1

Anti derivatives DEF – 1: Let f(x) be a given function. If there is a differentiable function F(x) such that

d F x = f xdx

then F(x) is called an “Anti derivative” or “Primitive” of f(x).

DEF – 2: If F(x) is an anti derivative of f(x), then the expression F(x) + C is called indefinite integral of f(x) and is denoted by

f x dx = F x + C ,

Where f(x) is called “Integrand” and F(x) is called an integral. “x” is the variable of integration and “C” is an arbitrary constant of integration but usually omitted in practice.

NOTE: The process of finding anti derivative is called an “Integration”.

Theorem: If F(x) and G(x) are two anti derivatives of f(x), then F(x) – G(x) = A constant.

Proof: By definition of anti derivative, we have

F x = f xandG x = f x

Consider F(x) – G(x) = H(x), then

d F x G x H xdