦2𝑧 + 𝑐 𝑧2𝑥)

= 2𝑎 𝑥𝑦 + 𝑐 𝑧2 𝑖 + 𝑎 𝑥2 + 2𝑏 𝑦𝑧 𝑗 + (𝑏 𝑦2 + 2𝑐 𝑧𝑥)𝑘

At (1, 1, 1), ∇ ∅ = 2𝑎 + 𝑐 𝑖 + 𝑎 + 2𝑏 𝑗 + (𝑏 + 2𝑐)𝑘

We know that directional derivative of ∅ is maximum in the direction of its normal vector ∇ ∅, but it

is given to be maximum in the direction of the line 𝑥−1

2=

𝑦−3

−2=

𝑧

1 .

Therefore, the line and normal vector are parallel to each other, which results as:

2𝑎+𝑐

2=

𝑎+2𝑏

−2=

𝑏+2𝑐

1 … (1)

Taking first two members of (1), 3𝑎 + 2𝑏 + 𝑐 = 0

and by last two members of (1), 𝑎 + 4𝑏 + 4𝑐 = 0

Solving the two obtained equations, 𝑎

4=

𝑏

−11=

𝑐

10 = 𝜆 𝐿𝑒𝑡

=> 𝑎 = 4𝜆, 𝑏 = −11𝜆 𝑎𝑛𝑑 𝑐 = 10𝜆 … (2)

Also given that maximum magnitude of directional derivative is 15 units i.e. ∇∅ = 15

So , 2𝑎 + 𝑐 2 + 𝑎 + 2𝑏 2 + 𝑏 + 2𝑐 2 = 15 2 …(3)

Putting the values of a, b and c from (2),

22

8𝜆 + 10𝜆 2 + 4𝜆 − 22𝜆 2 + −11𝜆 + 20𝜆 2 = 15 2 => 𝜆 = ± 5

9

Hence 𝑎 = ± 20

9 , 𝑏 = ∓

55

9 , 𝑐 = ±

50

9.

Example 24: In what direction from (3, 1, -2) is the directional derivative of ∅ = 𝒙𝟐𝒚𝟐𝒛𝟒

maximum? Find also the magnitude of this maximum. [KUK 2010, 2007, 2006]

Solution: The vector normal to the given surface is

𝑛 = ∇ ∅ = ∇ 𝑥2𝑦2𝑧4 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝑥2𝑦2𝑧4

= 2𝑥𝑦2𝑧4 𝑖 + 2𝑥2𝑦𝑧4 𝑗 + 4𝑥2𝑦2𝑧3 𝑘

At point (3, 1, -2) 𝑛 = 2 3 1 2 −2 4 𝑖 + 2 3 2 1 −2 4 𝑗 + 4 3 2 1 2 −2 3 𝑘

= 96 𝑖 + 288 𝑗 − 288 𝑘

Also 𝑛 = 96 2 + 288 2 + −288 2 = 96 19

So the directional derivative of given surface will be maximum in the direction of

96 𝑖 + 288 𝑗 − 288 𝑘 and the magnitude of this maximum is 96 19.

Example 25: Find the angle between the surfaces 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 = 𝟗 and 𝒛 = 𝒙𝟐 + 𝒚𝟐 − 𝟑 at

(2, -1, 2). [KUK 2008]

Solution: Given surfaces are

∅1 = 𝑥2 + 𝑦2 + 𝑧2 − 9 = 0 … (1)

and ∅2 = 𝑥2 + 𝑦2 − 𝑧 − 3 = 0 … (2)

We know that gradient of a surface gives the vector normal to the surface. Let 𝑛 1 and 𝑛 2 are the

vectors normal to the surfaces (1) and (2) respectively.

Now ∇∅1 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑥2 + 𝑦2 + 𝑧2 = 9 = 2𝑥 𝑖 + 2𝑦 𝑗 + 2𝑧 𝑘 … (3)

∇∅2 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑥2 + 𝑦2 − 𝑧 − 3 = 2𝑥 𝑖 + 2𝑦 𝑗 − 𝑘 … (4)

So, 𝑛 1 = ∇∅1 at (2,−1,2) = 4 𝑖 − 2 𝑗 + 4 𝑘 and 𝑛 2 = ∇∅2 at (2,−1,2) = 4 𝑖 − 2 𝑗 − 𝑘

Let 𝜃 be the angle between the given surfaces at point (2, -1, 2), then 𝜃 will also be an angle between

their normals 𝑛 1 and 𝑛 2.

Therefore, cos 𝜃 = 𝑛 1∙𝑛 2

𝑛 1 𝑛 2 =

4 𝑖 −2 𝑗 +4 𝑘 ∙ 4 𝑖 −2 𝑗 − 𝑘

4 𝑖 −2 𝑗 +4 𝑘 4 𝑖 −2 𝑗 − 𝑘 =

16+4−4

36 21=

8

3 21.

Example 26: Find the constants a and b so that the surface 𝒂𝒙𝟐 − 𝒃𝒚𝒛 = 𝒂 + 𝟐 𝒙 is orthogonal

to the surface 𝟒𝒙𝟐𝒚 + 𝒛𝟑 = 𝟒 at the point (1, -1, 2).

Solution: Given surfaces are

𝑓 = 𝑎𝑥2 − 𝑏𝑦𝑧 − 𝑎 + 2 𝑥 = 0 … (1)

and 𝑔 = 4𝑥2𝑦 + 𝑧3 − 4 = 0 …(2)

23

Let 𝑛1 and 𝑛2 are the vectors normal to the surfaces (1) and (2) at (1, -1, 2), respectively.

Consider ∇𝑓 = 𝑖 𝜕𝑓

𝜕𝑥+ 𝑗

𝜕𝑓

𝜕𝑦+ 𝑘

𝜕𝑓

𝜕𝑧 = 𝑖 2𝑎𝑥 − 𝑎 + 2 + 𝑗 (−𝑏𝑧) + 𝑘 (−𝑏𝑦) … (3)

And ∇𝑔 = 𝑖 𝜕𝑔

𝜕𝑥+ 𝑗

𝜕𝑔

𝜕𝑦+ 𝑘

𝜕𝑔

𝜕𝑧 = 𝑖 8𝑥𝑦 + 𝑗 (4𝑥2) + 𝑘 (3𝑧2) … (4)

So, 𝑛1 = ∇𝑓 (1, −1, 2) = 𝑖 𝑎 − 2 + 𝑗 −2𝑏 + 𝑘 𝑏 = 𝑎 − 2 𝑖 − 2𝑏𝑗 + 𝑏𝑘

𝑛2 = ∇𝑔 (1, −1, 2) = 𝑖 −8 + 𝑗 4 + 𝑘 12 = −8𝑖 + 4𝑗 + 12𝑘

Given that surfaces (1) and (2) cut orthogonally at (1, -1, 2), so their normal vectors i.e. 𝑛1 and 𝑛2

should also be orthogonal to each other.

Therefore, 𝑛1 . 𝑛2 = 0

=> 𝑎 − 2 𝑖 − 2𝑏𝑗 + 𝑏𝑘 . −8𝑖 + 4𝑗 + 12𝑘 = 0

=> −8 𝑎 − 2 − 8𝑏 + 12𝑏 = 0

=> 2𝑎 − 𝑏 = 4 … (5)

Also the point (1, -2, 1) lies on the surface (1), so we have

𝑎 + 2𝑏 − 𝑎 + 2 = 0 𝑜𝑟 2𝑏 − 2 = 0 𝑜𝑟 𝒃 = 𝟏

Putting value of b in (3), we get 2𝑎 − 1 = 4 𝑜𝑟 𝒂 =𝟓

𝟐

Example 27: Show that the components of a vector 𝒓 along and normal (perpendicular) to a

vector 𝒂 , in the plane of 𝒓 and 𝒂 , are 𝒓 ∙𝒂

𝒂 𝟐 𝒂 and

𝐚 × 𝒓 ×𝒂

𝒂 𝟐.

Solution: Let 𝑂𝐴 = 𝑎 and 𝑂𝐵 = 𝑟 and 𝑂𝑀 be the projection

of 𝑟 on 𝑎 (Fig. 17.8)

∴ Component of 𝑟 along 𝑎 = OM (unit vector along 𝑎 )

= 𝑟 ∙ 𝑎 𝑎 = 𝑟 ∙𝑎

𝑎

𝑎

𝑎 =

𝑟 ∙𝑎

𝑎 2 𝑎

Also the component of 𝑟 normal to 𝑎 = 𝑀𝐵 = 𝑂𝐵 − 𝑂𝑀

= 𝑟 − 𝑟 ∙𝑎

𝑎 2 𝑎 =

𝑎 ∙𝑎 𝑟 − 𝑟 ∙𝑎 𝑎

𝑎 2=

a × 𝑟 ×𝑎

𝑎 2

Example 28: If f and 𝑭 are point functions, prove that the components of the latter normal and

tangential to the surface f = 0 are 𝑭 ∙𝛁𝒇 𝛁𝒇

𝛁𝒇 𝟐 and

𝛁𝒇× 𝑭 ×𝛁𝒇

𝛁𝒇 𝟐.

Solution: We know that for the given surface f = 0, the vector normal to the surface is given by the

gradient i.e. ∇𝑓.

Now the component of 𝐹 normal to the given surface is = 𝐹 ∙ ∇𝑓

∇𝑓

∇𝑓

∇𝑓 =

𝐹 ∙∇𝑓 ∇𝑓

∇𝑓 2=

𝐹 ∙∇𝑓 ∇𝑓

∇𝑓 2

And the component of 𝐹 tangential to the given surface is = 𝐹 − 𝑡𝑕𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹

24

= 𝐹 − 𝐹 ∙∇𝑓 ∇𝑓

∇𝑓 2=

𝐹 ∇𝑓 2− 𝐹 ∙∇𝑓 ∇𝑓

∇𝑓 2=

𝐹 ∇𝑓∙∇𝑓 − 𝐹 ∙∇𝑓 ∇𝑓

∇𝑓 2=

∇𝑓× 𝐹 ×∇𝑓

∇𝑓 2

ASSIGNMENT 3

1. Find ∇∅, if ∅ = log 𝑥2 + 𝑦2 + 𝑧2 .

2. Show that 1

𝑟= −

𝑟

𝑟3.

3. What is the directional derivative of ∅ = 𝑥𝑦2 + 𝑦𝑧3 at the point (2, -1, 1) in the direction of the

normal to the surface 𝑥 log 𝑧 − 𝑦2 = −4 at (-1, 2, 1)? [JNTU 2005; VTU 2004]

4. What is the directional derivative of ∅ = 𝑥2𝑦𝑧 + 4𝑥𝑧2at the point (1, -2, 1) in the direction of the

vector 2𝑖 − 𝑗 − 2𝑘 . [VTU 2007; UP Tech, JNTU 2006]

5. The temperature of points in a space is given by 𝑇 𝑥, 𝑦, 𝑧 = 𝑥2 + 𝑦2 − 𝑧. A mosquito located at

(1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what

direction should it move?

6. What is the greatest rate of increase of 𝑢 = 𝑥2 + 𝑦𝑧2 at the point (1, -1, 3)?

7. Find the angle between the tangent planes to the surfaces 𝑥 log 𝑧 = 𝑦2 − 1 and 𝑥2𝑦 = 2 − 𝑧 at the

point (1, 1, 1). [JNTU 2003]

8. Calculate the angle between the normals to the surface 𝑥𝑦 = 𝑧2 at the points (4, 1, 2) and

(3, 3, -3).

9. Find the angle between the surfaces 𝑥2 + 𝑦2 + 𝑧2 = 9 and 𝑧 = 𝑥2 + 𝑦2 − 3 at (2, -1, 2).

10. Find the values of 𝜆 and 𝜇 so that the surface 𝜆𝑥2𝑦 + 𝜇𝑧3 = 4 may cut the surface

5𝑥2 = 2𝑦𝑧 + 9𝑥 orthogonally at (1, -1, 2)

17.6 DEL APPLIED TO VECTOR POINT FUNCTIONS (Divergence & Curl)

1. Divergence: Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑓1(𝑥, 𝑦, 𝑧)𝑖 + 𝑓2(𝑥, 𝑦, 𝑧)𝑗 + 𝑓3(𝑥, 𝑦, 𝑧)𝑘 be a continuously differentiable

vector point function. Divergence 𝑓 (𝑥, 𝑦, 𝑧) of is a scalar which is denoted by ∇ ∙ 𝑓 and is defined as

∇ ∙ 𝑓 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3𝑘 =

𝜕𝑓1

𝜕𝑥+

𝜕𝑓2

𝜕𝑦+

𝜕𝑓3

𝜕𝑧.

t is also denoted by 𝑑𝑖𝑣 𝑓 .

Physical interpretation of Divergence

Consider the case of fluid flow.

Let 𝑣 = 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 + 𝑣𝑧𝑘 be the velocity of the fluid at a

point 𝑃(𝑥, 𝑦, 𝑧) . Consider a small parallelopiped with

edges 𝛿𝑥, 𝛿𝑦 and 𝛿𝑧 parallel to the x, y and z axis

25

respectively in the mass of fluid, with one of its corner at point 𝑃.

So, the mass of fluid flowing in through the face 𝑃𝑄𝑅𝑆 per unit time = 𝑣𝑦 𝛿𝑧 𝛿𝑥

and the mass of fluid flowing out of the face 𝑃′𝑄′𝑅′𝑆′ per unit time

= 𝑣𝑦+𝛿𝑦 𝛿𝑧 𝛿𝑥 = 𝑣𝑦 +𝜕𝑣𝑦

𝜕𝑦 𝛿𝑦 𝛿𝑧 𝛿𝑥

∴ The net decrease in fluid mass in the parallelopiped corresponding to flow along y-axis

= 𝑣𝑦 +𝜕𝑣𝑦

𝜕𝑦 𝛿𝑦 𝛿𝑧 𝛿𝑥 − 𝑣𝑦 𝛿𝑧 𝛿𝑥 =

𝜕𝑣𝑦

𝜕𝑦 𝛿𝑥 𝛿𝑦 𝛿𝑧

Similarly, the net decrease in fluid mass in the parallelopiped corresponding to the flow along x-axis

and z-axis is 𝜕𝑣𝑥

𝜕𝑥 𝛿𝑥 𝛿𝑦 𝛿𝑧 and

𝜕𝑣𝑧

𝜕𝑧 𝛿𝑥 𝛿𝑦 𝛿𝑧 respectively.

So, total decrease in mass of fluid mass in the parallelopiped per unit time

= 𝜕𝑣𝑥

𝜕𝑥+

𝜕𝑣𝑦

𝜕𝑦+

𝜕𝑣𝑧

𝜕𝑧 𝛿𝑥 𝛿𝑦 𝛿𝑧

Thus, the rate of loss of fluid per unit volume =𝜕𝑣𝑥

𝜕𝑥+

𝜕𝑣𝑦

𝜕𝑦+

𝜕𝑣𝑧

𝜕𝑧

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 + 𝑣𝑧𝑘

= ∇ ∙ 𝑣 = 𝑑𝑖𝑣 𝑣

Hence, 𝑑𝑖𝑣 𝑣 gives the rate at which fluid is originating or diminishing at a point per unit volume.

If the fluid is incompressible, there can be no loss or gain in the volume element i.e. 𝑑𝑖𝑣 𝑣 = 0.

Observations:

(i) if 𝑣 represent the electric flux, then 𝑑𝑖𝑣 𝑣 is the amount of flux which diverges per unit volume in unit

time.

(ii) if 𝑣 represent the heat flux, then 𝑑𝑖𝑣 𝑣 is the rate at which the heat is issuing from a point per unit

volume.

(iii) If the flux entering any element of space is the same as that leaving it i.e.𝑑𝑖𝑣 𝑣 = 0 everywhere, then

such a vector point function is called Solenoidal.

Example 29: If 𝒓 = 𝒙𝒊 + 𝒚𝒋 + 𝒛𝒌 then show that 𝒅𝒊𝒗 𝒓 = 𝟑.

Solution: 𝑑𝑖𝑣 𝑟 = ∇ ∙ 𝑟 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘

=𝜕

𝜕𝑥 𝑥 +

𝜕

𝜕𝑦 𝑦 +

𝜕

𝜕𝑧 𝑧 = 1 + 1 + 1 = 3

Example 30: Evaluate 𝒅𝒊𝒗 𝒇 where 𝒇 = 𝟐𝒙𝟐𝒛 𝒊 − 𝒙𝒚𝟐𝒛 𝒋 + 𝟑𝒚𝟐𝒙 𝒌 at (1, 1, 1).

Solution: 𝑑𝑖𝑣 𝑓 = 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 ∙ 2𝑥2𝑧 𝑖 − 𝑥𝑦2𝑧 𝑗 + 3𝑦2𝑥 𝑘

26

=𝜕

𝜕𝑥 2𝑥2𝑧 +

𝜕

𝜕𝑦 −𝑥𝑦2𝑧 +

𝜕

𝜕𝑧 3𝑦2𝑥

= 4𝑥𝑧 − 2𝑥𝑦𝑧 + 0

At (1, 1, 1), 𝑑𝑖𝑣 𝑓 = 4 1 1 − 2 1 1 1 = 2

Example 31: Determine the constant a so that the vector 𝒇 = 𝒙 + 𝟑𝒚 𝒊 + 𝒚 − 𝟐𝒛 𝒋 + (𝒙 + 𝒂𝒛) 𝒌

is solenoidal.

Solution: Given that the vector 𝑓 is solenoidal, so 𝑑𝑖𝑣 𝑓 = 0

𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑥 + 3𝑦 𝑖 + 𝑦 − 2𝑧 𝑗 + (𝑥 + 𝑎𝑧) 𝑘 = 0

𝜕

𝜕𝑥 𝑥 + 3𝑦 +

𝜕

𝜕𝑦 𝑦 − 2𝑧 +

𝜕

𝜕𝑧 𝑥 + 𝑎𝑧 = 0

1 + 1 + 𝑎 = 0 => 𝑎 = −2

2. Curl: Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑓1(𝑥, 𝑦, 𝑧)𝑖 + 𝑓2(𝑥, 𝑦, 𝑧)𝑗 + 𝑓3(𝑥, 𝑦, 𝑧)𝑘 be a continuously differentiable vector

point function. Curl of 𝑓 (𝑥, 𝑦, 𝑧) of is a vector which is denoted by ∇ × 𝑓 and is defined as

∇ × 𝑓 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 × 𝑓 = 𝑖 ×

𝜕𝑓

𝜕𝑥 +𝑗 ×

𝜕𝑓

𝜕𝑦 +𝑘 ×

𝜕𝑓

𝜕𝑧

Also in component form, curl of 𝑓 (𝑥, 𝑦, 𝑧) is

∇ × 𝑓 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 × 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3𝑘

=

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑓1 𝑓2 𝑓3

= 𝑖 𝜕𝑓3

𝜕𝑦−

𝜕𝑓2

𝜕𝑧 + 𝑗

𝜕𝑓1

𝜕𝑧−

𝜕𝑓3

𝜕𝑥 + 𝑘

𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦

Physical Interpretation of Curl

Consider the motion of a rigid body rotating with

angular velocity 𝜔 about an axis OA, where O is a

fixed point in the body. Let 𝑟 be the position vector of

any point P of the body. The point P describing a circle

whose center is M and radius is PM = 𝑟 sin 𝜃 where 𝜃 is

the angle between 𝜔 and 𝑟 , then the velocity of P is

𝜔𝑟 sin 𝜃. This velocity is normal to the plane POM i.e.

normal to the plane of 𝜔 and 𝑟 .

So, if 𝑣 is the linear velocity of P, then 𝑣 =

𝜔 𝑟 sin 𝜃 𝑛 = 𝜔 × 𝑟

27

(𝑛 𝑏𝑒𝑖𝑛𝑔 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝜔 𝑎𝑛𝑑 𝑟 )

Now, if 𝜔 = 𝜔1 𝑖 + 𝜔2 𝑗 + 𝜔3𝑘 and 𝑟 = 𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 , then

𝐶𝑢𝑟𝑙 𝑣 = 𝜔 × 𝑟 = 𝑖 𝑗 𝑘

𝜔1 𝜔2 𝜔3

𝑥 𝑦 𝑧 = 𝑖 𝜔2𝑧 − 𝜔3𝑦 + 𝑗 𝜔3𝑥 − 𝜔1𝑧 + 𝑘 𝜔1𝑦 − 𝜔2𝑥

And

𝐶𝑢𝑟𝑙 𝑣 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝜔2𝑧 − 𝜔3𝑦 𝜔3𝑥 − 𝜔1𝑧 𝜔1𝑦 − 𝜔2𝑥

= 𝑖 𝜔1 + 𝜔1 + 𝑗 𝜔2 + 𝜔2 + 𝑘 𝜔3 + 𝜔3

= 2𝜔1 𝑖 + 2𝜔2 𝑗 + 2𝜔3 𝑘 = 2 𝜔

Hence 𝜔 =1

2𝐶𝑢𝑟𝑙 𝑣

Thus the angular velocity of rotation at any point is equal to half the curl of the velocity vector.

Observations:

(i) The curl of a vector point function gives the measure of the angular velocity at a point.

(ii) If the curl of a vector point function becomes zero i.e. ∇ × 𝑓 = 0, then 𝑓 is called an irrotational

vector.

Example 32: If 𝒓 = 𝒙𝒊 + 𝒚𝒋 + 𝒛𝒌 then show that 𝒄𝒖𝒓𝒍 𝒓 = 𝟎 .

Solution: 𝑐𝑢𝑟𝑙 𝑟 = ∇ × 𝑟 =

𝑖 𝑗 𝑘 𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑥 𝑦 𝑧

= 𝑖 𝜕𝑧

𝜕𝑦−

𝜕𝑦

𝜕𝑧 + 𝑗

𝜕𝑥

𝜕𝑧−

𝜕𝑧

𝜕𝑥 + 𝑘

𝜕𝑦

𝜕𝑥−

𝜕𝑥

𝜕𝑦

= 𝑖 0 − 0 + 𝑗 0 − 0 + 𝑘 0 − 0 = 0 .

Example 33: Find a so that the vector 𝒇 = 𝒂𝒙𝒚 − 𝒛𝟑 𝒊 + 𝒂 − 𝟐 𝒙𝟐 𝒋 + 𝟏 − 𝒂 𝒙𝒛𝟐 𝒌 is

irrotational.

Solution: Given that 𝑓 is irrotational, therefore 𝑐𝑢𝑟𝑙 𝑓 = 0 … (1)

But 𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑎𝑥𝑦 − 𝑧3 𝑎 − 2 𝑥2 1 − 𝑎 𝑥𝑧2

= 𝑖 0 − 0 + 𝑗 −3𝑧2 − 1 − 𝑎 𝑧2 + 𝑘 𝑎 − 2 2𝑥 − 𝑎𝑥

= 0 𝑖 + −4 + 𝑎 𝑧2 𝑗 + −4 + 𝑎 𝑥 𝑘

28

Using (1), 0 𝑖 + −4 + 𝑎 𝑧2 𝑗 + −4 + 𝑎 𝑥 𝑘 = 0 𝑖 + 0 𝑗 + 0 𝑘

Comparing the corresponding components both sides,

−4 + 𝑎 = 0 => 𝑎 = 4.

Example 34: If 𝒇 = 𝒙𝒚𝟐 𝒊 + 𝟐𝒙𝟐𝒚𝒛 𝒋 − 𝟑𝒚𝒛𝟐 𝒌 , find the 𝒄𝒖𝒓𝒍 𝒇 at the point (1, -1, 1).

Solution: 𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑥𝑦2 2𝑥2𝑦𝑧 −3𝑦𝑧2

= 𝑖 −3𝑧2 − 2𝑥2𝑦 + 𝑗 0 − 0 + 𝑘 4𝑥𝑦𝑧 − 2𝑥𝑦

At (1, -1, 1),

𝑐𝑢𝑟𝑙 𝑓 = 𝑖 −3(1)2 − 2 1 2(−1) + 𝑗 0 − 0 + 𝑘 4 1 −1 (1) − 2(1)(−1)

= −𝑖 − 2 𝑘 .

17.7 DEL APPLIED TO THE PRODUCT OF POINT FUNCTIONS

Let 𝜙, 𝜓 are two scalar point functions and 𝑓 , 𝑔 are two vector point functions, then

1. ∇ ϕ ψ = ϕ ∇ψ + ψ ∇ϕ

2. ∇ ∙ ϕ 𝑓 = ∇ϕ ∙ 𝑓 + ϕ ∇ ∙ 𝑓

3. ∇ × ϕ 𝑓 = ∇ϕ × 𝑓 + ϕ ∇ × 𝑓

4. ∇ 𝑓 ∙ 𝑔 = 𝑓 ∙ ∇ 𝑔 + 𝑔 ∙ ∇ 𝑓 + 𝑓 × ∇ × 𝑔 + 𝑔 × ∇ × 𝑓 [KUK 2007]

5. ∇ ∙ 𝑓 × 𝑔 = 𝑔 ∙ ∇ × 𝑓 − 𝑓 ∙ ∇ × 𝑔

6. ∇ × 𝑓 × 𝑔 = 𝑓 ∇ ∙ 𝑔 − 𝑔 ∇ ∙ 𝑓 + 𝑔 ∙ ∇ 𝑓 − 𝑓 ∙ ∇ 𝑔 [KUK 2011]

Proof 1: Consider ∇ ϕ ψ = 𝑖 𝜕

𝜕𝑥 ϕ ψ = 𝑖 ϕ

𝜕ψ

𝜕𝑥+ ψ

𝜕ϕ

𝜕𝑥 = 𝑖 ϕ

𝜕ψ

𝜕𝑥 + 𝑖 ψ

𝜕ϕ

𝜕𝑥

= ϕ 𝑖 𝜕ψ

𝜕𝑥 + ψ 𝑖

𝜕ϕ

𝜕𝑥 = ϕ ∇ψ + ψ ∇ϕ

Proof 2: Consider ∇ ∙ ϕ 𝑓 = 𝑖 ∙ 𝜕

𝜕𝑥 ϕ 𝑓 = 𝑖 ∙

𝜕ϕ

𝜕𝑥𝑓 + ϕ

𝜕𝑓

𝜕𝑥 = 𝑖 ∙

𝜕ϕ

𝜕𝑥𝑓 + 𝑖 ∙ ϕ

𝜕𝑓

𝜕𝑥

= 𝑖 𝜕ϕ

𝜕𝑥 ∙ 𝑓 + ϕ 𝑖 ∙

𝜕𝑓

𝜕𝑥 = ∇ϕ ∙ 𝑓 + ϕ ∇ ∙ 𝑓

Proof 3: Consider ∇ × ϕ 𝑓 = 𝑖 × 𝜕

𝜕𝑥 ϕ 𝑓 = 𝑖 ×

𝜕ϕ

𝜕𝑥𝑓 + ϕ

𝜕𝑓

𝜕𝑥 = 𝑖 ×

𝜕ϕ

𝜕𝑥𝑓 + 𝑖 × ϕ

𝜕𝑓

𝜕𝑥

= 𝑖 𝜕ϕ

𝜕𝑥 × 𝑓 + ϕ 𝑖 ×

𝜕𝑓

𝜕𝑥 = ∇ϕ × 𝑓 + ϕ ∇ × 𝑓

Proof 4: Consider ∇ 𝑓 ∙ 𝑔 = 𝑖 𝜕

𝜕𝑥 𝑓 ∙ 𝑔 = 𝑖

𝜕𝑓

𝜕𝑥∙ 𝑔 + 𝑓 ∙

𝜕𝑔

𝜕𝑥 = 𝑖

𝜕𝑓

𝜕𝑥 ∙ 𝑔 + 𝑖 𝑓 ∙

𝜕𝑔

𝜕𝑥

… (1)

29

But, 𝑔 × 𝑖 ×𝜕𝑓

𝜕𝑥 = 𝑔 ∙

𝜕𝑓

𝜕𝑥 𝑖 − 𝑔 ∙ 𝑖

𝜕𝑓

𝜕𝑥

or 𝑔 ∙𝜕𝑓

𝜕𝑥 𝑖 = 𝑔 × 𝑖 ×

𝜕𝑓

𝜕𝑥 + 𝑔 ∙ 𝑖

𝜕𝑓

𝜕𝑥

So, 𝑔 ∙𝜕𝑓

𝜕𝑥 𝑖 = 𝑔 × 𝑖 ×

𝜕𝑓

𝜕𝑥 + 𝑔 ∙ 𝑖

𝜕𝑓

𝜕𝑥= 𝑔 × ∇ × 𝑓 + 𝑔 ∙ ∇ 𝑓 … (2)

Interchanging 𝑓 and 𝑔 in (2)

𝑓 ∙𝜕𝑔

𝜕𝑥 𝑖 = 𝑓 × 𝑖 ×

𝜕𝑔

𝜕𝑥 + 𝑓 ∙ 𝑖

𝜕𝑔

𝜕𝑥= 𝑓 × ∇ × 𝑔 + 𝑓 ∙ ∇ 𝑔 … (3)

Using (2) and (3) in (1), we get

∇ 𝑓 ∙ 𝑔 = 𝑓 ∙ ∇ 𝑔 + 𝑔 ∙ ∇ 𝑓 + 𝑓 × ∇ × 𝑔 + 𝑔 × ∇ × 𝑓

Proof 5: Consider ∇ ∙ 𝑓 × 𝑔 = 𝑖 ∙ 𝜕

𝜕𝑥 𝑓 × 𝑔 = 𝑖 ∙

𝜕𝑓

𝜕𝑥× 𝑔 + 𝑓 ×

𝜕𝑔

𝜕𝑥

= 𝑖 ∙ 𝜕𝑓

𝜕𝑥× 𝑔 + 𝑖 ∙ 𝑓 ×

𝜕𝑔

𝜕𝑥 = 𝑔 ∙ 𝑖 ×

𝜕𝑓

𝜕𝑥 − 𝑓 ∙ 𝑖 ×

𝜕𝑔

𝜕𝑥

= 𝑔 ∙ ∇ × 𝑓 − 𝑓 ∙ ∇ × 𝑔

[using 𝑎 ∙ 𝑏 × 𝑐 = 𝑐 ∙ 𝑎 × 𝑏 = −𝑏 ∙ 𝑎 × 𝑐 ]

Proof 6: Consider ∇ × 𝑓 × 𝑔 = 𝑖 × 𝜕

𝜕𝑥 𝑓 × 𝑔 = 𝑖 ×

𝜕𝑓

𝜕𝑥× 𝑔 + 𝑓 ×

𝜕𝑔

𝜕𝑥

= 𝑖 × 𝜕𝑓

𝜕𝑥× 𝑔 + 𝑖 × 𝑓 ×

𝜕𝑔

𝜕𝑥

= 𝑖 ∙ 𝑔 𝜕𝑓

𝜕𝑥− 𝑖 ∙

𝜕𝑓

𝜕𝑥 𝑔 + 𝑖 ∙

𝜕𝑔

𝜕𝑥 𝑓 − 𝑖 ∙ 𝑓

𝜕𝑔

𝜕𝑥

[using 𝑎 × 𝑏 × 𝑐 = 𝑎 ∙ 𝑐 𝑏 − 𝑎 ∙ 𝑏 𝑐 ]

= 𝑔 ∙ 𝑖 𝜕𝑓

𝜕𝑥− 𝑔 𝑖 ∙

𝜕𝑓

𝜕𝑥 + 𝑓 𝑖 ∙

𝜕𝑔

𝜕𝑥 − 𝑓 ∙ 𝑖

𝜕𝑔

𝜕𝑥

= 𝑔 ∙ ∇ 𝑓 − 𝑔 ∇ ∙ 𝑓 + 𝑓 ∇ ∙ 𝑔 − 𝑓 ∙ ∇ 𝑔

= 𝑓 ∇ ∙ 𝑔 − 𝑔 ∇ ∙ 𝑓 + 𝑔 ∙ ∇ 𝑓 − 𝑓 ∙ ∇ 𝑔

17.8 DEL APPLIED TWICE TO POINT FUNCTIONS

Let ∅ be a scalar point function and 𝑓 be a vector point function, then ∇∅ and ∇ × 𝑓 being the vector

point functions, we can find their divergence and curl; whereas ∇ ∙ 𝑓 being the scalar point function,

we can find its gradient only. Thus we have following formulae:

1. 𝑑𝑖𝑣 𝑔𝑟𝑎𝑑 ∅ = ∇ ∙ ∇∅ = ∇2∅ =𝜕2∅

𝜕𝑥2 +𝜕2∅

𝜕𝑦2 +𝜕2∅

𝜕𝑧2

2. 𝑐𝑢𝑟𝑙 𝑔𝑟𝑎𝑑 ∅ = ∇ × ∇∅ = 0

3. 𝑑𝑖𝑣 𝑐𝑢𝑟𝑙 𝑓 = ∇ ∙ ∇ × 𝑓 = 0

4. 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑓 = ∇ × ∇ × 𝑓 = ∇ ∇ ∙ 𝑓 − ∇2𝑓 = 𝑔𝑟𝑎𝑑 𝑑𝑖𝑣 𝑓 − ∇2𝑓 [KUK 2006]

5. 𝑔𝑟𝑎𝑑 𝑑𝑖𝑣 𝑓 = ∇ ∇ ∙ 𝑓 = 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑓 + ∇2𝑓 = ∇ × ∇ × 𝑓 + ∇2𝑓

30

Proof 1: ∇2∅ = ∇ ∙ ∇∅ = ∇ ∙ 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧

=𝜕

𝜕𝑥 𝜕∅

𝜕𝑥 +

𝜕

𝜕𝑦

𝜕∅

𝜕𝑦 +

𝜕

𝜕𝑧

𝜕∅

𝜕𝑧 =

𝜕2∅

𝜕𝑥2+

𝜕2∅

𝜕𝑦2+

𝜕2∅

𝜕𝑧2

Here ∇2 is called the Laplacian Operator and ∇2∅ = 0 is called the Laplace’s Equation.

Proof 2: 𝑐𝑢𝑟𝑙 𝑔𝑟𝑎𝑑 ∅ = ∇ × ∇∅ = ∇ × 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧

=

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝜕∅

𝜕𝑥

𝜕∅

𝜕𝑦

𝜕∅

𝜕𝑧

= 𝑖

𝜕2∅

𝜕𝑦𝜕𝑧−

𝜕2∅

𝜕𝑧𝜕𝑦 = 0

Proof 3: 𝑑𝑖𝑣 𝑐𝑢𝑟𝑙 𝑓 = ∇ ∙ ∇ × 𝑓 = 𝑖 𝜕

𝜕𝑥 ∙ 𝑖 ×

𝜕𝑓

𝜕𝑥+ 𝑗 ×

𝜕𝑓

𝜕𝑦+ 𝑘 ×

𝜕𝑓

𝜕𝑧

= 𝑖 ∙ 𝑖 ×𝜕2𝑓

𝜕𝑥2+ 𝑗 ×

𝜕2𝑓

𝜕𝑥𝜕𝑦+ 𝑘 ×

𝜕2𝑓

𝜕𝑥𝜕𝑧

= 𝑖 × 𝑖 ∙𝜕2𝑓

𝜕𝑥2 + 𝑖 × 𝑗 ∙𝜕2𝑓

𝜕𝑥𝜕𝑦+ 𝑖 × 𝑘 ∙

𝜕2𝑓

𝜕𝑥𝜕𝑧

= 𝑘 ∙𝜕2𝑓

𝜕𝑥𝜕𝑦− 𝑗 ∙

𝜕2𝑓

𝜕𝑥𝜕𝑧 = 0

Proof 4: 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑓 = ∇ × ∇ × 𝑓 = 𝑖 𝜕

𝜕𝑥 × 𝑖 ×

𝜕𝑓

𝜕𝑥+ 𝑗 ×

𝜕𝑓

𝜕𝑦+ 𝑘 ×

𝜕𝑓

𝜕𝑧

= 𝑖 × 𝑖 ×𝜕2𝑓

𝜕𝑥2 + 𝑗 ×𝜕2𝑓

𝜕𝑥𝜕𝑦+ 𝑘 ×

𝜕2𝑓

𝜕𝑥𝜕𝑧

= 𝑖 × 𝑖 ×𝜕2𝑓

𝜕𝑥2 + 𝑖 × 𝑗 ×𝜕2𝑓

𝜕𝑥𝜕𝑦 + 𝑖 × 𝑘 ×

𝜕2𝑓

𝜕𝑥𝜕𝑧

= 𝑖 ∙𝜕2𝑓

𝜕𝑥2 𝑖 − 𝑖 ∙ 𝑖 𝜕2𝑓

𝜕𝑥2 + 𝑖 ∙𝜕2𝑓

𝜕𝑥𝜕𝑦 𝑗 − 𝑖 ∙ 𝑗

𝜕2𝑓

𝜕𝑥𝜕𝑦 + 𝑖 ∙

𝜕2𝑓

𝜕𝑥𝜕𝑧 𝑘 − 𝑖 ∙ 𝑘

𝜕2𝑓

𝜕𝑥𝜕𝑧

= 𝑖 ∙𝜕2𝑓

𝜕𝑥2 𝑖 + 𝑖 ∙𝜕2𝑓

𝜕𝑥𝜕𝑦 𝑗 + 𝑖 ∙

𝜕2𝑓

𝜕𝑥𝜕𝑧 𝑘 −

𝜕2𝑓

𝜕𝑥2

= 𝑖 𝜕

𝜕𝑥 𝑖 ∙

𝜕𝑓

𝜕𝑥+ 𝑗 ∙

𝜕𝑓

𝜕𝑦+ 𝑘 ∙

𝜕𝑓

𝜕𝑧 −

𝜕2𝑓

𝜕𝑥2 = ∇ ∇ ∙ 𝑓 − ∇2𝑓

Proof 5: To get this formula we are to re-arrange the terms in last proof.

Example 35: Show that 𝛁𝟐 𝒓𝒏 = 𝒏(𝒏 + 𝟏)𝒓𝒏−𝟐.

Solution: We know that 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2

On differentiation w. r. t. x, 𝜕𝑟

𝜕𝑥=

𝑥

𝑟

Similarly 𝜕𝑟

𝜕𝑦=

𝑦

𝑟 and

𝜕𝑟

𝜕𝑧=

𝑧

𝑟

Now ∇2 𝑟𝑛 = 𝜕2

𝜕𝑥2+

𝜕2

𝜕𝑦2+

𝜕2

𝜕𝑧2 𝑟𝑛 =

𝜕

𝜕𝑥

𝜕𝑟𝑛

𝜕𝑥 +

𝜕

𝜕𝑦

𝜕𝑟𝑛

𝜕𝑦 +

𝜕

𝜕𝑧

𝜕𝑟𝑛

𝜕𝑧 … (1)

And 𝜕

𝜕𝑥

𝜕𝑟𝑛

𝜕𝑥 =

𝜕

𝜕𝑥 𝑛 𝑟𝑛−1 𝜕𝑟

𝜕𝑥 =

𝜕

𝜕𝑥 𝑛 𝑟𝑛−1 𝑥

𝑟 = 𝑛

𝜕

𝜕𝑥 𝑟𝑛−2𝑥

31

= 𝑛 𝑟𝑛−2 + 𝑥 (𝑛 − 2)𝑟𝑛−3 𝜕𝑟

𝜕𝑥 = 𝑛 𝑟𝑛−2 + 𝑥2(𝑛 − 2)𝑟𝑛−4

Similarly 𝜕

𝜕𝑦 𝜕𝑟𝑛

𝜕𝑦 = 𝑛 𝑟𝑛−2 + 𝑦2(𝑛 − 2)𝑟𝑛−4

𝜕

𝜕𝑧

𝜕𝑟𝑛

𝜕𝑧 = 𝑛 𝑟𝑛−2 + 𝑧2(𝑛 − 2)𝑟𝑛−4

Using all these values in (1),

∇2 𝑟𝑛 = 𝑛 𝑟𝑛−2 + 𝑥2(𝑛 − 2)𝑟𝑛−4 ) + 𝑛 𝑟𝑛−2 + (𝑛 − 2)𝑟𝑛−4 + 𝑛 𝑟𝑛−2 + 𝑧2(𝑛 − 2)𝑟𝑛−4

= 3𝑛 𝑟𝑛−2 + 𝑛 𝑛 − 2 𝑟𝑛−4 𝑥2 + 𝑦2 + 𝑧2

= 3𝑛 𝑟𝑛−2 + 𝑛 𝑛 − 2 𝑟𝑛−2 = 𝑛 𝑛 + 1 𝑟𝑛−2

Example 36: Show that *(i) 𝛁𝟐𝒇 𝒓 = 𝒇 ′′ 𝒓 +𝟐

𝒓 𝒇 ′ (𝒓) (ii) 𝛁 ∙ 𝝓 𝛁𝝍 − 𝝍 𝛁𝝓 = 𝝓 𝛁𝟐𝝍 − 𝝍 𝛁𝟐𝝓

*[KUK 2008]

Solution: (i) ∇2𝑓 𝑟 = 𝜕2

𝜕𝑥2 +𝜕2

𝜕𝑦2 +𝜕2

𝜕𝑧2 𝑓 𝑟 =𝜕

𝜕𝑥

𝜕

𝜕𝑥𝑓 𝑟 +

𝜕

𝜕𝑦

𝜕

𝜕𝑦𝑓 𝑟 +

𝜕

𝜕𝑧

𝜕

𝜕𝑧𝑓 𝑟 …(1)

And 𝜕

𝜕𝑥

𝜕

𝜕𝑥𝑓 𝑟 =

𝜕

𝜕𝑥 𝑓 ′(𝑟)

𝜕𝑟

𝜕𝑥 =

𝜕

𝜕𝑥 𝑓 ′(𝑟)

𝑥

𝑟 = 𝑓 ′ (𝑟)

𝜕

𝜕𝑥 𝑥

𝑟 +

𝑥

𝑟 𝑓 ′′(𝑟)

𝜕𝑟

𝜕𝑥

= 𝑓 ′ (𝑟) 1

𝑟−

𝑥

𝑟2

𝜕𝑟

𝜕𝑥 +

𝑥

𝑟 𝑓 ′′(𝑟)

𝜕𝑟

𝜕𝑥

= 𝑓 ′ 𝑟

𝑟− 𝑥2 𝑓 ′ 𝑟

𝑟3+ 𝑥2 𝑓 ′′ (𝑟)

𝑟2

Similarly 𝜕

𝜕𝑦

𝜕

𝜕𝑦𝑓 𝑟 =

𝑓 ′ 𝑟

𝑟− 𝑦2 𝑓 ′ 𝑟

𝑟3+ 𝑦2 𝑓 ′′ (𝑟)

𝑟2

𝜕

𝜕𝑧

𝜕

𝜕𝑧𝑓 𝑟 =

𝑓 ′ 𝑟

𝑟− 𝑧2 𝑓 ′ 𝑟

𝑟3+ 𝑧2 𝑓 ′′ (𝑟)

𝑟2

Using all these values in (1)

∇2𝑓 𝑟 = 𝑓 ′ 𝑟

𝑟− 𝑥2 𝑓 ′ 𝑟

𝑟3+ 𝑥2 𝑓 ′′

𝑟

𝑟2 +

𝑓 ′ 𝑟

𝑟− 𝑦2 𝑓 ′ 𝑟

𝑟3+ 𝑦2 𝑓 ′′

𝑟

𝑟2

+ 𝑓 ′ 𝑟

𝑟− 𝑧2 𝑓 ′ 𝑟

𝑟3+ 𝑧2 𝑓 ′′ (𝑟)

𝑟2

=3 𝑓 ′ 𝑟

𝑟− 𝑥2 + 𝑦2 + 𝑧2

𝑓 ′ 𝑟

𝑟3+

𝑓 ′′ 𝑟

𝑟2 𝑥2 + 𝑦2 + 𝑧2

=3 𝑓 ′ (𝑟)

𝑟−

𝑓 ′ (𝑟)

𝑟+ 𝑓 ′′ (𝑟)

=2 𝑓 ′ (𝑟)

𝑟+ 𝑓 ′′ (𝑟)

Hence ∇2𝑓 𝑟 = 𝑓 ′′ 𝑟 +2

𝑟 𝑓 ′ (𝑟) .

(ii) Consider ∇ ∙ 𝜙 ∇𝜓 − 𝜓 ∇𝜙 = ∇ ∙ 𝜙 ∇𝜓 − ∇ ∙ 𝜓 ∇𝜙

= ∇𝜙 ∙ ∇𝜓 + 𝜙 ∇ ∙ ∇𝜓 − ∇𝜓 ∙ ∇𝜙 + 𝜓 ∇ ∙ ∇𝜙

= ∇𝜙 ∙ ∇𝜓 + 𝜙 ∇2𝜓 − ∇𝜙 ∙ ∇𝜓 − 𝜓 ∇2𝜙

= 𝜙 ∇2𝜓 − 𝜓 ∇2𝜙

32

Hence ∇ ∙ 𝜙 ∇𝜓 − 𝜓 ∇𝜙 = 𝜙 ∇2𝜓 − 𝜓 ∇2𝜙.

Example 37: Find the value of n for which the vector 𝒓𝒏 𝒓 is solenoidal, where

𝒓 = 𝒙𝒊 + 𝒚𝒋 + 𝒛𝒌 .

Solution: Consider 𝑑𝑖𝑣 𝑟𝑛 𝑟 = ∇ ∙ 𝑟𝑛 𝑟 = ∇𝑟𝑛 ∙ 𝑟 + 𝑟𝑛 ∇ ∙ 𝑟 … (1)

But ∇𝑟𝑛 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝑟𝑛 = 𝑛 𝑟𝑛−1 𝑖

𝜕𝑟

𝜕𝑥+ 𝑗

𝜕𝑟

𝜕𝑦+ 𝑘

𝜕𝑟

𝜕𝑧

= 𝑛 𝑟𝑛−1 𝑥

𝑟𝑖 +

𝑦

𝑟𝑗 +

𝑧

𝑟𝑘 = 𝑛 𝑟𝑛−2 𝑟 … (2)

And ∇ ∙ 𝑟 = 3 … (3)

Therefore, 𝑑𝑖𝑣 𝑟𝑛 𝑟 = 𝑛 𝑟𝑛−2 𝑟 ∙ 𝑟 + 3 𝑟𝑛 = 𝑛 𝑟𝑛−2 𝑟 ∙ 𝑟 + 3𝑟𝑛

= 𝑛 𝑟𝑛−2 𝑟2 + 3𝑟𝑛 = (𝑛 + 3)𝑟𝑛 … (4)

As given the vector 𝑟𝑛 𝑟 is solenoidal, so 𝑑𝑖𝑣 𝑟𝑛 𝑟 = 0

So using (4), 𝑛 + 3 𝑟𝑛 = 0 implies that 𝑛 = −3 (since 𝑟 ≠ 0)

Example 38: If 𝒂 and 𝒃 are irrotational, prove that 𝒂 × 𝒃 is solenoidal.

Solution: Given 𝑎 and 𝑏 are irrotational, so ∇ × 𝑎 = 0 = ∇ × 𝑏 … (1)

Consider 𝐷𝑖𝑣 𝑎 × 𝑏 = 𝑏 ∙ ∇ × 𝑎 − 𝑎 ∙ ∇ × 𝑏 = 𝑏 ∙ 0 − 𝑎 ∙ 0 = 0 [using (1)]

Thus 𝑎 × 𝑏 is solenoidal.

Example 39: Show that the vector field 𝒇 = 𝒛𝟐 + 𝟐𝒙 + 𝟑𝒚 𝒊 + 𝟑𝒙 + 𝟐𝒚 + 𝒛 𝒋 + (𝒚 + 𝟐𝒛𝒙)𝒌 is

irrotational but not solenoidal. Also obtain a scalar function ∅ such that 𝛁∅ = 𝒇 .

Solution: Consider 𝐶𝑢𝑟𝑙 𝑓 = ∇ × 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑧2 + 2𝑥 + 3𝑦 3𝑥 + 2𝑦 + 𝑧 𝑦 + 2𝑧𝑥

= 𝑖 1 − 1 − 𝑗 2𝑧 − 2𝑧 + 𝑘 3 − 3 = 0

So 𝑓 is irrotational vector field.

Also consider 𝑑𝑖𝑣 𝑓 = ∇ ∙ 𝑓 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 ∙ 𝑓 = 2 + 2 + 2𝑥 = 2(𝑥 + 2) ≠ 0

So 𝑓 is not solenoidal vector field.

Now 𝑑∅ =𝜕∅

𝜕𝑥 𝑑𝑥 +

𝜕∅

𝜕𝑦 𝑑𝑦 +

𝜕∅

𝜕𝑧 𝑑𝑧 = 𝑖

𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 ∙ 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘

= 𝑔𝑟𝑎𝑑 ∅ ∙ 𝑑𝑟 = 𝑓 ∙ 𝑑𝑟 (as 𝑓 = 𝑔𝑟𝑎𝑑 ∅ )

= 𝑧2 + 2𝑥 + 3𝑦 𝑖 + 3𝑥 + 2𝑦 + 𝑧 𝑗 + (𝑦 + 2𝑧𝑥)𝑘 ∙ 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘

= 𝑧2 + 2𝑥 + 3𝑦 𝑑𝑥 + 3𝑥 + 2𝑦 + 𝑧 𝑑𝑦 + (𝑦 + 2𝑧𝑥)𝑑𝑧

= 𝑧2𝑑𝑥 + 2𝑧𝑥 𝑑𝑧 + 3𝑦 𝑑𝑥 + 3𝑥 𝑑𝑦 + 𝑧 𝑑𝑦 + 𝑦 𝑑𝑧 + 2𝑥 𝑑𝑥 + 2𝑦 𝑑𝑦

= 𝑑 𝑥𝑧2 + 3𝑑 𝑥𝑦 + 𝑑 𝑦𝑧 + 𝑑 𝑥2 + 𝑑(𝑦2)

Integrating both sides, we get

33

∅ = 𝑥𝑧2 + 3𝑥𝑦 + 𝑦𝑧 + 𝑥2 + 𝑦2 + 𝑐

Example 40: If 𝑽 𝟏 𝐚𝐧𝐝 𝑽 𝟐 be the vectors joining the fixed points 𝒙𝟏, 𝒚𝟏, 𝒛𝟏 and 𝒙𝟐, 𝒚𝟐, 𝒛𝟐

respectively to a variable point (𝒙, 𝒚, 𝒛), prove that *[KUK 2010]

(i) 𝒅𝒊𝒗 𝑽 𝟏 × 𝑽 𝟐 = 𝟎 *(ii) 𝒄𝒖𝒓𝒍 𝑽 𝟏 × 𝑽 𝟐 = 𝟐(𝑽 𝟏 − 𝑽 𝟐) (iii) 𝒈𝒓𝒂𝒅 𝑽 𝟏 ∙ 𝑽 𝟐 = 𝑽 𝟏 + 𝑽 𝟐.

Solution: Here, 𝑉 1 = 𝑥 − 𝑥1 𝑖 + 𝑦 − 𝑦1 𝑗 + 𝑧 − 𝑧1 𝑘 and 𝑉 2 = 𝑥 − 𝑥2 𝑖 + 𝑦 − 𝑦2 𝑗 + 𝑧 − 𝑧2 𝑘

(i) 𝑉 1 × 𝑉 2 = 𝑖 𝑗 𝑘

𝑥 − 𝑥1 𝑦 − 𝑦1 𝑧 − 𝑧1

𝑥 − 𝑥2 𝑦 − 𝑦2 𝑧 − 𝑧2

= 𝑦 − 𝑦1 𝑧 − 𝑧2 − 𝑦 − 𝑦2 𝑧 − 𝑧1 𝑖 + 𝑥 − 𝑥2 𝑧 − 𝑧1 − 𝑥 − 𝑥1 𝑧 − 𝑧2 𝑗

+ 𝑥 − 𝑥1 𝑦 − 𝑦2 − 𝑥 − 𝑥2 𝑦 − 𝑦1 𝑘

So 𝑑𝑖𝑣 𝑉 1 × 𝑉 2 =𝜕

𝜕𝑥 𝑦 − 𝑦1 𝑧 − 𝑧2 − 𝑦 − 𝑦2 𝑧 − 𝑧1

+𝜕

𝜕𝑦 𝑥 − 𝑥2 𝑧 − 𝑧1 − 𝑥 − 𝑥1 𝑧 − 𝑧2

+𝜕

𝜕𝑧 𝑥 − 𝑥1 𝑦 − 𝑦2 − 𝑥 − 𝑥2 𝑦 − 𝑦1 = 0

(ii) 𝑐𝑢𝑟𝑙 𝑉 1 × 𝑉 2 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑦 − 𝑦1 𝑧 − 𝑧2 − 𝑦 − 𝑦2 𝑧 − 𝑧1 𝑥 − 𝑥2 𝑧 − 𝑧1 − 𝑥 − 𝑥1 𝑧 − 𝑧2 𝑥 − 𝑥1 𝑦 − 𝑦2 − 𝑥 − 𝑥2 𝑦 − 𝑦1

= 𝑥 − 𝑥1 − 𝑥 − 𝑥2 − 𝑥 − 𝑥2 + 𝑥 − 𝑥1 𝑖

+[ 𝑦 − 𝑦1 − 𝑦 − 𝑦2 − 𝑦 − 𝑦2 + 𝑦 − 𝑦1 ]𝑗

+[ 𝑧 − 𝑧1 − 𝑧 − 𝑧2 − 𝑧 − 𝑧2 + 𝑧 − 𝑧1 ]𝑘

= 2 𝑥 − 𝑥1 𝑖 + (𝑦 − 𝑦1)𝑗 + 𝑧 − 𝑧1 𝑘 − 2[ 𝑥 − 𝑥2 𝑖 + 𝑦 − 𝑦2 𝑗 + 𝑧 − 𝑧2 𝑘 ]

= 2(𝑉 1 − 𝑉 2)

(iii) 𝑉 1 ∙ 𝑉 2 = 𝑥 − 𝑥1 𝑥 − 𝑥2 + 𝑦 − 𝑦1 𝑦 − 𝑦2 + 𝑧 − 𝑧1 𝑧 − 𝑧2

So 𝑔𝑟𝑎𝑑 𝑉 1 ∙ 𝑉 2 = 𝑖 𝜕

𝜕𝑥 𝑥 − 𝑥1 𝑥 − 𝑥2 + 𝑗

𝜕

𝜕𝑦 𝑦 − 𝑦1 𝑦 − 𝑦2 + 𝑘

𝜕

𝜕𝑧 𝑧 − 𝑧1 𝑧 − 𝑧2

= 𝑖 𝑥 − 𝑥1 + 𝑥 − 𝑥2 + 𝑗 𝑦 − 𝑦1 + 𝑦 − 𝑦2 + 𝑘 𝑧 − 𝑧1 + 𝑧 − 𝑧2

= 𝑥 − 𝑥1 𝑖 + 𝑦 − 𝑦1 𝑗 + 𝑧 − 𝑧1 𝑘 + 𝑥 − 𝑥2 𝑖 + 𝑦 − 𝑦2 𝑗 + 𝑧 − 𝑧2 𝑘

= 𝑉 1 + 𝑉 2

34

Example 41: If 𝒂 is a constant vector and 𝒓 = 𝒙 𝒊 + 𝒚 𝒋 + 𝒛 𝒌 , prove that [KUK 2009]

(i) 𝒈𝒓𝒂𝒅 𝒂 ∙ 𝒓 = 𝒂 (ii) 𝒅𝒊𝒗 𝒂 × 𝒓 = 𝟎 (iii) ) 𝒄𝒖𝒓𝒍 𝒂 × 𝒓 = 𝟐𝒂 (iv) 𝒄𝒖𝒓𝒍 𝒂 ∙ 𝒓 𝒓 = 𝒂 × 𝒓

Solution: Let 𝑎 = 𝑎1𝑖 + 𝑎2𝑗 + 𝑎3 𝑘 is the constant vector.

(i) 𝑎 ∙ 𝑟 = 𝑎1𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 ∙ 𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 = 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧

So 𝑔𝑟𝑎𝑑 𝑎 ∙ 𝑟 = 𝑖 𝜕

𝜕𝑥 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 + 𝑗

𝜕

𝜕𝑦 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 + 𝑘

𝜕

𝜕𝑧 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧

= 𝑎1𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 = 𝑎

(ii) 𝑎 × 𝑟 = 𝑖 𝑗 𝑘

𝑎1 𝑎2 𝑎3

𝑥 𝑦 𝑧 = 𝑖 𝑎2𝑧 − 𝑎3𝑦 + 𝑗 𝑎3𝑥 − 𝑎1𝑧 + 𝑘 𝑎1𝑦 − 𝑎2𝑥

𝑑𝑖𝑣 𝑎 × 𝑟 =𝜕

𝜕𝑥 𝑎2𝑧 − 𝑎3𝑦 +

𝜕

𝜕𝑦 𝑎3𝑥 − 𝑎1𝑧 +

𝜕

𝜕𝑧 𝑎1𝑦 − 𝑎2𝑥 = 0

(iii) 𝑐𝑢𝑟𝑙 𝑎 × 𝑟 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑎2𝑧 − 𝑎3𝑦 𝑎3𝑥 − 𝑎1𝑧 𝑎1𝑦 − 𝑎2𝑥

= 𝑖 𝑎1 + 𝑎1 + 𝑗 𝑎2 + 𝑎2 + 𝑘 𝑎3 + 𝑎3 = 2 𝑎1𝑖 + 𝑎2𝑗 + 𝑎3 𝑘 = 2𝑎

(iv) 𝑎 ∙ 𝑟 𝑟 = 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑥𝑖 + 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑦𝑗 + 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑧 𝑘

𝐶𝑢𝑟𝑙 𝑎 ∙ 𝑟 𝑟 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑥 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑦 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑧 𝑧

= 𝑖 𝑎2𝑧 − 𝑎3𝑦 + 𝑗 𝑎3𝑥 − 𝑎1𝑧 + 𝑘 𝑎1𝑦 − 𝑎2𝑥 = 𝑎 × 𝑟 [using part (ii)]

Example 42: Find 𝒇 × 𝛁 × 𝒈 at the point (1, -1, 2),

if 𝒇 = 𝒙𝒛𝟐 𝒊 + 𝟐𝒚 𝒋 − 𝟑𝒙𝒛 𝒌 , 𝒈 = 𝟑𝒙𝒛 𝒊 + 𝟐𝒚𝒛 𝒋 − 𝒛𝟐 𝒌 .

Solution: ∇ × 𝑔 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

3𝑥𝑧 2𝑦𝑧 −𝑧2

= 𝑖 0 − 2𝑦 + 𝑗 3𝑥 − 0 + 𝑘 0 − 0 = −2𝑦 𝑖 + 3𝑥 𝑗 + 0 𝑘

Now 𝑓 × ∇ × 𝑔 = 𝑖 𝑗 𝑘

𝑥𝑧2 2𝑦 −3𝑥𝑧−2𝑦 3𝑥 0

= 9𝑥2𝑧 𝑖 + 6𝑥𝑦𝑧 𝑗 + 3𝑥2𝑧2 + 4𝑦2 𝑘

At (1, -1, 2), 𝑓 × ∇ × 𝑔 = 9 1 2(2) 𝑖 + 6 1 −1 (2) 𝑗 + 3(1)2(2)2 + 4(−1)2 𝑘

= 18 𝑖 − 12 𝑗 + 16 𝑘

Example 43: If 𝒇 = 𝒚𝒛𝟐 𝒊 − 𝟑𝒙𝒛𝟐 𝒋 + 𝟐𝒙𝒚𝒛 𝒌 , 𝒈 = 𝟑𝒙 𝒊 + 𝟒𝒛 𝒋 − 𝒙𝒚 𝒌 and ∅ = 𝒙𝒚𝒛; find

(i) 𝒇 × 𝛁∅ (ii) 𝒇 × 𝛁 ∅ (iii) 𝛁 × 𝒇 × 𝐠 (iv) 𝐠 ∙ 𝛁 × 𝒇

Solution: ∇∅ = 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 = 𝑦𝑧 𝑖 + 𝑧𝑥 𝑗 + 𝑥𝑦 𝑘

35

(i) 𝑓 × ∇∅ = 𝑖 𝑗 𝑘

𝑦𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧𝑦𝑧 𝑥𝑧 𝑥𝑦

= −5𝑥2𝑦𝑧2 𝑖 + 𝑥𝑦2𝑧2 𝑗 + 4𝑥𝑦𝑧3 𝑘

(ii) 𝑓 × ∇=

𝑖 𝑗 𝑘

𝑦𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

= 𝑖 −3𝑥𝑧2 𝜕

𝜕𝑧− 2𝑥𝑦𝑧

𝜕

𝜕𝑦 + 𝑗 2𝑥𝑦𝑧

𝜕

𝜕𝑥− 𝑦𝑧2 𝜕

𝜕𝑧 + 𝑘 𝑦𝑧2 𝜕

𝜕𝑦+ 3𝑥𝑧2 𝜕

𝜕𝑥

Now 𝑓 × ∇ ∅ = 𝑖 −3𝑥𝑧2 𝜕∅

𝜕𝑧− 2𝑥𝑦𝑧

𝜕∅

𝜕𝑦 + 𝑗 2𝑥𝑦𝑧

𝜕∅

𝜕𝑥− 𝑦𝑧2 𝜕∅

𝜕𝑧 + 𝑘 𝑦𝑧2 𝜕∅

𝜕𝑦+ 3𝑥𝑧2 𝜕∅

𝜕𝑥

= 𝑖 −3𝑥𝑧2 𝑦𝑥 − 2𝑥𝑦𝑧 𝑥𝑧 + 𝑗 2𝑥𝑦𝑧 𝑦𝑧 − 𝑦𝑧2 𝑥𝑦 + 𝑘 𝑦𝑧2 𝑥𝑧 + 3𝑥𝑧2 𝑦𝑧

= −5𝑥2𝑦𝑧2 𝑖 + 𝑥𝑦2𝑧2𝑗 + 4𝑥𝑦𝑧3 𝑘

(iii) ∇ × 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑦𝑧2 −3𝑥𝑧2 2𝑥𝑦𝑧

= 2𝑥𝑧 + 6𝑥𝑧 𝑖 + 2𝑦𝑧 − 2𝑦𝑧 𝑗 + (−3𝑧2 − 𝑧2) 𝑘

= 8𝑥𝑧 𝑖 + 0 𝑗 − 4𝑧2 𝑘

Now ∇ × 𝑓 × g = 𝑖 𝑗 𝑘

8𝑥𝑧 0 −4𝑧2

3𝑥 4𝑧 −𝑥𝑦

= 0 + 16𝑧3 𝑖 + −12𝑥𝑧2 + 8𝑥2𝑦𝑧 𝑗 + (32𝑥𝑧2 − 0) 𝑘

= 16𝑧3 𝑖 + −12𝑥𝑧2 + 8𝑥2𝑦𝑧 𝑗 + 32𝑥𝑧2 𝑘

g ∙ ∇ × 𝑓 = 3𝑥 𝑖 + 4𝑧 𝑗 − 𝑥𝑦 𝑘 ∙ 8𝑥𝑧 𝑖 + 0 𝑗 − 4𝑧2 𝑘 = 24𝑥2𝑧 + 4𝑥𝑦𝑧2

Example 44: Find the directional derivative of 𝛁 ∙ 𝛁∅ at the point (1, -2, 1) in the direction of

the normal to the surface 𝒙𝒚𝟐𝒛 = 𝟑𝒙 + 𝒛𝟐 where ∅ = 𝟐𝒙𝟑𝒚𝟐𝒛𝟒. [Raipur 2005]

Solution: Given ∅ = 2𝑥3𝑦2𝑧4

So ∇∅ = 𝑖 𝜕∅

𝜕𝑥+ 𝑗

𝜕∅

𝜕𝑦+ 𝑘

𝜕∅

𝜕𝑧 = 𝑖 6𝑥2𝑦2𝑧4 + 𝑗 (4𝑥3𝑦𝑧4) + 𝑘 (8𝑥3𝑦2𝑧3)

And 𝑓 = ∇ ∙ ∇∅ = 𝜕

𝜕𝑥 6𝑥2𝑦2𝑧4 +

𝜕

𝜕𝑦(4𝑥3𝑦𝑧4) +

𝜕

𝜕𝑧(8𝑥3𝑦2𝑧3)

= 12𝑥𝑦2𝑧4 + 4𝑥3𝑧4 + 24 𝑥3𝑦2𝑧2

Consider 𝑔𝑟𝑎𝑑 𝑓 = ∇𝑓 = 𝑖 𝜕𝑓

𝜕𝑥+ 𝑗

𝜕𝑓

𝜕𝑦+ 𝑘

𝜕𝑓

𝜕𝑧

= 𝑖 12𝑦2𝑧4 + 12𝑥2𝑧4 + 72𝑥3𝑦2𝑧2 + 𝑗 24𝑥𝑦𝑧4 + 48𝑥3𝑦𝑧2

+ 𝑘 (48𝑥𝑦2𝑧3 + 16𝑥3𝑧3 + 48𝑥3𝑦2𝑧)

At point (1, -2, 1)

𝑔𝑟𝑎𝑑 𝑓 = 𝑖 (48 + 12 + 288) + 𝑗 −48 − 96 + 𝑘 192 + 16 + 192

= 348 𝑖 − 144 𝑗 + 400 𝑘

Now consider the surface 𝑔 = 𝑥𝑦2𝑧 − 3𝑥 − 𝑧2 = 0

36

So, ∇𝑔 = 𝑖 𝜕𝑔

𝜕𝑥+ 𝑗

𝜕𝑔

𝜕𝑦+ 𝑘

𝜕𝑔

𝜕𝑧 = 𝑖 𝑦2𝑧 − 3 + 𝑗 2𝑥𝑦𝑧 + 𝑘 𝑥𝑦2 − 2𝑧

The vector normal to the surface (1) at point (1, -2, 1) is given by

𝑛 = 𝑖 − 4𝑗 + 2𝑘

And 𝑛 = 1 + 16 + 4 = 21

So the direction derivative of 𝑓 = ∇ ∙ ∇∅ at the point (1, -2, 1) in the direction of the normal to the

surface (1) is 𝑔𝑟𝑎𝑑 𝑓 .𝑛

𝑛

= 348 𝑖 − 144 𝑗 + 400 𝑘 .1

21 𝑖 − 4𝑗 + 2𝑘

=1

21 348 + 576 + 800

= 𝟏𝟕𝟐𝟒/ 𝟐𝟏

Example 45: If r is the distance of a point (x, y, z) from the origin, prove that

𝒄𝒖𝒓𝒍 𝒌 × 𝒈𝒓𝒂𝒅 𝟏

𝒓 + 𝒈𝒓𝒂𝒅 𝒌 ∙ 𝒈𝒓𝒂𝒅

𝟏

𝒓 = 𝟎 where 𝒌 is the unit vector in the direction of OZ.

Solution: Here 𝑟 = 𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 so that 𝑟 = 𝑥2 + 𝑦2 + 𝑧2

Now 𝑔𝑟𝑎𝑑 1

𝑟= 𝑖

𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧

1

𝑟= 𝑖

𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 𝑥2 + 𝑦2 + 𝑧2 −

1

2

= −1

2 𝑥2 + 𝑦2 + 𝑧2 −

3

2(2𝑥 𝑖 + 2𝑦 𝑗 + 2𝑧 𝑘 ) = −1

𝑟3 𝑟

𝑔𝑟𝑎𝑑 𝑘 ∙ 𝑔𝑟𝑎𝑑 1

𝑟 = 𝑔𝑟𝑎𝑑 −

1

𝑟3 𝑧 = −𝑔𝑟𝑎𝑑 𝑧

𝑥2+𝑦2+𝑧2 3/2

=3𝑧𝑥

𝑥2+𝑦2+𝑧2 5/2𝑖 +

3𝑧𝑦

𝑥2+𝑦2+𝑧2 5/2𝑗 +

𝑥2+𝑦2−2𝑧2

𝑥2+𝑦2+𝑧2 5/2𝑘 … (1)

And 𝑐𝑢𝑟𝑙 𝑘 × 𝑔𝑟𝑎𝑑 1

𝑟 = 𝑐𝑢𝑟𝑙 𝑘 × −

1

𝑟3 𝑟 = 𝑐𝑢𝑟𝑙 1

𝑟3 𝑦 𝑖 − 𝑥 𝑗

= −3𝑧𝑥

𝑥2+𝑦2+𝑧2 5/2𝑖 −

3𝑧𝑦

𝑥2+𝑦2+𝑧2 5/2𝑗 −

𝑥2+𝑦2−2𝑧2

𝑥2+𝑦2+𝑧2 5/2𝑘 … (2)

Adding (1) and (2), 𝑐𝑢𝑟𝑙 𝑘 × 𝑔𝑟𝑎𝑑 1

𝑟 + 𝑔𝑟𝑎𝑑 𝑘 ∙ 𝑔𝑟𝑎𝑑

1

𝑟 = 0 .

Example 46: Prove that 𝒂 ∙ 𝛁 𝒃 ∙ 𝛁𝟏

𝒓 =

𝟑(𝒂 ∙𝒓 )(𝒃 ∙𝒓 )

𝒓𝟓−

𝒂 ∙𝒃

𝒓𝟑, where 𝒂 and 𝒃 are constant vectors.

Solution: From example 45, ∇1

𝑟= −

1

𝑟3 𝑟 = − 𝑥2 + 𝑦2 + 𝑧2 −3

2(𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 )

Let the constant vectors are 𝑎 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 and 𝑏 = 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3 𝑘

So ∇ 𝑏 ∙ ∇1

𝑟 = ∇ − 𝑥2 + 𝑦2 + 𝑧2 −

3

2 𝑏1𝑥 + 𝑏2𝑦 + 𝑏3𝑧

= − 𝑥2 + 𝑦2 + 𝑧2 −3

2 ∇ 𝑏1𝑥 + 𝑏2𝑦 + 𝑏3𝑧 − 𝑏1𝑥 + 𝑏2𝑦 + 𝑏3𝑧 ∇ 𝑥2 + 𝑦2 + 𝑧2 −3

2

= − 𝑥2 + 𝑦2 + 𝑧2 −3

2 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3 𝑘 +3 𝑏1𝑥 + 𝑏2𝑦 + 𝑏3𝑧 𝑥2 + 𝑦2 + 𝑧2 −5

2(𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 )

= −𝑏

𝑟3+

3 𝑏 ∙𝑟 𝑟

𝑟5

37

Now 𝑎 ∙ ∇ 𝑏 ∙ ∇1

𝑟 = 𝑎 ∙ −

𝑏

𝑟3 +3 𝑏 ∙𝑟 𝑟

𝑟5 =3(𝑎 ∙𝑟 )(𝑏 ∙𝑟 )

𝑟5 −𝑎 ∙𝑏

𝑟3

Hence Proved.

ASSIGNMENT 4

1. If 𝑓 = 𝑥 + 𝑦 + 1 𝑖 + 𝑗 − (𝑥 + 𝑦)𝑘 , show that 𝑓 ∙ 𝑐𝑢𝑟𝑙 𝑓 = 0.

2. Evaluate (a) 𝑑𝑖𝑣 3𝑥2𝑖 + 5𝑥𝑦2𝑗 + 𝑥𝑦𝑧3𝑘 at the point (1, 2, 3).

(b) 𝑐𝑢𝑟𝑙 [𝑒𝑥𝑦𝑧 𝑖 + 𝑗 + 𝑘 ].

3. Find the value of ‘a’ if the vector 𝑎𝑥2𝑦 + 𝑦𝑧 𝑖 + 𝑥𝑦2 − 𝑥𝑧2 𝑗 + 2𝑥𝑦𝑧 − 2𝑥2𝑦2 𝑘 has zero

divergence. Find the curl of above vector which has zero divergence .

4. If 𝑣 = 𝑟 / 𝑥2 + 𝑦2 + 𝑧2, show that ∇ ∙ 𝑣 = 2/ 𝑥2 + 𝑦2 + 𝑧2 and ∇ × 𝑣 = 0 .

5. If 𝑢 = 𝑥2 + 𝑦2 + 𝑧2 and 𝑣 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 , show that 𝑑𝑖𝑣 𝑢 𝑣 = 5𝑢.

6. Show that each of following vectors are solenoidal:

(a) 𝑥 + 3𝑦 𝑖 + 𝑦 − 3𝑧 𝑗 + 𝑥 − 2𝑧 𝑘 (b) 3𝑦4𝑧2𝑖 + 4𝑥3𝑧2𝑗 + 3𝑥2𝑦2𝑘 (c) ∇u × ∇𝑣

7. If 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 and 𝑟 = 𝑟 ≠ 0, show that

(a) ∇ 1/𝑟2 = −2𝑟 /𝑟4; ∇ ∙ 𝑟 /𝑟2 = 1/𝑟2 (b) ∇ ∙ 𝑟𝑛𝑟 = (𝑛 + 3)𝑟𝑛 ; ∇ × 𝑟𝑛𝑟 = 0 .

(c) ∇ ∇ ∙𝑟

𝑟 = −

2

𝑟3 𝑟

8. Prove that (a) ∇𝑎 2 = 2 𝑎 ∙ ∇ 𝑎 + 2𝑎 × ∇ × 𝑎 , where 𝑎 is the constant vector.

(b) ∇ × 𝑟 × 𝑢 = 𝑟 ∇ ∙ u − 2𝑢 − 𝑟 ∙ ∇ 𝑢 .

9. (a) If ∅ = 𝑥2 + 𝑦2 + 𝑧2 −𝑛 , find the 𝑑𝑖𝑣 𝑔𝑟𝑎𝑑 ∅ and determine n if 𝑑𝑖𝑣 𝑔𝑟𝑎𝑑 ∅ = 0.

(b) Show that 𝑔𝑟𝑎𝑑 𝑟𝑛 = 𝑛(𝑛 + 1)𝑟𝑛−2 , where 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2.

10. In electromagnetic theory, we have ∇ ∙ 𝐷 = 𝜌, ∇ ∙ 𝐻 = 0, ∇ × 𝐷 = −1

𝑐

𝜕𝐻

𝜕𝑡, ∇ × 𝐻 =

1

𝑐 𝜌𝑉 +

𝜕𝐷

𝜕𝑡 .

Prove that ∇2𝐻 −1

𝑐2

𝜕2𝐻

𝜕𝑡 2 = −1

𝑐∇ × 𝜌𝑉 and ∇2𝐷 −

1

𝑐2

𝜕2𝐷

𝜕𝑡 2 = ∇𝜌 +1

𝑐2

𝜕

𝜕𝑡 𝜌𝑉 .

11. If 𝑢 = 𝑥2𝑦𝑧, 𝑣 = 𝑥𝑦 − 3𝑧2 , find (i) ∇ ∇𝑢 ∙ ∇𝑣 (ii) ∇ ∙ ∇𝑢 × ∇𝑣 .

12. For a solenoidal vector 𝑓 , show that 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑐𝑢𝑟𝑙 𝑓 = ∇4𝑓 .

13. Calculate (i) 𝑐𝑢𝑟𝑙 𝑔𝑟𝑎𝑑 𝑓 , given 𝑓 𝑥, 𝑦, 𝑧 = 𝑥2 + 𝑦2 − 𝑧 [BPTU 2006]

(ii) 𝑐𝑢𝑟𝑙(𝑐𝑢𝑟𝑙 𝑎 ), given 𝑎 = 𝑥2𝑦 𝑖 + 𝑦2𝑧 𝑗 + 𝑧2𝑦 𝑘

14. Show that each of the following vectors are solenoidal:

(i) −𝑥2 + 𝑦𝑧 𝑖 + 4𝑦 − 𝑧2𝑥 𝑗 + (2𝑥𝑧 − 4𝑧) 𝑘

(ii) 3𝑦4𝑧2 𝑖 + 4𝑥3𝑧2 𝑗 + 3𝑥2𝑦2 𝑘

(iii) ∇𝜙 × ∇𝜓

INTEGRAL VECTOR CALCULUS

38

17.9 INTEGRATION OF VECTORS

If two vector functions 𝑓 (𝑡) and 𝑔 (𝑡) be such that 𝑑

𝑑𝑡 𝑔 (𝑡) = 𝑓 (𝑡), then 𝑔 (𝑡) is called an integral

of 𝑓 (𝑡) with respect to a scalar variable t and we can write 𝑓 (𝑡)𝑑𝑡 = 𝑔 (𝑡).

Indefinite Integral

If 𝑐 be an arbitrary constant vector and 𝑓 𝑡 =𝑑

𝑑𝑡 𝑔 (𝑡) =

𝑑

𝑑𝑡 𝑔 𝑡 + 𝑐 , then 𝑓 (𝑡)𝑑𝑡 = 𝑔 𝑡 + 𝑐 .

This is called the indefinite integral of 𝑓 (𝑡).

Definite Integral

If 𝑑

𝑑𝑡 𝑔 (𝑡) = 𝑓 𝑡 for all values of 𝑡 in the interval 𝑎, 𝑏 , then the definite integral of 𝑓 𝑡 between

𝑎 and 𝑏 is defined and denoted by 𝑓 𝑡 𝑑𝑡𝑏

𝑎= 𝑔 (𝑡) 𝑎

𝑏 = 𝑔 𝑏 − 𝑔 𝑎 .

Example 47: If 𝒅𝟐𝒇

𝒅𝒕𝟐= 𝟔𝒕 𝒊 − 𝟏𝟐𝒕𝟐 𝒋 + 𝟒 𝐜𝐨𝐬 𝒕 𝒌 , find 𝒇 , given that

𝒅𝒇

𝒅𝒕= − 𝒊 − 𝟑 𝒌 and

𝒇 = 𝟐 𝒊 + 𝒋 at 𝒕 = 𝟎.

Solution: Given that 𝑑2𝑓

𝑑𝑡2= 6𝑡 𝑖 − 12𝑡2 𝑗 + 4 cos 𝑡 𝑘 … (1)

Integrating (1) with respect to t,

𝑑2𝑓

𝑑𝑡2 𝑑𝑡 = 6𝑡 𝑖 − 12𝑡2 𝑗 + 4 cos 𝑡 𝑘 𝑑𝑡

Implying 𝑑𝑓

𝑑𝑡 = 3𝑡2 𝑖 − 4𝑡3 𝑗 + 4 sin 𝑡 𝑘 + 𝑐 1 … (2)

Now, integrating (2) with respect to t,

𝑑𝑓

𝑑𝑡𝑑𝑡 = 3𝑡2 𝑖 − 4𝑡3 𝑗 + 4 sin 𝑡 𝑘 + 𝑐 1 𝑑𝑡

Implying 𝑓 = 𝑡3 𝑖 − 𝑡4 𝑗 − 4 cos 𝑡 𝑘 + 𝑐 1𝑡 + 𝑐 2 … (3)

Also we are given that at 𝑡 = 0, 𝑑𝑓

𝑑𝑡= − 𝑖 − 3 𝑘 … (4)

𝑓 = 2 𝑖 + 𝑗 … (5)

Using (2) and (4), 𝑐 1 = − 𝑖 − 3 𝑘

Using (3) and (5), −4 𝑘 + 𝑐 2 = 2 𝑖 + 𝑗 => 𝑐 2 = 2 𝑖 + 𝑗 + 4 𝑘

Putting the values of the constant vectors 𝑐 1 & 𝑐 2 in (3), we get

𝑓 = 𝑡3 𝑖 − 𝑡4 𝑗 − 4 cos 𝑡 𝑘 + − 𝑖 − 3 𝑘 𝑡 + 2 𝑖 + 𝑗 + 4 𝑘

= 𝑡3 − 𝑡 + 2 𝑖 + 1 − 𝑡4 𝑗 − 4 cos 𝑡 + 3 𝑡 + 4 𝑘

ASSIGNMENT 5

1. For given 𝑓 𝑡 = 5𝑡2 − 3𝑡 𝑖 + 6𝑡3 𝑗 − 7𝑡 𝑘 , evaluate 𝑓 𝑡 𝑑𝑡4

2.

2. Given 𝑟 𝑡 = 3𝑡2 𝑖 + 𝑡 𝑗 − 𝑡3 𝑘 , evaluate 𝑟 ×𝑑2𝑟

𝑑𝑡 2 𝑑𝑡1

0.

39

X O

Y

Z

3. If 𝑟 𝑡 = 2 𝑖 − 𝑗 + 2 𝑘 , 𝑤𝑕𝑒𝑛 𝑡 = 1

3 𝑖 − 2𝑗 + 4 𝑘 , 𝑤𝑕𝑒𝑛 𝑡 = 2 , show that 𝑟 ∙

𝑑𝑟

𝑑𝑡 𝑑𝑡 = 10

2

1.

4. The acceleration of a particle at any time 𝑡 ≥ 0 is given by 12 cos 2𝑡 𝑖 − 8 sin 2𝑡 𝑗 + 16𝑡 𝑘 , the

displacement and velocity are initially zero. Find the velocity and displacement at any time.

17.10 LINE INTEGRAL

Let kuzjuyiuxur

)()()()( defines a curve C joining points P1 and P2 where )(ur

is the

position vector of ),,( zyx and the value of u at P1 and P2 is u1 and u2, respectively.

Now if kAjAiAzyxA

321),,( be vector function of defined position and continuous along C,

then the integral of the tangential component of A along C from P1 to P2 written as 2

1

.P

PrdA

is

known as the line integral. Also in terms of Cartesian components, we have

2

1

2

1

)(.)(. 321

P

P

P

PkdzjdyidxkAjAiArdA

)()( 321321

2

1

C

P

PdzAdyAdxAdzAdyAdxA

If A

(vector function of the position) represents the force F

on a particle moving along C, then the

line integral represents the work done by the force F

. If C is a simple closed curve, then the integral

around C is generally written as

1P 2P

C

Fig 17.11

)(. 321 dzAdyAdxArdA

In Fluid Mechanics and Aerodynamics, the above integral is called circulation of A

about C, where

A

represents the velocity of the fluid.

Let gradA , then we have

Q

P

Q

PrdrdA

.)(.

Q

Pdzkdyjdxi

zk

yj

xi )(.

40

Q

Pdz

zdy

ydx

x

PQ

Q

P

Q

Pd ][ … (1)

We see that the integral Q

PrdA

. depends on the value of at the end P and Q and not on the

particular path. In case is single valued and the integral is taken round a closed curve, the terminal

points P and Q coincide and PB .

[Because function is uniform]

The integration along a closed curve is denoted by the sign of circle in the mid of the integral sign

i.e. for a uniform function, we have

C

rd 0.)(

… (2)

The converse of the above result is also true i.e. if there exists a vector A

and its integral round

every closed curve in the region under consideration vanishes, then there exist a point function

such that gradA

.

To prove this consider any closed curve ABCD such that the integral round it is zero, so integral along

ABC must be equal to that along ADC. Similarly, the integral along ABC must be equal to that along

any curve joining A to C, i.e. independent of the path from A to C with A be a fixed point and C a

variable point. Then due to the fact that line integral is independent of the path chosen, the value of

the line integral from A to C must be a scalar point function, say i.e. C

ArdA

.

Now if d is the increment in due to a small displacement 𝑑𝑟 of 𝑟 , then we have rdAd

.

But we already know that rdd

. , so ,.)(. rdrdA

,0.)( rdA

which is true for all rd

and hence A

.

The vector A

is called a potential vector (or gradient vector), and in cartesian component; the

condition that dzAdyAdxArdA 321.

be a perfect differential can be thrown easily into the

form

0,0,0 211332

x

A

y

A

z

A

x

A

y

A

z

A … (3)

Circulation: If A

represents the velocity of a fluid particle, then the line integral CrdA

. is called

the circulation of A

along C.

The vector point function A

, is said to be irrotational in a region, if its circulation along every

closed curve in the region is zero i.e. 0. CrdA

41

Theorem: The necessary and sufficient condition for a vector point function A

to be irrotaional in a

simply connected region is the 0Acurl

at every point of the region.

Work: If A

represents the force acting on a particle moving along an arc PQ then the work done

during the small displacement rd

is equal to rdA

. . Therefore, the total work done by A

during the

displacement from P to Q is given by the line integral Q

PrdA

. .

Example 48: Evaluate the line integral (𝒙𝟐 + 𝒙𝒚)𝒅𝒙 + (𝒙𝟐 + 𝒚𝟐)𝒅𝒚 , where 𝑪 is the square

formed by the lines 𝒚 = ±𝟏 and 𝒙 = ±𝟏.

Solution: Curve 𝐶 is square in the 𝑥𝑦 plane where 𝑧 = 0

∴ 𝑟 = 𝑥𝑖 + 𝑦𝑗 in 𝑥𝑦 plane

𝑑𝑟 = 𝑑𝑥𝑖 + 𝑑𝑦𝑗 … (1)

Now 𝐹. 𝑑𝑟 𝐶

= 𝑥2 + 𝑥𝑦 𝑑𝑥 + 𝑥2 + 𝑦2 𝑑𝑦

Path of the integration is shown in figure 17.12, it consists of lines

AB, BC, CD and DA. As curve C is a square, then

On AB, 𝑦 = −1 ⇒ 𝑑𝑦 = 0 and 𝑥 varies from −1 to 1

On BC, 𝑥 = 1 ⇒ 𝑑𝑥 = 0 and 𝑦 varies from −1 to 1

On CD, 𝑦 = 1 ⇒ 𝑑𝑦 = 0 and 𝑥 varies from 1 to −1

On DA, 𝑥 = −1 ⇒ 𝑑𝑥 = 0 and 𝑦 varies from 1 to −1

𝐹 . 𝑑𝑟 𝐶

= 𝑥2 − 𝑥 𝑑𝑥𝐴𝐵

+ 1 + 𝑦2 𝑑𝑦𝐵𝐶

+ 𝑥2 + 𝑥 𝑑𝑥𝐶𝐷

+ 1 + 𝑦2 𝑑𝑦𝐷𝐴

= 𝑥2 − 𝑥 𝑑𝑥1

−1+ 1 + 𝑦2 𝑑𝑦

1

−1+ 𝑥2 + 𝑥 𝑑𝑥

−1

1+ 1 + 𝑦2 𝑑𝑦

−1

1

= 𝑥3

3−

𝑥2

2 −1

1

+ 1 + 𝑦2 𝑑𝑦1

−1+

𝑥3

3+

𝑥2

2

1

−1

− 1 + 𝑦2 𝑑𝑦1

−1

= 1

3−

1

2+

1

3+

1

2 + −

1

3+

1

2−

1

3−

1

2 = 0

Example 49: If 𝑭 = 𝟑𝒙𝒚 𝒊 − 𝒚𝟐𝒋 , evaluate 𝑭 . 𝒅𝒓 𝑪

where 𝑪 is the arc of the parabola 𝒚 = 𝟐𝒙𝟐

from (𝟎, 𝟎) to (𝟏, 𝟐).

Solution: Because the integration is performed in the 𝑥𝑦-plane (𝑧 = 0), we take

𝑟 = 𝑥𝑖 + 𝑦𝑗 so that 𝑑𝑟 = 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗

∴ 𝐹 . 𝑑𝑟 = 3𝑥𝑦 𝑖 − 𝑦2𝑗 . 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 = 3𝑥𝑦 𝑑𝑥 − 𝑦2𝑑𝑦

On the curve C: 𝑦 = 2𝑥2 from (0, 0) to (1, 2)

42

𝐹 . 𝑑𝑟 = 3𝑥 2𝑥2 𝑑𝑥 − 2𝑥2 24𝑥 𝑑𝑥 = (6𝑥3 − 16𝑥5)𝑑𝑥

Also x varies from 0 to 1.

𝐹 . 𝑑𝑟 𝐶

= (6𝑥3 − 16𝑥5)𝑑𝑥1

0=

6𝑥4

4−

16𝑥6

6

0

1

=3

2−

8

3= −

7

6

Note: If the curve is traversed in the opposite direction, that is from (1, 2) to (0, 0), the value of the integral

would be 7

6.

Example 50: A vector field is given by 𝑭 = 𝐬𝐢𝐧𝒚 𝒊 + 𝒙(𝟏 + 𝐜𝐨𝐬 𝒚)𝒋 . Evaluate the line integral

over the circular path given by 𝒙𝟐 + 𝒚𝟐 = 𝒂𝟐, 𝒛 = 𝟎. [PTU 2003]

Solution: The parametric equations of the circular path are 𝑥 = 𝑎 cos 𝑡 , 𝑦 = 𝑎 sin 𝑡 , 𝑧 = 0 where 𝑡

varies from 0 to 2𝜋. Since the particle moves in the 𝑥𝑦-plane (𝑧 = 0), we can take 𝑟 = 𝑥 𝑖 + 𝑦 𝑗 .

𝐹 . 𝑑𝑟 𝐶

= sin 𝑦 𝑖 + 𝑥(1 + cos 𝑦)𝑗 𝐶

. 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗

= sin 𝑦 𝑑𝑥 + 𝑥 1 + cos 𝑦 𝑑𝑦 𝐶

= sin 𝑦 𝑑𝑥 + 𝑥 cos 𝑦 𝑑𝑦 + 𝑥 𝑑𝑦 𝐶

= 𝑑(𝑥 sin 𝑦)𝐶

+ 𝑥 𝑑𝑦𝐶

= 𝑑 a cos 𝑡 sin a sin 𝑡 𝑑𝑡 + a cos 𝑡 . a cos 𝑡 𝑑𝑡2𝜋

0

2𝜋

0

= a cos 𝑡 sin(a sin 𝑡) 02𝜋 + 𝑎2 cos2 𝑡 𝑑𝑡

2𝜋

0

=𝑎2

2 (1 + cos 2𝑡)𝑑𝑡

2𝜋

0=

𝑎2

2 𝑡 +

sin 2𝑡

2

0

2𝜋

=𝑎2

2 2𝜋 = 𝜋𝑎2

Example 51: Compute the line integral (𝒚𝟐𝒅𝒙 − 𝒙𝟐𝒅𝒚)𝑪

about the triangle whose vertices are

(𝟏, 𝟎), (0, 1) and (−𝟏, 𝟎). [NIT Uttrakhand 2011]

Solution: Here the closed curve C is a triangle ABC.

On AB: Equation of line AB is

𝑦 − 0 =1−0

0−1(𝑥 − 1) ⇒ 𝑦 = 1 − 𝑥

∴ 𝑑𝑦 = −𝑑𝑥 and 𝑥 varies from 1 to 0.

On BC: Equation of line BC is

𝑦 − 1 =0−1

−1−0(𝑥 − 0) ⇒ 𝑦 = 1 + 𝑥

∴ 𝑑𝑦 = 𝑑𝑥 and 𝑥 varies from 0 to -1.

On CA: 𝑦 = 0. Therefore, 𝑑𝑦 = 0 and 𝑥 varies from -1 to 1.

𝑦2𝑑𝑥 − 𝑥2𝑑𝑦 = 𝑦2𝑑𝑥 − 𝑥2𝑑𝑦 𝐴𝐵𝐶

+ 𝑦2𝑑𝑥 − 𝑥2𝑑𝑦 𝐵𝐶

+ 𝑦2𝑑𝑥 − 𝑥2𝑑𝑦 𝐶𝐴

= (1 − 𝑥)2𝑑𝑥 − 𝑥2(−𝑑𝑥) 0

1+ (1 + 𝑥)2𝑑𝑥 − 𝑥2𝑑𝑥 + 0 𝑑𝑥

1

−1

−1

0

43

= 2𝑥2 − 2𝑥 + 1 𝑑𝑥0

1+ 2𝑥 + 1 𝑑𝑥

−1

0+ 0

= 2𝑥3

3−

2𝑥2

2+ 𝑥

1

0

+ 2𝑥2

2+ 𝑥

0

−1

= −2

3+ 1 − 1 + 1 − 1 = −

2

3

Example 52: If 𝑭 = 𝟑𝒙𝟐 + 𝟔𝒚 𝒊 − 𝟏𝟒𝒚𝒛𝒋 + 𝟐𝟎𝒙𝒛𝟐𝒌 , evaluate 𝑭 . 𝒅𝒓 𝑪

where

(i) 𝑪 is the line joining the point (𝟎, 𝟎, 𝟎) to (𝟏, 𝟏, 𝟏)

(ii) 𝑪 is given by 𝒙 = 𝒕, 𝒚 = 𝒕𝟐, 𝒛 = 𝒕𝟑 from the point (𝟎, 𝟎, 𝟎) to (𝟏, 𝟏, 𝟏).

Solution: (i) Equation of line joining (0,0,0) to (1,1,1) is

𝑥−0

1−0=

𝑦−0

1−0=

𝑧−0

1−0= 𝑡 (say)

∴ Parametric equations of the line C are 𝑥 = 𝑡, 𝑦 = 𝑡, 𝑧 = 𝑡; 0 ≤ 𝑡 ≤ 1

∴ 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = 𝑡𝑖 + 𝑡𝑗 + 𝑡𝑘

⇒ 𝒅𝑟

𝒅𝒕= 𝑖 + 𝑗 + 𝑘

Now 𝐹 = 3𝑥2 + 6𝑦 𝑖 − 14𝑦𝑧𝑗 + 20𝑥𝑧2𝑘 = 3𝑡2 + 6𝑡 𝑖 − 14𝑡2𝑗 + 20𝑡3𝑘

𝐹 . 𝑑𝑟 𝐶

= 𝐹 .𝑑𝑟

𝑑𝑡 𝑑𝑡

𝐶

= 3𝑡2 + 6𝑡 𝑖 − 14𝑡2𝑗 + 20𝑡3𝑘 . 𝑖 + 𝑗 + 𝑘 𝑑𝑡𝐶

= 3𝑡2 + 6𝑡 − 14𝑡2 + 20𝑡3 𝑑𝑡1

0

= 3𝑡3

3+

6𝑡2

2−

14𝑡3

3+

20𝑡4

4

0

1

= 1 + 3 −14

3+ 5 =

13

3

(ii) Here the curve C is given by 𝑥 = 𝑡, 𝑦 = 𝑡2 , 𝑧 = 𝑡3 from the point (0,0,0) to (1,1,1)

∴ 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = 𝑡𝑖 + 𝑡2𝑗 + 𝑡3𝑘

⇒ 𝒅𝑟

𝒅𝒕= 𝑖 + 2𝑡 𝑗 + 3𝑡2 𝑘

Now 𝐹 = 3𝑥2 + 6𝑦 𝑖 − 14𝑦𝑧𝑗 + 20𝑥𝑧2𝑘 = 9𝑡2𝑖 − 14𝑡5𝑗 + 20𝑡7𝑘

𝐹 . 𝑑𝑟 𝐶

= 𝐹 .𝑑𝑟

𝑑𝑡 𝑑𝑡

𝐶

= 9𝑡2𝑖 − 14𝑡5𝑗 + 20𝑡7𝑘 . 𝑖 + 2𝑡 𝑗 + 3𝑡2 𝑘 𝑑𝑡𝐶

= 9𝑡2 − 28𝑡6 + 60𝑡9 𝑑𝑡1

0

= 9𝑡3

3−

28𝑡7

7+

60𝑡10

10

0

1

= 3 − 4 + 6 = 5

Example 53: Find the circulation of 𝑭 around the curve 𝑪 where 𝑭 = 𝒚𝒊 + 𝒛𝒋 + 𝒙𝒌 and 𝑪 is the

circle 𝒙𝟐 + 𝒚𝟐 = 𝟏, 𝒛 = 𝟎.

44

Solution: Circulation of 𝐹 along the curve 𝐶 is 𝐹 . 𝑑𝑟 𝐶

Equation of circle is 𝑥2 + 𝑦2 = 1, 𝑧 = 0

Its parameteric equations are 𝑥 = cos θ , 𝑦 = sin θ , 𝑧 = 0

Now 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = cos θ 𝑖 + sin θ 𝑗 + 0𝑘 so that

𝑑𝑟 = − sin θ 𝑖 + cos θ 𝑗 + 0𝑘 𝑑θ

Also 𝐹 = 𝑦𝑖 + 𝑧𝑗 + 𝑥𝑘 = sin θ 𝑖 + 0𝑗 + cos θ𝑘

∴ 𝐹 . 𝑑𝑟 𝐶

= sin θ 𝑖 + 0𝑗 + cos θ𝑘 .𝐶

− sin θ 𝑖 + cos θ 𝑗 + 0𝑘 𝑑θ

= − sin2 θ2𝜋

0 𝑑𝜃 (since along the circle, 𝜃 varies from 0 to 2𝜋)

= − 1−cos 2θ

2

2𝜋

0𝑑θ = −

θ

2−

sin 2θ

4

0

2π

= − 2π

2− 0 − (0 − 0) = −π

Example 54: Find the work done in moving a particle once round a circle 𝑪 in the 𝒙𝒚 plane, if

the circle has its centre at the origin and radius 2 units and the force field is given as

𝑭 = 𝟐𝒙 − 𝒚 + 𝟐𝒛 𝒊 + 𝒙 + 𝒚 − 𝒛 𝒋 + (𝟑𝒙 − 𝟐𝒚 − 𝟓𝒛)𝒌 .

Solution: Equation of a circle having centre (0, 0) with radius 2 in 𝑥𝑦 plane is 𝑥2 + 𝑦2 = 4 .

Parametric equations of this circle are 𝑥 = 2 cos 𝑡 , 𝑦 = 2 sin 𝑡 , 𝑧 = 0.

Since integration is to be performed around a circle in 𝑥𝑦 plane,

∴ 𝑟 = 𝑥𝑖 + 𝑦𝑗 = 2 cos 𝑡 𝑖 + 2 sin 𝑡 𝑗 ⇒ 𝑑𝑟

𝑑𝑡= −2 sin 𝑡 𝑖 + 2 cos 𝑡 𝑗

Work done, 𝐹 .𝑑𝑟

𝑑𝑡 𝑑𝑡

𝐶

= 2𝑥 − 𝑦 + 2𝑧 𝑖 + 𝑥 + 𝑦 − 𝑧 𝑗 + (3𝑥 − 2𝑦 − 5𝑧)𝑘 . −2 sin 𝑡 𝑖 + 2 cos 𝑡 𝑗 𝑑𝑡

= 4 cos 𝑡 − 2 sin 𝑡 𝑖 + 2 cos 𝑡 + 2 sin 𝑡 𝑗 + (6 cos 𝑡 − 4 sin 𝑡)𝑘 . −2 sin 𝑡 𝑖 + 2 cos 𝑡 𝑗 𝑑𝑡

In moving round the circle, 𝑡 varies from 0 to 2𝜋

∴ Work done = 4 cos 𝑡 − 2 sin 𝑡 (−2 sin 𝑡) + 2 cos 𝑡 + 2 sin 𝑡 (2 cos 𝑡) 2𝜋

0𝑑𝑡

= −8 cos 𝑡 sin 𝑡 + 4 sin2 𝑡 + 4 cos2 𝑡 + 4 sin 𝑡 cos 𝑡 2𝜋

0𝑑𝑡

= 4 − 4 sin 𝑡 cos 𝑡 2𝜋

0𝑑𝑡 = 4𝑡 − 4

sin 2 𝑡

2

0

2𝜋

= 8𝜋 − 2 sin2 2𝜋 − (0 − 0) = 8𝜋

ASSIGNMENT 6

45

1. Using the line integral, compute the work done by the force 𝐹 = 2𝑦 + 3 𝑖 + 𝑥𝑧 𝑗 + (𝑦𝑧 − 𝑥)𝑘 ,

when it moves a particle from (0,0,0) to (2,1,1) along the curve 𝑥 = 2𝑡2 , 𝑦 = 𝑡, 𝑧 = 𝑡3.

2. Find the work done in moving a particle in the force field 𝐹 = 3𝑥2𝑖 + 2𝑥𝑧 − 𝑦 𝑗 + 𝑧𝑘 , along

(a) the straight line from (0,0,0) to (2,1,3).

(b) the curve defined by 𝑥2 = 4𝑦, 3𝑥3 = 8𝑧 from 𝑥 = 0 to 𝑥 = 2.

3. If C is a simple closed curve in the 𝑥𝑦 plane not enclosing the origin, show that 𝐹 . 𝑑𝑟 𝐶

= 0,

where 𝐹 =𝑦𝑖 −𝑥𝑗

𝑥2+𝑦2.

4. If 𝑓 = 5𝑥𝑦 − 6𝑥2 𝑖 + 2𝑦 − 4𝑥 𝑗 , evaluate 𝑓 . 𝑑𝑟 𝐶

along the curve C in xy-plane, 𝑦 = 𝑥3

from the point 1, 1 to (2, 8).

5. Evaluate (𝑥𝑦 + 𝑧2)𝑑𝑠𝐶

where C is the arc of the helix 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 𝑧 = 𝑡 which joins

the points 1, 0, 0 𝑎𝑛𝑑 (−1, 0, 𝜋).

6. If 𝐹 = 2𝑦𝑖 − 𝑧𝑗 + 𝑥𝑘 , evaluate 𝐹 𝐶

× 𝑑𝑟 along the curve 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 2 cos 𝑡 from

𝑡 = 0 to 𝑡 =𝜋

2.

17.11 SURFACE INTEGRALS AND FLUX

An integral which is to be evaluated over a surface is called a surface integral. Suppose 𝑆 is a surface

of finite area. Divide the area 𝑆 into n sub-areas 𝛿𝑆1, 𝛿𝑆2, …… , 𝛿𝑆𝑛 .

In each area 𝛿𝑆𝑖 , choose an arbitrary point 𝑃𝑖 𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 . Let 𝜑 define

a scalar point function over the area 𝑆.

Now from the sum 𝜑 𝑃𝑖 𝛿𝑆𝑖𝑛𝑖=1 , where 𝜑 𝑃𝑖 = 𝜑 𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖

Now let us take the limit of the sum as 𝑛 → ∞, each sub-area 𝛿𝑆𝑖

reduces to a point and the limit if it exists is called the surface

integral of 𝜑 over 𝑆 and is denoted by 𝑑𝑆𝜑

.

Note: If 𝑆 is piecewise smooth then the function 𝜑 𝑥, 𝑦, 𝑧 is continuous over 𝑆 and then the limit exists and is

independent of sub-divisions and choice of the point 𝑃𝑖 .

Flux: Suppose 𝑆 is a piecewise smooth surface so that the vector function 𝐹 defined over 𝑆 is

continuous over 𝑆. Let 𝑃 be any point of the surface 𝑆 and suppose 𝑛 is a unit vector at 𝑃 in the

direction of outward drawn normal to the surface 𝑆 at 𝑃. Then the component of 𝐹 along 𝑛 is 𝐹 . 𝑛 and

the integral of 𝐹 . 𝑛 over 𝑆 is called the surface integral of 𝐹 over 𝑆 and is denoted by 𝐹 . 𝑛 𝑑𝑆𝑆

. It is

also called flux of 𝐹 over 𝑆.

Different Forms of Surface Integral

(i) Flux of 𝐹 over 𝑆 = 𝐹 . 𝑛 𝑑𝑆𝑆

… (1)

46

Now let 𝑑𝑆 denote a vector (called vector area) whose magnitude is that of differential of

surface area i.e., 𝑑𝑆 and whose direction is that of 𝑛 . Then clearly

𝑑𝑆 = 𝑛 𝑑𝑆

From (1), flux of 𝐹 over 𝑆 = 𝐹 . 𝑑𝑆 𝑆

…. (2)

(ii) Suppose outward drawn normal to the surface 𝑆 at 𝑃 makes angles 𝛼, 𝛽, 𝛾 with the

positive direction of axes and if 𝑙, 𝑚, 𝑛 denote the direction cosines of this outward drawn

normal, then

𝑙 = cos 𝛼 , 𝑚 = cos 𝛽 , 𝑛 = cos 𝛾

Therefore, 𝑛 = cos 𝛼 𝑖 + cos 𝛽 𝑗 + cos 𝛾 𝑘

If 𝐹 = 𝐹1𝑖 + 𝐹2𝑗 + 𝐹3𝑘 then 𝐹 . 𝑛 = 𝐹1 cos 𝛼 + 𝐹2 cos 𝛽 + 𝐹3 cos 𝛾

∴ From (1), flux of 𝐹 over 𝑆 = 𝐹1 cos 𝛼 + 𝐹2 cos 𝛽 + 𝐹3 cos 𝛾 𝑑𝑆𝑆

… (3)

Now 𝑑𝑆 cos 𝛼 is the projection of area 𝑑𝑆 on the 𝑦𝑧 plane, therefore 𝑑𝑆 cos 𝛼 = 𝑑𝑦𝑑𝑧.

Similarly 𝑑𝑆 cos 𝛽 and 𝑑𝑆 cos 𝛾 are the projections of the area 𝑑𝑆 on the 𝑧𝑥 and 𝑥𝑦 plane

respectively and therefore 𝑑𝑆 cos 𝛽 = 𝑑𝑧𝑑𝑥, 𝑑𝑆 cos 𝛾 = 𝑑𝑥𝑑𝑦.

∴ From (3), 𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 1𝑑𝑦𝑑𝑧 + 𝐹 2𝑑𝑧𝑑𝑥 + 𝐹 3𝑑𝑥𝑑𝑦𝑆

… (4)

Note: In order to evaluate surface integral it is convenient to express them as double integrals by

taking the projection of surface 𝑆 on one of the coordinate planes. This will happen only if any line

perpendicular to co-ordinate plane chosen meets the surface in one point and not more than one

point. Surface S is divided into sub surfaces, if above requirement is not met, so that sub surfaces

may satisfy the above requirement.

(iii) Suppose surface 𝑆 is such that any line

perpendicular to 𝑥𝑦 plane does not meet 𝑆 in more

than one point. Let the equation of surface 𝑆 be

𝑍 = 𝑕(𝑥, 𝑦).

Let 𝑅1 denotes the orthogonal projection of 𝑆 on the

𝑥𝑦 plane. Then projection of 𝑑𝑆 on the 𝑥𝑦 plane

= 𝑑𝑆 cos 𝛾 , where 𝛾 is the acute angle which the

normal to the surface 𝑆 makes with positive

direction of Z-axis.

∴ 𝑑𝑆 cos 𝛾 = 𝑑𝑥𝑑𝑦 … (5)

But cos 𝛾 = 𝑛 .𝑘

𝑛 = 𝑛 . 𝑘

Therefore from (5), 𝑑𝑆 =𝑑𝑥 𝑑𝑦

𝑛 .𝑘 … (6)

Thus 𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑅1

𝑑𝑥 𝑑𝑦

𝑛 .𝑘 … (7)

Similarly we have, 𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑅2

𝑑𝑦 𝑑𝑧

𝑛 .𝑖 … (8)

𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑅3

𝑑𝑧 𝑑𝑥

𝑛 .𝑗 … (9)

where 𝑅2, 𝑅3 are the projections of 𝑆 on 𝑧𝑥 and 𝑥𝑦 planes, respectively.

47

Example 55: Evaluate 𝑭 . 𝒏 𝒅𝑺𝑺

where 𝑭 = 𝒛 𝒊 + 𝒙 𝒋 + 𝟑𝒚𝟐𝒛 𝒌 and S is the surface of the

cylinder 𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔 included in the first octant between 𝒛 = 𝟎 and 𝒛 = 𝟓.

Solution: A vector normal to the surface 𝑆 is given by

𝑛 = ∇ 𝑥2 + 𝑦2 = 2𝑥 𝑖 + 2𝑦 𝑗 … (1)

∴ 𝑛 = unit vector normal to surface 𝑆 at any point 𝑥, 𝑦, 𝑧 =2𝑥𝑖 +2𝑦𝑗

4𝑥2+4𝑦2 … (2)

∵ 𝑥2 + 𝑦2 = 16, therefore 𝑛 =2𝑥𝑖 +2𝑦𝑗

4(𝑥2+𝑦2)=

2𝑥𝑖 +2𝑦𝑗

8=

𝑥

4𝑖 +

𝑦

4𝑗 … (3)

Now 𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑑𝑥 𝑑𝑧

𝑛 .𝑗 𝑅

(projection on 𝑥𝑦 plane can’t be taken as the surface 𝑠 is peerpendicular to 𝑥𝑦 plane)

= 𝑥𝑧

4+

𝑥𝑦

4

𝑑𝑥 𝑑𝑧𝑦

4𝑅

= 𝑥𝑧

𝑦+ 𝑥 𝑑𝑥 𝑑𝑧

𝑅, since from 3 , 𝑛 . 𝑗 =

𝑦

4

= 𝑥𝑧

16−𝑥2+ 𝑥 𝑑𝑥 𝑑𝑧

4

𝑥=0

5

𝑧=0=

−𝑧

2(−2𝑥)

16−𝑥2+ 𝑥 𝑑𝑥 𝑑𝑧

4

𝑥=0

5

𝑧=0

= −𝑧

2

16−𝑥2

1 2 +

𝑥2

2 𝑥=0

4

𝑑𝑧5

𝑧=0= (4𝑧 + 8)𝑑𝑧

5

𝑧=0=

4𝑧2

2+ 8𝑧

𝑧=0

5

= 90

Example 56: Evaluate 𝑭 . 𝒏 𝒅𝑺𝑺

where 𝑭 = 𝟔𝒛 𝒊 − 𝟒 𝒋 + 𝒚 𝒌 and S is the portion of the plane

𝟐𝒙 + 𝟑𝒚 + 𝟔𝒛 = 𝟏𝟐 in the first octant.

Solution: Vector normal to surface 𝑆 is given by

∇ 2𝑥 + 3𝑦 + 6𝑧 = 2𝑖 + 3𝑗 + 6𝑘

∴ 𝑛 = unit vector normal to surface 𝑆 at any point (𝑥, 𝑦, 𝑧) =2𝑖 +3𝑗 +6𝑘

4+9+36=

2

7𝑖 +

3

7𝑗 +

6

7𝑘

Now 𝐹 . 𝑛 = 6𝑧 𝑖 − 4 𝑗 + 𝑦 𝑘 . 2

7𝑖 +

3

7𝑗 +

6

7𝑘 =

12

7𝑧 −

12

7+

6

7𝑦

Taking projection on 𝑥𝑦 plane

𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑑𝑥 𝑑𝑦

𝑛 .𝑘 𝑅= 𝐹 . 𝑛

𝑑𝑥 𝑑𝑦

6 7 𝑅 … (1)

where 𝑅 is the region of projection of 𝑆 on 𝑥𝑦 plane. 𝑅 is bounded by x-axis, y-axis and the line

2𝑥 + 3𝑦 = 12, 𝑧 = 0. In order to evaluate double integral in (1), y varies from 0 to 4 and x varies

from 0 to 12−3𝑦

2. Therefore from (1)

𝐹 . 𝑛 𝑑𝑆𝑆

= 2𝑧 − 2 + 𝑦 𝑑𝑥 𝑑𝑦,12−3𝑦

2𝑥=0

4

𝑦=0 Find 𝑧 from 2𝑥 + 3𝑦 + 6𝑧 = 12

= 2 2 −𝑥

3−

𝑦

2 − 2 + 𝑦 𝑑𝑥 𝑑𝑦

12−3𝑦

2𝑥=0

4

𝑦=0

48

= 2 −2𝑥

3 𝑑𝑥 𝑑𝑦

12−3𝑦

2𝑥=0

4

𝑦=0= 2𝑥 −

𝑥2

3 𝑥=0

12−3𝑦

2𝑑𝑦

4

0

= 2 ×12−3𝑦

2−

1

3

12−3𝑦

2

2

𝑑𝑦4

0

= 12−3𝑦 2

−6+

12−3𝑦 3

108 𝑦=0

4

=144

6−

1728

108= 24 − 16 = 8

Example 57: 𝝋𝑺

𝒏 𝒅𝑺 where 𝝋 =𝟑

𝟖𝒙𝒚𝒛 and 𝑺 is the surface of cylinder 𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔

included in the first octant between 𝒛 = 𝟎 to 𝒛 = 𝟓.

Solution: A vector normal to the surface 𝑆 is given by

𝑛 = ∇ 𝑥2 + 𝑦2 = 2𝑥𝑖 + 2𝑦𝑗

∴ 𝑛 = unit vector normal to surface 𝑆 at any point (𝑥, 𝑦, 𝑧) =2𝑥𝑖 +2𝑦𝑗

4𝑥2+4𝑦2

𝑛 =2𝑥𝑖 +2𝑦𝑗

4(𝑥2+𝑦2)=

2𝑥𝑖 +2𝑦𝑗

8=

𝑥

4𝑖 +

𝑦

4𝑗 ∵ 𝑥2 + 𝑦2 = 16

Now 𝜑 𝑛 𝑑𝑆𝑆

= 3

8𝑥𝑦𝑧

𝑥

4𝑖 +

𝑦

4𝑗

𝑑𝑥 𝑑𝑧

𝑛 .𝑗 𝑅 …(1)

where 𝑅 is the region of projection of 𝑆 on 𝑧𝑥 plane. Therefore, from (1)

𝜑 𝑛 𝑑𝑆𝑆

= 3

32𝑥2𝑦𝑧 𝑖 +

3

32𝑥𝑦2𝑧 𝑗

4

𝑥=0

5

𝑧=0

𝑑𝑥 𝑑𝑧

𝑦 4 , where 𝑦2 = 16 − 𝑥2

= 3

8𝑥2𝑧 𝑖 +

3

8𝑥𝑧 16 − 𝑥2 𝑗

4

𝑥=0

5

𝑧=0𝑑𝑥 𝑑𝑧

=3

8

𝑥3𝑧

3 𝑖 −

𝑧

2

16−𝑥2 3 2

3 2 𝑗

𝑥=0

45

𝑧=0𝑑𝑍

=3

8

64

3𝑧 𝑖 +

64

3𝑧 𝑗

5

𝑧=0𝑑𝑧 = 8

𝑧2

2𝑖 +

𝑧2

2𝑗

𝑧=0

5

= 100 𝑖 + 100 𝑗

Example 58: Evaluate 𝑭 . 𝒅𝑺𝑺

where 𝑭 = 𝒙𝒊 − 𝒛𝟐 − 𝒛𝒙 𝒋 − 𝒙𝒚𝒌 and 𝑺 is the triangular

surface with vertices 𝟐, 𝟎, 𝟎 , (𝟎, 𝟐, 𝟎) and (𝟎, 𝟎, 𝟒).

Solution:The triangular surface 𝑆 with vertices 2,0,0 , (0,2,0), and (0,0,4) is given by the equation

𝑥

2+

𝑦

2+

𝑧

4= 1 ⇒ 2𝑥 + 2𝑦 + 𝑧 = 4 … (1)

Vector normal to surface 𝑆 is given by ∇ 2𝑥 + 2𝑦 + 𝑧 = 2𝑖 + 2𝑗 + 𝑘

∴ 𝑛 = unit vector normal to surface 𝑆 at any point (𝑥, 𝑦, 𝑧) =2𝑖 +2𝑗 +𝑘

4+4+1=

2

3𝑖 +

2

3𝑗 +

1

3𝑘

Now 𝐹 . 𝑛 = 𝑥𝑖 − 𝑧2 − 𝑧𝑥 𝑗 − 𝑥𝑦𝑘 . 2

3𝑖 +

2

3𝑗 +

1

3𝑘 =

2

3𝑥 −

2

3 𝑧2 − 𝑧𝑥 −

1

3𝑥𝑦

Taking projection on 𝑥𝑦 plane

49

𝐹 . 𝑛 𝑑𝑆𝑆

= 𝐹 . 𝑛 𝑑𝑥 𝑑𝑦

𝑛 .𝑘 𝑅= 𝐹 . 𝑛

𝑑𝑥 𝑑𝑦

1 3 𝑅 … (2)

where 𝑅 is the region of projection of 𝑆 on 𝑥𝑦 plane. 𝑅 is bounded by x-axis, y-axis and the line

2𝑥 + 2𝑦 = 4 i.e., 𝑥 + 𝑦 = 2, 𝑧 = 0. In order to integrate double integral in (2), y varies from 0 to 2

and x varies from 0 to 2 − 𝑦. Therefore from (2)

𝐹 . 𝑛 𝑑𝑆𝑆

= 2

3𝑥 −

2

3 𝑧2 − 𝑧𝑥 −

1

3𝑥𝑦 𝑑𝑥 𝑑𝑦

2−𝑦

𝑥=0

2

𝑦=0 Find 𝑧 from 2𝑥 + 2𝑦 + 𝑧 = 4

= 2

3𝑥 −

2

3 4 − 2𝑥 − 2𝑦 2 − 4 − 2𝑥 − 2𝑦 𝑥 −

1

3𝑥𝑦 𝑑𝑥 𝑑𝑦

2−𝑦

𝑥=0

2

𝑦=0

=1

3 2𝑥 − 2 16 + 4𝑥2 + 4𝑦2 − 16𝑥 + 8𝑥𝑦 − 16𝑦 − 4𝑥 + 2𝑥2 + 2𝑥𝑦 − 𝑥𝑦 𝑑𝑥 𝑑𝑦

2−𝑦

𝑥=0

2

𝑦=0

=1

3 −12𝑥2 − 8𝑦2 + 42𝑥 + 32𝑦 − 21𝑥𝑦 − 32 𝑑𝑥 𝑑𝑦

2−𝑦

𝑥=0

2

𝑦=0

=1

3 −4𝑥3 − 8𝑥𝑦2 + 21𝑥2 + 32𝑥𝑦 − 21

𝑥2𝑦

2− 32𝑥

𝑥=0

2−𝑦

𝑑𝑦2

𝑦=0

=1

3 −4 2 − 𝑦 3 − 8 2 − 𝑦 𝑦2 + 21 2 − 𝑦 2 + 32 2 − 𝑦 𝑦 − 21

2−𝑦 2𝑦

2 − 32 2 − 𝑦 𝑑𝑦

2

𝑦=0

=1

3

3

2𝑦3 + 9𝑦2 + 18𝑦 − 12 𝑑𝑦

2

𝑦=0= 38

Example 59: Evalutate 𝒇 ∙ 𝒏 𝑺

𝒅𝒔 where 𝒇 = 𝟐𝒙𝟐𝒚 𝒊 − 𝒚𝟐 𝒋 + 𝟒𝒙𝒛𝟐 𝒌 and S is closed surface of

the region in the first octant bounded by the cylinder 𝒚𝟐 + 𝒛𝟐 = 𝟗 and the planes 𝒙 = 𝟎, 𝒙 = 𝟐,

𝒚 = 𝟎 𝐚𝐧𝐝 𝒛 = 𝟎.

Solution: The given closed surface S is piecewise smooth and is

comprised of S1 –the rectangular face OAEB in xy-plane; S2 –the

rectangular face OADC in xz-plane; S3 – the circular quadrant ABC

in yz-plane; S4 – the circular quadrant AED and S5 – the curved

surface BCDE of the cylinder in the first octant (see Fig. 17.16).

∴ 𝑓 ∙ 𝑛 𝑆

𝑑𝑠 = 𝑓 ∙ 𝑛 𝑆1

𝑑𝑠 + 𝑓 ∙ 𝑛 𝑆2

𝑑𝑠 + 𝑓 ∙ 𝑛 𝑆3

𝑑𝑠

+ 𝑓 ∙ 𝑛 𝑆4

𝑑𝑠 + 𝑓 ∙ 𝑛 𝑆5

𝑑𝑠 … (1)

Now 𝑓 ∙ 𝑛 𝑆1

𝑑𝑠 = 2𝑥2𝑦 𝑖 − 𝑦2 𝑗 + 4𝑥𝑧2 𝑘 ∙ −𝑘 𝑆1

𝑑𝑠 = −4 𝑥𝑧2𝑆1

𝑑𝑠 = 0

(𝑎𝑠 𝑧 = 0 𝑖𝑛 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒)

Similarly, 𝑓 ∙ 𝑛 𝑆2

𝑑𝑠 = 0 and 𝑓 ∙ 𝑛 𝑆3

𝑑𝑠

𝑓 ∙ 𝑛 𝑆4

𝑑𝑠 = 2𝑥2𝑦 𝑖 − 𝑦2 𝑗 + 4𝑥𝑧2 𝑘 ∙ 𝑖 𝑆4

𝑑𝑠 = 2𝑥2𝑦𝑆4

𝑑𝑠

= 8𝑦 𝑑𝑦𝑑𝑧 9−𝑧2

0

3

0= 4 (9 − 𝑧2)𝑑𝑧

3

0= 72

To find 𝑛 in S5 , we note that ∇ 𝑦2 + 𝑧2 − 9 = 2𝑦 𝑗 + 2𝑧𝑘 ,

Implying 𝑛 =2𝑦 𝑗 +2𝑧𝑘

4 𝑦2+𝑧2 =

𝑦 𝑗 +𝑧𝑘

3 and 𝑛 ∙ 𝑘 =

𝑧

3 so that 𝑑𝑠 = 𝑑𝑥𝑑𝑦/(𝑧/3) (𝑎𝑠 𝑦2 + 𝑧2 = 9)

50

𝑓 ∙ 𝑛 𝑆5

𝑑𝑠 = −𝑦3+4𝑥𝑧3

3 𝑑𝑥𝑑𝑦/(𝑧/3)

3

0

2

0=

−𝑦3

𝑧+ 4𝑥𝑧2 𝑑𝑥𝑑𝑦

3

0

2

0

Now putting 𝑦 = 3 sin 𝜃 , 𝑧 = 3 cos 𝜃 ∴ 𝑑𝑦 = 3 cos 𝜃 𝑑𝜃

−27 sin 3 𝜃

3 cos 𝜃+ 4𝑥 (9 cos2 𝜃) 3 cos 𝜃 𝑑𝜃𝑑𝑥

𝜋

20

2

0

ASSIGNMENT 7

1. If velocity vector is 𝐹 = 𝑦 𝑖 + 2 𝑗 + 𝑥𝑧 𝑘 m/sec., show that the flux of water through the

parabolic cylinder 𝑦 = 𝑥2, 0 ≤ 𝑥 ≤ 3, 0 ≤ 𝑧 ≤ 2 is 69 𝑚3/𝑠𝑒𝑐.

2. Evaluate 𝐹 . 𝑛 𝑑𝑆𝑆

where 𝐹 = 𝑥 + 𝑦2 𝑖 − 2𝑥 𝑗 + 2𝑦𝑧 𝑘 and S is the surface of the plane

2𝑥 + 𝑦 + 2𝑧 = 6 in the first octant.

3. If 𝐹 = 4𝑥𝑧 𝑖 − 𝑦2 𝑗 + 𝑦𝑧 𝑘 ; evaluate 𝐹 . 𝑛 𝑑𝑆𝑆

, where S is the surface of the cube bounded by

𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1.

4. If 𝐹 = 2𝑦 𝑖 − 3 𝑗 + 𝑥2 𝑘 and S is the surface of the parabolic cylinder 𝑦2 = 8𝑥 in the first

octant bounded by the planes 𝑦 = 4 and 𝑧 = 6, show that 𝐹 ∙ 𝑛 𝑆

𝑑𝑠 = 132.

5. Evaluate 𝐹 ∙ 𝑛 𝑆

𝑑𝑠 where 𝐹 = 6𝑧 𝑖 − 4 𝑗 + 𝑦 𝑘 and S is the portion of the plane

2𝑥 + 3𝑦 + 6𝑧 = 12 in the first octant.

17.12 VOLUME INTEGRALS

Suppose 𝑉 is the volume bounded by a surface 𝑆. Divide the volume 𝑉 into sub-volumes 𝛿𝑉1, 𝛿𝑉2,

…, 𝛿𝑉𝑛 . In each 𝛿𝑉𝑖 , choose an arbitrary point 𝑃𝑖 whose coordinates are (𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖). Let 𝜑 be a single

valued function defined over 𝑉. Form the sum 𝜑 𝑃𝑖 𝛿𝑉𝑖 , where 𝜑 𝑃𝑖 = 𝜑 𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 .

Now let us take the limit of the sum as 𝑛 → ∞, then the limit, if exists, is called the volume integral

of 𝜑 over 𝑉 and is denoted as 𝜑𝑉

𝑑𝑉.

Likewise if 𝐹 is a vector point function defined in the given region of volume 𝑉 then vector volume

integral of 𝐹 over 𝑉 is 𝐹 𝑉

𝑑𝑉.

Note: Above volume integral becomes 𝜑𝑉

𝑑𝑥 𝑑𝑦 𝑑𝑧 if we subdivide the volume 𝑉 into small

cuboids by drawing lines parallel to three co-ordinate axes because in that case 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧.

Example 60: If 𝑭 = 𝟐𝒙𝟐 − 𝟑𝒛 𝒊 − 𝟐𝒙𝒚𝒋 − 𝟒𝒙𝒌 , evualuate 𝛁 × 𝑭 𝒅𝑽𝑽

where 𝑽 is the region

bounded by the co-ordinate planes and the plane 𝟐𝒙 + 𝟐𝒚 + 𝒛 = 𝟒.

Solution: Consider

51

∇ × 𝐹 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

2𝑥2 − 3𝑧 −2𝑥𝑦 −4𝑥

= 0 𝑖 + 𝑗 − 2𝑦 𝑘

Region bounded by 2𝑥 + 2𝑦 + 𝑧 = 4 and coordinate planes such that

2𝑥 ≤ 4, 2𝑥 + 2𝑦 ≤ 4, 2𝑥 + 2𝑦 + 𝑧 ≤ 4

i.e. 𝑥 ≤ 2, 𝑦 ≤ 2 − 𝑥, 𝑧 ≤ 4 − 2𝑥 − 2𝑦

∴ ∇ × 𝐹 𝑑𝑉𝑉

= 𝑗 − 2𝑦 𝑘 𝑑𝑧 𝑑𝑦 𝑑𝑥4−2𝑥−2𝑦

𝑧=0

2−𝑥

𝑦=0

2

𝑥=0

= 𝑧 𝑗 − 2𝑦𝑧 𝑘 𝑧=0

𝑧=4−2𝑥−2𝑦 𝑑𝑦 𝑑𝑥

2−𝑥

𝑦=0

2

𝑥=0

= (4 − 2𝑥 − 2𝑦) 𝑗 − 2𝑦(4 − 2𝑥 − 2𝑦) 𝑘 𝑑𝑦 𝑑𝑥2−𝑥

𝑦=0

2

𝑥=0

= 4𝑦 − 2𝑥𝑦 − 𝑦2 𝑗 − 4𝑦2 − 2𝑥𝑦2 −4

3𝑦3 𝑘

𝑦=0

𝑦=2−𝑥2

𝑥=0𝑑𝑥

= 8 − 4𝑥 − 2𝑥 2 − 𝑥 − 2 − 𝑥 2 𝑗 −

4(2 − 𝑥)2 − 2𝑥(2 − 𝑥)2 −4

3(2 − 𝑥)3 𝑘

2

𝑥=0𝑑𝑥

= 4 − 4𝑥 + 𝑥2 𝑗 − −2

3𝑥3 + 4𝑥2 − 8𝑥 +

16

3 𝑘

2

𝑥=0𝑑𝑥

= 4𝑥 − 2𝑥2 +𝑥3

3 𝑥=0

𝑥=2

𝑗 − −1

6𝑥4 +

4

3𝑥3 − 4𝑥2 +

16

3𝑥

𝑥=0

𝑥=2

𝑘

=8

3𝑗 −

8

3𝑘 =

8

3(𝑗 − 𝑘 )

ASSIGNMENT 8

1. Evaluate φ 𝑑𝑉𝑉

where 𝜑 = 45𝑥2𝑦 and 𝑉 is the region bounded by the planes 4𝑥 + 2𝑦 + 𝑧 =

8, 𝑥 = 0, 𝑦 = 0, 𝑧 = 0.

2. If 𝐹 = 2𝑥𝑧 𝑖 − 𝑥 𝑗 + 𝑦2 𝑘 ; evaluate F 𝑑𝑉𝑉

where 𝑉 is the region bounded by the planes

𝑥 = 0, 𝑦 = 0, 𝑥 = 2, 𝑦 = 6, 𝑧 = 𝑥2 , 𝑧 = 4.

17.13 STOKE’S THEOREM (Relation between Line and Surface Integral)

Statement: Let S be a piecewise smooth open surface bounded by a piecewise smooth simple curve

𝐶 . If 𝑓 (𝑥, 𝑦, 𝑧) be a continuous vector function which has continuous first partial derivative in a

region of space which contains 𝑆 , then 𝑓 . 𝑑𝑟 𝐶

= 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 𝐶

𝑑𝑆 , where 𝑛 is the unit normal

vector at any point of 𝑆 and 𝐶 is trasversed in positive direction.

Direction of 𝐶 is positive if an observer walking on the boundary of 𝑆 in this direction with its head

pointing in the direction of outward normal 𝑛 to 𝑆 has the surface on the left.

We may put the statement of Stoke’s theorem in words as under:

52

The line integral of the tangential component of a vector 𝑓 taken around a simple closed curve 𝐶 is

equal to the surface integral of normal component of curl of 𝑓 taken over 𝑆 having 𝐶 as its boundary.

Stoke’s Theorem in Cartesian Form:

a) Cartesian Form of Stoke’s Theorem in Plane (or Green’s Theorem in Plane)

Choose system of coordinate axes such that the plane of the surface is in 𝑥𝑦 plane and normal to the

surface 𝑆 lies along the z-axis. Normal vector is constant in this case.

Suppose 𝑓 = 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3 𝑘

∴ 𝑓 . 𝑑𝑟 𝐶

= 𝑓 .𝑑𝑟

𝑑𝑠 𝑑𝑠

𝐶= 𝑓 . 𝑡 𝑑𝑠

𝐶 where 𝑡 =

𝑑𝑟

𝑑𝑠 is unit vector tangent to 𝐶.

∴ 𝑓 . 𝑑𝑟 𝐶

= 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3 𝑘 . 𝑖 𝑑𝑥

𝑑𝑠+ 𝑗

𝑑𝑦

𝑑𝑠+ 𝑘

𝑑𝑧

𝑑𝑠 𝑑𝑠

𝐶

= 𝑓1 𝑑𝑥

𝑑𝑠+ 𝑓2

𝑑𝑦

𝑑𝑠+ 𝑓3

𝑑𝑧

𝑑𝑠 𝑑𝑠

𝐶

But tangent at any point lies in the 𝑥𝑦 plane, so 𝑑𝑧

𝑑𝑠= 0

∴ 𝑓 . 𝑑𝑟 𝐶

= 𝑓1 𝑑𝑥

𝑑𝑠+ 𝑓2

𝑑𝑦

𝑑𝑠 𝑑𝑠

𝐶 … (1)

Now 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 𝑆

𝑑𝑆 = 𝑐𝑢𝑟𝑙 𝑓 . 𝑘 𝑆

𝑑𝑆 (Here normal is along Z-axis)

= 𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦

𝑆𝑑𝑥𝑑𝑦 … (2)

Using (1) and (2), Stoke’s theorem is

𝑓1𝑑𝑥 + 𝑓2𝑑𝑦 𝐶

= 𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦

𝑆𝑑𝑥𝑑𝑦

b) Cartesian Form of Stoke’s Theorem in Space

Suppose 𝑓 = 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3𝑘 and 𝑛 is an outward drawn normal unit vector of 𝑆 making angles 𝛼,

𝛽, 𝛾 with positive direction of axes.

∴ 𝑛 = cos 𝛼 𝑖 + cos 𝛽 𝑗 + cos 𝛾 𝑘

Now, ∇ × 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑓1 𝑓2 𝑓3

𝑖. 𝑒. 𝑐𝑢𝑟𝑙 𝑓 = 𝜕𝑓3

𝜕𝑦−

𝜕𝑓2

𝜕𝑧 𝑖 +

𝜕𝑓1

𝜕𝑧−

𝜕𝑓3

𝜕𝑥 𝑗 +

𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦 𝑘

∴ 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 = 𝜕𝑓3

𝜕𝑦−

𝜕𝑓2

𝜕𝑧 cos 𝛼 +

𝜕𝑓1

𝜕𝑧−

𝜕𝑓3

𝜕𝑥 cos 𝛽 +

𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦 cos 𝛾 … (1)

Also 𝑓 . 𝑑𝑟 = 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3𝑘 . 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘

53

or 𝑓 . 𝑑𝑟 = 𝑓1 𝑑𝑥 + 𝑓2 𝑑𝑦 + 𝑓3 𝑑𝑧

Then Stoke’s theorem is

𝑓1 𝑑𝑥 + 𝑓2 𝑑𝑦 + 𝑓3 𝑑𝑧 𝐶

= 𝜕𝑓3

𝜕𝑦−

𝜕𝑓2

𝜕𝑧 cos 𝛼 +

𝜕𝑓1

𝜕𝑧−

𝜕𝑓3

𝜕𝑥 cos 𝛽 +

𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦 cos 𝛾

𝑆𝑑𝑆

Example 61: Verify Stoke’s theorem for 𝒇 = 𝟐𝒙 − 𝒚 𝒊 − 𝒚𝒛𝟐𝒋 − 𝒚𝟐𝒛𝒌 when 𝑺 is the upper half

of the surface of the sphere 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 = 𝟏 and 𝑪 is its boundary.

Solution: The boundary 𝐶 of the upper half of the sphere 𝑆 is circle in the 𝑥𝑦 plane. Therefore,

parametric equations of 𝐶 are 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 𝑧 = 0 when 0 ≤ t ≤ 2π

Now, 𝑓 . 𝑑𝑟 𝐶

= 𝑓1 𝑑𝑥 + 𝑓2 𝑑𝑦 + 𝑓3 𝑑𝑧 𝐶

= 2𝑥 − 𝑦 𝑑𝑥 − 𝑦𝑧2𝑑𝑦 − 𝑦2𝑧 𝑑𝑧𝐶

= 2 cos 𝑡 − sin 𝑡 (− sin 𝑡) 𝑑𝑡𝐶

∵ 𝑥 = cos 𝑡 , ∴ 𝑑𝑥 = − sin 𝑡 𝑑𝑡 and other terms of integrand become zero as 𝑧 = 0

= 2 cos 𝑡 − sin 𝑡 + sin2 𝑡 𝑑𝑡2𝜋

𝑡=0= 2

cos 2 𝑡

2 𝑡=0

2𝜋

+ sin2 𝑡 𝑑𝑡2𝜋

𝑡=0

= 1 − 1 + 4 sin2 𝑡 𝑑𝑡𝜋 2

𝑡=0 (Property of definite integral)

= 0 + 4.1

2.𝜋

2= 𝜋 … (1)

Now, 𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

2𝑥 − 𝑦 −𝑦𝑧2 −𝑦2𝑧

= −2𝑦𝑧 + 2𝑦𝑧 𝑖 + 0 − 0 𝑗 + 0 + 1 𝑘 = 𝑘

Now, 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 𝑑𝑆 = 𝑘 𝑆

. 𝑛 𝑑𝑆

= 𝑘 𝑅

. 𝑛 𝑑𝑥 𝑑𝑦

𝑛 .𝑘 [where 𝑅 is the projection of S on 𝑥𝑦 − plane]

= 𝑑𝑥 𝑑𝑦𝑅

Now projection of 𝑆 on 𝑥𝑦 plane is circle 𝑥2 + 𝑦2 = 1.

= 𝑑𝑥 1−𝑥2

𝑦=− 1−𝑥2 𝑑𝑦1

𝑥=−1= 4 𝑑𝑥

1−𝑥2

𝑦=0𝑑𝑦

1

𝑥=0 [By definite integral]

= 4 1 − 𝑥21

𝑥=0𝑑𝑥 = 4

𝑥 1−𝑥2

2+

1

2sin−1 𝑥

𝑥=0

1

= 4 1

2sin−1 1 = 4 ×

𝜋

4 = 𝜋 … (2)

From (1) and (2), Stoke’s theorem is verified.

Example 62: Verify Stoke’s theorem for the function 𝒇 = 𝒙𝟐 + 𝒚𝟐 𝒊 − 𝟐𝒙𝒚 𝒋 taken round the

rectangle bounded by 𝒙 = ±𝒂, 𝒚 = 𝟎, 𝒚 = 𝒃. [KUK 2006]

54

O B A

D C

𝑦 = 0

𝑦 = 𝑏

𝑥 = 𝑎 𝑥 = −𝑎

X

Y

Solution: Given 𝑓 = 𝑥2 + 𝑦2 𝑖 − 2𝑥𝑦 𝑗

Therefore, 𝑓 . 𝑑𝑟 = 𝑥2 + 𝑦2 𝑖 − 2𝑥𝑦 𝑗 . 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 = 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦

Fig. 17.17

∴ 𝑓 . 𝑑𝑟 𝐶

= 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐶

= 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐷𝐴

+ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐴𝐵

+ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐵𝐶

+ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐶𝐷

… (1)

On DA: 𝑥 = −𝑎, ∴ 𝑑𝑥 = 0

∴ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐷𝐴

= −2 −𝑎 𝑦 𝑑𝑦𝐷𝐴

= 2𝑎𝑦 𝑑𝑦0

𝑦=𝑏= 𝑎𝑦2 𝑏

0 = −𝑎𝑏2

On AB: 𝑦 = 0, ∴ 𝑑𝑦 = 0

∴ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐴𝐵

= 𝑥2 𝑑𝑥𝐴𝐵

= 𝑥2𝑎

−𝑎𝑑𝑥 =

2

3𝑎3

On BC: 𝑥 = 𝑎, ∴ 𝑑𝑥 = 0

∴ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐵𝐶

= −2𝑎𝑦 𝑑𝑦𝐵𝐶

= −2𝑎𝑦 𝑑𝑦𝑏

𝑦=0= −𝑎𝑏2

On CD: 𝑦 = 𝑏, ∴ 𝑑𝑦 = 0

∴ 𝑥2 + 𝑦2 𝑑𝑥 − 2𝑥𝑦 𝑑𝑦𝐶𝐷

= 𝑥2 + 𝑏2 𝑑𝑥𝐶𝐷

= 𝑥2 + 𝑏2 𝑑𝑥−𝑎

𝑥=𝑎

= 𝑥3

3+ 𝑏2𝑥

𝑥=𝑎

−𝑎

= −2 𝑎3

3+ 𝑏2𝑎

Substituting these values in (1), we get

𝑓 . 𝑑𝑟 𝐶

= −𝑎𝑏2 +2

3𝑎3 − 𝑎𝑏2 − 2

𝑎3

3+ 𝑏2𝑎 = −4𝑎𝑏2 … (2)

Now, 𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑥2 + 𝑦2 −2𝑥𝑦 0

= 0 𝑖 + 0 𝑗 + −2𝑦 − 2𝑦 𝑘 = −4𝑦 𝑘

Since Surface lies in xy-plane, therefore 𝑛 = 𝑘 .

55

X

Y

B (1,1,0)

O (0,0,0) X

∴ 𝑐𝑢𝑟𝑙 𝑓 𝑆

. 𝑛 𝑑𝑆 = −4𝑦 𝑘 . 𝑘 𝑆

𝑑𝑆 = −4𝑦𝑎

𝑥=−𝑎𝑑𝑦

𝑏

𝑦=0𝑑𝑥 = −4𝑎𝑏2 … (3)

Hence from (2) and (3), theorem is verified.

Example 63: Evaluate by Stoke’s theorem 𝒚𝒛 𝒅𝒙 + 𝒙𝒛 𝒅𝒚 + 𝒙𝒚𝒅𝒛 𝑪

, where 𝑪 is the curve

𝒙𝟐 + 𝒚𝟐 = 𝟏, 𝒛 = 𝒚𝟐.

Solution: 𝑦𝑧 𝑑𝑥 + 𝑥𝑧 𝑑𝑦 + 𝑥𝑦𝑑𝑧 𝐶

= 𝑦𝑧 𝑖 + 𝑥𝑧 𝑗 + 𝑥𝑦 𝑘 𝐶

. 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘

= 𝑓 𝐶

. 𝑑𝑟 , where 𝑓 = 𝑦𝑧 𝑖 + 𝑥𝑧 𝑗 + 𝑥𝑦 𝑘

Now, 𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑦𝑧 𝑧𝑥 𝑥𝑦

= 𝑥 − 𝑥 𝑖 + (𝑦 − 𝑦) 𝑗 + 𝑧 − 𝑧 𝑘 = 0

∴ 𝑓 𝐶

. 𝑑𝑟 = 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 𝑆

𝑑𝑆 = 0 ∵ 𝑐𝑢𝑟𝑙 𝑓 = 0

Example 64: Evaluate 𝒇 𝑪

. 𝒅𝒓 by Stoke’s theorem, where 𝒇 = 𝒚𝟐 𝒊 + 𝒙𝟐 𝒋 − 𝒙 + 𝒛 𝒌 and 𝑪 is

the boundary of triangle with vertices at (𝟎, 𝟎, 𝟎), (𝟏, 𝟎, 𝟎), (𝟏, 𝟏, 𝟎).

Solution: Here,

𝑐𝑢𝑟𝑙 𝑓 =

𝑖 𝑗 𝑘

𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

𝑦2 𝑥2 −(𝑥 + 𝑧)

= 0 𝑖 + 𝑗 + 2 𝑥 − 𝑦 𝑘

Fig. 17.18

Here triangle is in the xy plane as z co-ordinate of each vertex of the triangle is zero.

∴ 𝑛 = 𝑘

∴ 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 = 0 𝑖 + 𝑗 + 2 𝑥 − 𝑦 𝑘 . 𝑘 = 2 𝑥 − 𝑦

By Stoke’s theorem, 𝑓 𝐶

. 𝑑𝑟 = 𝑐𝑢𝑟𝑙 𝑓 . 𝑛 𝑆

𝑑𝑆

= 2 𝑥 − 𝑦 𝑆

𝑑𝑦 𝑑𝑥

Note here the equation of OB is 𝑦 = 𝑥, thus for 𝑆, x varies from 0 to 1 and 𝑦 from 0 to 𝑥.

A (1,0,0)

56

∴ 𝑓 𝐶

. 𝑑𝑟 = 2(𝑥 − 𝑦)𝑥

𝑦=0𝑑𝑦

1

𝑥=0𝑑𝑥 = 2 𝑥𝑦 −

𝑦2

2 𝑦=0

𝑥

𝑑𝑥1

𝑥=0

= 2 𝑥2 −𝑥2

2 𝑑𝑥

1

𝑥=0= 𝑥2𝑑𝑥

1

𝑥=0=

1

3

Note: Green’s Theorem in plane is special case of Stoke’s Theorem: If R is the region in xy plane

bounded by a closed curve 𝐶 then this is a special case of Stoke’s theorem. In this case 𝑛 = 𝑘 and it

is called vector form of Green’s theorem in plane. Vector form of Green’s theorem can be written as

∇ × 𝑓 𝑅

. 𝑘 𝑑𝑅 = 𝑓 𝐶

. 𝑑𝑟

Example 65: Evaluate 𝒚 − 𝐬𝐢𝐧 𝒙 𝒅𝒙 + 𝐜𝐨𝐬 𝒙𝒅𝒚 𝑪

, where 𝑪 is the triangle having vertices

𝟎, 𝟎 , 𝝅

𝟐, 𝟎 and

𝝅

𝟐, 𝟏 (i) directly (ii) by using Green’s theorem in plane [KUK 2011]

Solution:

(i) Here 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐶

= 𝑦 − sin 𝑥 𝑖 + cos 𝑥 𝑗 . (𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 𝐶

)

= 𝑓 𝐶

. 𝑑𝑟

where 𝑓 = 𝑦 − sin 𝑥 𝑖 + cos 𝑥 𝑗 and 𝑑𝑟 = 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 and

C is triangle OAB

Now 𝑓 𝐶

. 𝑑𝑟 = 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐶

= 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝑂𝐴

+ 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐴𝐵

+ 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐵𝑂

… (1)

On OA: 𝑦 = 0, ∴ 𝑑𝑦 = 0

∴ 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝑂𝐴

= − sin 𝑥 𝑑𝑥 𝑂𝐴

= − sin 𝑥 𝑑𝑥𝜋 2

𝑥=0= −1

On AB: 𝑥 =𝜋

2, ∴ 𝑑𝑥 = 0

∴ 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐴𝐵

= 0

On BO: 𝑦 =2

𝜋𝑥, ∴ 𝑑𝑦 =

2

𝜋𝑑𝑥

∴ 𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝐵𝑂

= 2

𝜋𝑥 − sin 𝑥 𝑑𝑥 + cos 𝑥

2

𝜋𝑑𝑥

0

𝑥=𝜋 2

= 2

𝜋

𝑥2

2+ cos 𝑥 +

2

𝜋sin 𝑥

𝑥=𝜋 2

0

= 1 − 𝜋

4+

2

𝜋

57

Substituting these values in (1), we get

𝑓 𝐶

. 𝑑𝑟 = −1 + 0 + 1 −𝜋

4−

2

𝜋= −

𝜋

4+

2

𝜋

(ii) By Green’s Theorem

𝑓1𝑑𝑥 + 𝑓2𝑑𝑦 𝐶

= 𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦

𝑆𝑑𝑥𝑑𝑦

= − sin 𝑥 − 1 𝑑𝑦2𝑥 𝜋

𝑦=0𝑑𝑥

𝜋 2

𝑥=0= − sin 𝑥 − 1 𝑦 𝑦=0

2𝑥 𝜋 𝑑𝑥

𝜋 2

𝑥=0

= −2

𝜋𝑥 sin 𝑥 + 1 𝑑𝑥

𝜋 2

𝑥=0= −

2

𝜋 𝑥 sin 𝑥 + 𝑥 𝑑𝑥

𝜋 2

𝑥=0

= −2

𝜋 𝑥 − cos 𝑥 + sin 𝑥 +

𝑥2

2 𝑥=0

𝜋 2

= −2

𝜋 −0 + 1 +

𝜋2

8 − (0 + 0 + 0)

= − 2

𝜋+

𝜋

4

Example 66: Verify Green’s theorem in the plane for 𝒙𝒚 + 𝒚𝟐 𝒅𝒙 + 𝒙𝟐𝒅𝒚𝑪

, where 𝑪 is the

closed curve of the region bounded by 𝒚 = 𝒙 and 𝒚 = 𝒙𝟐.

Solution:

𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦𝐶

= 𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦 𝑂𝐵𝐴

+ 𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦 𝑂𝐴

… (1)

Along curve OBA: 𝑦 = 𝑥2 ∴ 𝑑𝑦 = 2𝑥 𝑑𝑥

∴ 𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦 𝑂𝐵𝐴

= 𝑥3 + 𝑥4 𝑑𝑥 + 2𝑥3𝑑𝑥 1

𝑥=0=

19

20

Along curve AO: 𝑦 = 𝑥 ∴ 𝑑𝑦 = 𝑑𝑥

∴ 𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦 𝐴𝑂

= 𝑥2 + 𝑥2 𝑑𝑥 + 𝑥2𝑑𝑥 𝐴𝑂

= 3𝑥2 𝑑𝑥0

𝑥=1= −1

∴ from (1), 𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦𝐶

=19

20− 1 = −

1

20 … (2)

Here 𝑓1 = 𝑥𝑦 + 𝑦2, 𝑓2 = 𝑥2

⇒ 𝜕𝑓1

𝜕𝑦= 𝑥 + 2𝑦,

𝜕𝑓2

𝜕𝑥= 2𝑥

By Green’s theorem,

𝑓1 𝑑𝑥 + 𝑓2 𝑑𝑦 𝐶

= 𝜕𝑓2

𝜕𝑥−

𝜕𝑓1

𝜕𝑦

𝑆𝑑𝑦 𝑑𝑥

= 2𝑥 − 𝑥 − 2𝑦 𝑆

𝑑𝑦 𝑑𝑥

= 𝑥 − 2𝑦 𝑑𝑦𝑦=𝑥

𝑦=𝑥2 𝑑𝑥1

𝑥=0 = 𝑥𝑦 − 𝑦2

𝑦=𝑥2𝑦=𝑥

𝑑𝑥1

𝑥=0

= 𝑥4 − 𝑥3 𝑑𝑥1

𝑥=0=

x5

5−

x4

4

x=0

x=1

=1

5−

1

4= −

1

20 … (3)

Equation (2) and (3) verify the result.

58

ASSIGNMENT 9

1. Verify Green’s theorem for 3𝑥 − 8𝑦2 𝑑𝑥 + 4𝑦 − 6𝑥𝑦 𝑑𝑦 𝐶

where C is the boundary of the

region bounded by 0,0 yx and 1 yx . [KUK 2007]

2. Verify Green’s theorem in plane for 3𝑥2 − 8𝑦2 𝑑𝑥 + 4𝑦 − 6𝑥𝑦 𝑑𝑦 𝐶

, where C is the

boundary of the region defined by xy and 2xy . [KUK 2008]

3. Apply Green’s theorem to evaluate 2𝑥2 − 𝑦2 𝑑𝑥 + 𝑥2 + 𝑦2 𝑑𝑦 𝐶

, where C is the boundary

of the area enclosed by x-axis and the upper half of the circle 122 yx . [KUK 2010]

4. Evaluate the surface integral SdSnFcurl ˆ.

by transforming it into a line integral, S being that

part of the surface of the paraboloid 𝑧 = (1 − 𝑥2 − 𝑦2) for which 0z and kxjziyF ˆˆˆ

.

[KUK 2008]

5. Using Stoke’s theorem, evaluate 𝑥 + 𝑦 𝑑𝑥 + 2𝑥 − 𝑧 𝑑𝑦 + 𝑦 + 𝑧 𝑑𝑧 𝐶

, where C is the

boundary of the triangle with vertices )0,0,2( , )0,3,0( and )6,0,0( . [KUK 2009]

6. Verify Stoke’s theorem for a vector field defined by 𝑓 = −𝑦3 𝑖 + 𝑥3 𝑗 , in the region

𝑥2 + 𝑦2 ≤ 1, 𝑧 = 0.

17.14 GAUSS’S DIVERGENCE THEOREM (Relation between Volume and Surface Integral)

Statement: Suppose 𝑉 is the volume bounded by a closed piecewise smooth surface S. Suppose

𝑓 (𝑥, 𝑦, 𝑧) is a vector function which is continuous and has continuous first partial derivatives in

𝑉.Then

∇. 𝑓 𝑉

𝑑𝑉 = 𝑓 . 𝑛 𝑑𝑆

where 𝑛 is the outward unit normal to the surface 𝑆.

In other words: The surface integral of the normal component of a vector 𝑓 taken over a closed

surface is equal to the integral of the divergence of 𝑓 over the volume enclosed by the surface.

Divergence Theorem in Cartesian Coordinates

If 𝑓 = 𝑓1𝑖 + 𝑓2𝑗 + 𝑓3𝑘 then 𝑑𝑖𝑣 𝑓 = ∇. 𝑓 =𝜕𝑓1

𝜕𝑥+

𝜕𝑓2

𝜕𝑦+

𝜕𝑓3

𝜕𝑧 . Suppose 𝛼, 𝛽, 𝛾 are the angles made by

the outward drawn unit normal with the positive direction of axes, then

𝑛 = cos 𝛼 𝑖 + cos 𝛽 𝑗 + cos 𝛾 𝑘

Now, 𝑓 . 𝑛 = 𝑓1 cos 𝛼 + 𝑓2 cos 𝛽 + 𝑓3 cos 𝛾

Then divergence theorem is

𝜕𝑓1

𝜕𝑥+

𝜕𝑓2

𝜕𝑦+

𝜕𝑓3

𝜕𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧

𝑉= 𝑓1 cos 𝛼 + 𝑓2 cos 𝛽 + 𝑓3 cos 𝛾 𝑑𝑆

𝑆

= 𝑓1𝑑𝑦𝑑𝑧 + 𝑓2 𝑑𝑧𝑑𝑥 + 𝑓3𝑑𝑥𝑑𝑦 𝑆

∵ cos 𝛼 𝑑𝑆 = 𝑑𝑦𝑑𝑧 𝑒𝑡𝑐.

59

Proof: Let 𝑓 = 𝑓1𝑖 + 𝑓2𝑗 + 𝑓3𝑘 where 𝑓1, 𝑓2 and 𝑓3 and their derivatives in any direction are finite

and continuouos.

Suppose 𝑆 is a closed surface such that it is possible to

choose rectangular cartesian co-ordinate system such that

any line drawn parallel to coordinate axes does not cut 𝑆

in more than two points.

Let 𝑅 be the orthogonal projection of 𝑆 on the xy-plane.

Any line parallel to z-axis through a point of 𝑅 meets the

boundary of S in two points. Let 𝑆1 and 𝑆2 be the lower

and upper portions of 𝑆 . Let the equations of these

portions be

𝑧 = 𝛷1(𝑥, 𝑦) and 𝑧 = 𝛷2(𝑥, 𝑦)

where 𝛷1(𝑥, 𝑦) ≥ 𝛷2(𝑥, 𝑦)

Consider the volume integral

𝜕𝑓3

𝜕𝑧 𝑑𝑉 =

𝑉

𝜕𝑓3

𝜕𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧 =

𝜕𝑓3

𝜕𝑧𝑑𝑧

𝑧=𝛷1(𝑥 ,𝑦)

𝑧=𝛷2(𝑥 ,𝑦) 𝑑𝑥 𝑑𝑦

𝜕𝑓3

𝜕𝑧 𝑑𝑉 =

𝑉 𝑓3(𝑥, 𝑦, 𝑧) 𝛷2(𝑥 ,𝑦)

𝛷1(𝑥 ,𝑦)𝑑𝑥 𝑑𝑦 = 𝑓3 𝑥, 𝑦, 𝛷1 − 𝑓3 𝑥, 𝑦, 𝛷2 𝑑𝑥 𝑑𝑦 …(1)

Let 𝑛 1 be the unit outward drawn vector making an acute angle 𝛾1 with 𝑘 for the upper position 𝑆1 as

shown in the figure.

Now projection 𝑑𝑥 𝑑𝑦 of 𝑑𝑆1 on the 𝑥𝑦 plane is given as 𝑑𝑥𝑑𝑦 = 𝑑𝑆1 cos 𝛾 = 𝑑𝑆1 𝑘 . 𝑛 1 = 𝑘 . 𝑛 1𝑑𝑆1

Now 𝑓3 𝑥, 𝑦, 𝛷1 𝑅𝑑𝑥𝑑𝑦 = 𝑓3 𝑘 . 𝑛 1𝑑𝑆1𝑆1

… (2)

Similarly if 𝑛 2 be the unit outward drawn normal to the lower surface 𝑆2 making an angle 𝛾2 with 𝑘 .

Obviously 𝛾2 is an obtuse angle

∴ 𝑑𝑥𝑑𝑦 = 𝑑𝑆2 cos 𝜋 − 𝛾2 = −𝑑𝑆2 cos 𝛾2 = −𝑘 . 𝑛 2 𝑑𝑆2

∴ 𝑓3 𝑥, 𝑦, 𝛷2 𝑅𝑑𝑥 𝑑𝑦 = − 𝑓3 𝑘 . 𝑛 2𝑑𝑆2𝑆2

… (3)

From (1), (2) and (3), we have

𝜕𝑓3

𝜕𝑧 𝑑𝑉

𝑉= 𝑓3 𝑘 . 𝑛 1𝑑𝑆1𝑆1

+ 𝑓3 𝑘 . 𝑛 2𝑑𝑆2𝑆2= 𝑓3 𝑘 . 𝑛 𝑑𝑆

𝑆 … (4)

Similarly by projecting 𝑆 on the other coordinate planes

𝜕𝑓2

𝜕𝑦 𝑑𝑉

𝑉= 𝑓2 𝑗 . 𝑛 𝑑𝑆

𝑆 … (5)

And 𝜕𝑓1

𝜕𝑥 𝑑𝑉

𝑉= 𝑓1 𝑖 . 𝑛 𝑑𝑆

𝑆 … (6)

60

Adding (4), (5) and (6), we get

𝜕𝑓1

𝜕𝑥+

𝜕𝑓2

𝜕𝑦+

𝜕𝑓3

𝜕𝑧 𝑑𝑉

𝑉= 𝑓1 𝑖 + 𝑓2 𝑗 + 𝑓3 𝑘 . 𝑛 𝑑𝑆

𝑆 = 𝑓 . 𝑛 𝑑𝑆

𝑆

Note: With the help of this theorem we can express volume integral as surface integral or vice versa.

17.15 GREEN’S THEOREM (For Harmonic Functions)

Statement: If 𝛷 and 𝜓 are two scalar point functions having continuous second order derivatives in a

region 𝑉 bounded by a closed surface 𝑆, then

𝛷∇2𝜓 − 𝜓∇2𝛷 𝑉

𝑑𝑉 = 𝛷 ∇𝜓 − 𝜓 ∇𝛷 . 𝑛 𝑑𝑆𝑆

Proof: By Gauss’s divergence theorem,

∇. 𝑓 𝑑𝑉𝑉

= 𝑓 . 𝑛 𝑆

𝑑𝑆 … (1)

Take 𝑓 = 𝛷 ∇𝜓 so that ∇. 𝑓 = ∇. 𝛷 ∇𝜓

= 𝛷 ∇. ∇𝜓 + ∇𝛷. ∇𝜓

= 𝛷∇2𝜓 − ∇𝛷. ∇𝜓 … (2)

Now 𝑓 . 𝑛 = 𝛷 ∇𝜓 . 𝑛 . Using this and (2) in (1), we get

𝛷 ∇2𝜓 − ∇𝛷. ∇𝜓 𝑑𝑉𝑉

= 𝛷 ∇𝜓 . 𝑛 𝑆

𝑑𝑆 … (3)

Again starting as above by interchanging 𝛷 and 𝜓, we obtain as in (3)

𝜓 ∇2𝛷 − ∇𝜓. ∇𝛷 𝑑𝑉𝑉

= 𝜓 ∇𝛷 . 𝑛 𝑆

𝑑𝑆 … (4)

Subtracting (4) from (3), we get

𝛷∇2𝜓 − 𝜓∇2𝛷 𝑉

𝑑𝑉 = 𝛷 ∇𝜓 − 𝜓 ∇𝛷 . 𝑛 𝑑𝑆𝑆

… (5)

Another Form of Green’s Theorem:

𝜕𝛷

𝜕𝑛 and

𝜕𝜓

𝜕𝑛 denote the direction of derivative of 𝛷 and 𝜓 along the outward drawn normal at any point

of 𝑆.

∇𝛷 =𝜕𝛷

𝜕𝑛𝑛 and ∇𝜓 =

𝜕𝜓

𝜕𝑛𝑛

∴ 𝛷 ∇𝜓 − 𝜓 ∇𝛷 = 𝛷𝜕𝜓

𝜕𝑛 𝑛 − 𝜓

𝜕𝛷

𝜕𝑛 𝑛

or 𝛷 ∇𝜓 − 𝜓 ∇𝛷 . 𝑛 = 𝛷𝜕𝜓

𝜕𝑛𝑛 − 𝜓

𝜕𝛷

𝜕𝑛𝑛 . 𝑛 = 𝛷

𝜕𝜓

𝜕𝑛− 𝜓

𝜕𝛷

𝜕𝑛

∴ Equation (5) becomes

𝛷∇2𝜓 − 𝜓∇2𝛷 𝑉

𝑑𝑉 = 𝛷𝜕𝜓

𝜕𝑛− 𝜓

𝜕𝛷

𝜕𝑛

𝑆𝑑𝑆

61

Example 67: Evaluate 𝒇 . 𝒏 𝒅𝑺𝑺

where 𝒇 = 𝟒𝒙𝒚 𝒊 + 𝒚𝒛 𝒋 − 𝒛𝒙 𝒌 and 𝑺 is the surface of the

cube bounded by the planes 𝒙 = 𝟎, 𝒙 = 𝟐, 𝒚 = 𝟎, 𝒚 = 𝟐, 𝒛 = 𝟎, 𝒛 = 𝟐.

Solution: Here 𝑓 = 4𝑥𝑦 𝑖 + 𝑦𝑧 𝑗 − 𝑧𝑥 𝑘 . By Gauss’s divergence theorem

𝑓 . 𝑛 𝑑𝑆𝑆

= ∇. 𝑓 𝑑𝑉𝑉

= 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧

𝑉. 4𝑥𝑦 𝑖 + 𝑦𝑧 𝑗 − 𝑧𝑥 𝑘 𝑑𝑉

= 4𝑦 + 𝑧 − 𝑥 𝑑𝑉𝑉

= 4𝑦 + 𝑧 − 𝑥 𝑑𝑧 𝑑𝑦 𝑑𝑥2

𝑧=0

2

𝑦=0

2

𝑥=0

= 4𝑦𝑧 +𝑧2

2− 𝑥𝑧

𝑧=0

𝑧=22

𝑦=0

2

𝑥=0 𝑑𝑦 𝑑𝑥

= 8𝑦 + 2 − 2𝑥 2

𝑦=0𝑑𝑦

2

𝑥=0𝑑𝑥

= 4𝑦2 + 2𝑦 − 2𝑥𝑦 𝑦=0𝑦=2

𝑑𝑥2

𝑥=0= 20 − 4𝑥 𝑑𝑥

2

𝑥=0

= 20𝑥 − 2𝑥2 𝑥=0𝑥=2 = 32

Example 68: Use Gauss theorem to show that 𝒙𝟑 − 𝒚𝒛 𝒊 − 𝟐𝒙𝟐𝒚 𝒋 + 𝟐𝒌 . 𝒏 𝒅𝑺𝑺

=𝒂𝟓

𝟑

where S denotes the surface of the cube bounded by the planes, 𝒙 = 𝟎, 𝒙 = 𝒂, 𝒚 = 𝟎, 𝒚 = 𝒂,

𝒛 = 𝟎, 𝒛 = 𝒂 .

Solution: By Gauss’s divergence theorem

𝑥3 − 𝑦𝑧 𝑖 − 2𝑥2𝑦 𝑗 + 2𝑘 . 𝑛 𝑑𝑆𝑆

= ∇.𝑉

𝑥3 − 𝑦𝑧 𝑖 − 2𝑥2𝑦 𝑗 + 2𝑘 𝑑𝑉

= 3𝑥2 − 2𝑥2 𝑑𝑥𝑎

0𝑑𝑦

𝑎

0𝑑𝑧

𝑎

0

= 𝑥2𝑑𝑥𝑎

0 𝑑𝑦

𝑎

0𝑑𝑧

𝑎

0=

𝑥3

3

0

𝑎

𝑑𝑦𝑎

0𝑑𝑧

𝑎

0

= 𝑎3

3 𝑑𝑦

𝑎

0𝑑𝑧

𝑎

0=

𝑎3

3𝑦

0

𝑎

𝑑𝑧𝑎

0=

𝑎4

3𝑑𝑧

𝑎

0=

𝑎4

3 𝑧 0

𝑎 =𝑎5

3

Example 69: Evaluate 𝒇 . 𝒏 𝑺

𝒅𝑺 with the help of Gauss theorem for 𝒇 = 𝟔𝒛 𝒊 + 𝟐𝒙 + 𝒚 𝒋 −

𝒙 𝒌 taken over the region 𝑺 bounded by the surface of the cylinder 𝒙𝟐 + 𝒛𝟐 = 𝟗 included

between 𝒙 = 𝟎, 𝒚 = 𝟎, 𝒛 = 𝟎 and 𝒚 = 𝟖.

Solution: 𝑓 = 6𝑧 𝑖 + 2𝑥 + 𝑦 𝑗 − 𝑥 𝑘

∇. 𝑓 = 𝑖 𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧 . 6𝑧 𝑖 + 2𝑥 + 𝑦 𝑗 − 𝑥 𝑘 = 1

By Gauss’s divergence theorem,

𝑓 . 𝑛 𝑆

𝑑𝑆 = 1 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑉

62

= 𝑑𝑧 9−𝑥2

𝑧=0𝑑𝑦

8

𝑦=0𝑑𝑥

3

𝑥=0= 𝑧 𝑧=0

𝑧= 9−𝑥2𝑑𝑦

8

𝑦=0𝑑𝑥

3

𝑥=0

= 9 − 𝑥2 𝑑𝑦8

𝑦=0𝑑𝑥

3

𝑥=0

= 9 − 𝑥2 𝑦 𝑦=0

8𝑑𝑥

3

𝑥=0= 8 9 − 𝑥2 𝑑𝑥

3

𝑥=0

= 8 𝑥 9−𝑥2

2+

9

2sin−1 𝑥

3 𝑥=0

3

= 18 𝜋

Example 70: Evaluate (𝒙𝒅𝒚𝒅𝒛 + 𝒚𝒅𝒛𝒅𝒙 + 𝒛𝒅𝒙𝒅𝒚)𝑺

where 𝑺 is the surface of the sphere

𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 = 𝒂𝟐. [KUK 2011, 2009]

Solution: By Gauss’s divergence theorem

(𝑥𝑑𝑦𝑑𝑧 + 𝑦𝑑𝑧𝑑𝑥 + 𝑧𝑑𝑥𝑑𝑦)𝑆

= 𝜕𝑥

𝜕𝑥+

𝜕𝑦

𝜕𝑦+

𝜕𝑧

𝜕𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧 =

𝑉 3 𝑑𝑥 𝑑𝑦 𝑑𝑧

𝑉

= 3 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑉

= 3 × volume of the sphere x2 + y2 + z2 = a2

= 3 ×4

3πa3 = 4πa3

Example 71: Show that 𝒏 𝑺

𝒅𝑺 = 𝟎 for any closed surface 𝑺.

Solution: Let 𝐶 be any closed vector.

∴ 𝐶 𝑛 𝑆

𝑑𝑆 = 𝐶.𝑆

𝑛 𝑑𝑆 = 𝑑𝑖𝑣 𝐶𝑉

𝑑𝑉

∴ 𝐶 𝑛 𝑆

𝑑𝑆 = 0 ∵ 𝐶 is constant, therfore 𝑑𝑖𝑣 𝐶 = 0

⇒ 𝑛 𝑆

𝑑𝑆 = 0

Example 72: Prove that 𝟏

𝒓𝟐𝒅𝑽

𝑽=

𝒓 .𝒏

𝒓𝟐𝒅𝑺

𝑺, where 𝒓 = 𝒙 𝒊 + 𝒚 𝒋 + 𝒛 𝒌 and 𝒓 = 𝒓.

Solution: 𝑟 .𝑛

𝑟2 𝑑𝑆𝑆

= 𝑟

𝑟2 . 𝑛 𝑑𝑆𝑆

= ∇.𝑟

𝑟2 𝑉𝑑𝑉 … (1)

Now, ∇. 𝑟

𝑟2 =1

𝑟2 ∇. 𝑟 + 𝑟 . ∇

1

𝑟2 … (2)

Also, 𝑟 = 𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 ⇒ 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2

∴ 2𝑟𝜕𝑟

𝜕𝑥= 2𝑥; 2𝑟

𝜕𝑟

𝜕𝑦= 2𝑦; 2𝑟

𝜕𝑟

𝜕𝑧= 2𝑧

⇒ 𝜕𝑟

𝜕𝑥=

𝑥

𝑟;

𝜕𝑟

𝜕𝑦=

𝑦

𝑟;

𝜕𝑟

𝜕𝑧=

𝑧

𝑟 … (3)

Now, ∇ 1

𝑟2 = 𝑖

𝜕

𝜕𝑥+ 𝑗

𝜕

𝜕𝑦+ 𝑘

𝜕

𝜕𝑧

1

𝑟2 =

−2

𝑟3 𝑖

𝜕𝑟

𝜕𝑥+ 𝑗

𝜕𝑟

𝜕𝑦+ 𝑘

𝜕𝑟

𝜕𝑧 using (3)

=−2

𝑟4 𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 =

−2

𝑟4𝑟

63

Also, ∇. 𝑟 = 1 + 1 + 1 = 3

Substituting these values in (2), we have

∇. 𝑟

𝑟2 =1

𝑟2 . 3 + 𝑟 . −2

𝑟4 𝑟 =3

𝑟2 −2

𝑟4 𝑟2 =1

𝑟2 ∵ 𝑟 . 𝑟 = 𝑟2

∴ From 1 , 𝑟 .𝑛

𝑟2𝑑𝑆

𝑆=

1

𝑟2𝑑𝑉

𝑉

ASSIGNMENT 10

1. Find SdSnF ˆ.

where kzyjyxzizxF ˆ)2(ˆ)(ˆ)2( 2

and S is the surface of the sphere

having centre )2,1,3( and radius 3 units. [KUK 2006]

2. Use Divergence theorem to evaluate SSdF

. where kzjyixF ˆˆˆ 333

and S is the outer

surface of the sphere 1222 zyx . [KUK 2007]

3. Verify Divergence theorem for kxyzjzxyiyzxF ˆ)(ˆ)(ˆ)( 222

taken over rectangular

parallelopiped ,0 ax ,0 by .0 cz [KUK 2010]

ANSWERS

ASSIGNMENT 1

1. −4(𝑖 + 2𝑗 ) 3. 𝑥 − 𝑎/ 2 = 𝑦 − 𝑎/ 2 = 𝑧 −𝑎𝜋

4tan 𝛼 / 2 𝑡𝑎𝑛 𝛼

4. 𝑡 𝑖 + 2𝑗 − 2𝑡 − 3 𝑘 / (5𝑡2 − 12𝑡 + 13) ; 1

3(2 𝑖 + 2𝑗 + 𝑘 )

5. (a) 𝑢 𝑎2 sec 𝛼 (b) 𝑎3 tan 𝛼; (− cos 𝛼 sin 𝑡 𝑖 + cos 𝛼 cos 𝑡 𝑗 + sin 𝛼 𝑘 )

6. (a) 𝑝 𝑖 + 𝑝 + 2𝑞 𝑗 + 𝑝 + 𝑞 𝑘 ; 1

6 −𝑖 − 𝑗 + 2 𝑘 (b) 𝑝 + 𝑞 𝑖 + 𝑞 𝑗 + 2𝑞 𝑘 ;

1

5 2𝑗 − 𝑘

17. (a) 𝑎𝑏/ 𝑎2 sin2 𝑡 + 𝑏2 cos2 𝑡 3/2 (b) 1/4 2

ASSIGNMENT 2

1. 𝑣 = 37 and 𝑎 = 325 at 𝑡 = 0 3. 8

7 14 ;

1

7 14 4. 𝑎 = ±

1

6 7.

70

29;

436

29

8. 21.29 knots/hr. in the direction 740 47’ South of East 9. 17 meter per hour in the direction

tan−1 0.25 North of East

ASSIGNMENT 3

1. (a) 2(𝑥 𝑖 + 𝑦 𝑗 + 𝑧 𝑘 )/ (𝑥2 + 𝑦2 + 𝑧2) 3. 15

17 4.

37

3 5.

1

3(2 𝑖 + 2 𝑗 − 𝑘 ) 6. 11

7. cos−1 −1

30 8. cos−1 1/ 22 9. cos−1

8

3 21 10. 𝜆 = 4 𝑎𝑛𝑑 𝜇 = 1

64

ASSIGNMENT 4

2. (a) 80 (b) 𝑒𝑥𝑦𝑧 𝑥 𝑧 − 𝑦 𝑖 + 𝑦 𝑥 − 𝑧 𝑗 + 𝑧 𝑦 − 𝑥 𝑘

3. 𝑎 = −2; 4𝑥 𝑧 − 𝑥𝑦 𝑖 + 𝑦 1 − 2𝑧 + 4𝑥𝑦 𝑗 + (2𝑥2 + 𝑦2 − 𝑧2 − 𝑧)𝑘

9. (a) 2𝑛 2𝑛−1

𝑥2+𝑦2+𝑧2 𝑛+1; 𝑛 =

1

2

11. (i) 2 𝑦3 + 3𝑥2𝑦 − 6𝑥𝑦2 𝑧 𝑖 + 2 3𝑥𝑦2 + 𝑥3 − 6𝑥2𝑦 𝑧 𝑗 + 2 𝑥𝑦2 + 𝑥3 − 3𝑥2𝑦 𝑦 𝑘 (ii) Zero

13. (i) 0 (ii) 2 𝑥 + 𝑧 𝑗 + 2𝑦 𝑘

ASSIGNMENT 5

1. 226

3 𝑖 + 360 𝑗 − 42 𝑘 2. −2 𝑖 + 3 𝑗 − 3 𝑘

4. 𝑣 = 6 sin 2𝑡 𝑖 + 4(cos 2𝑡 − 1) 𝑗 + 8𝑡2 𝑘 and 𝑟 = 3 1 − cos 2𝑡 𝑖 + 2 sin 2𝑡 𝑗 +8𝑡3

3 𝑘

ASSIGNMENT 6

1. 88

35 2. 16, 16 4. 35 5.

𝜋3 2

3

6. 2 −𝜋

4 𝑖 − 𝜋 −

1

2 𝑗

ASSIGNMENT 7

2. 81 3. 3/2 5. 8

ASSIGNMENT 8

1. 128 2. 128𝑖 − 24𝑗 + 384𝑘

ASSIGNMENT 9

3. 4/3 5. 21

ASSIGNMENT 10

1. 108π 2. 56πa2/9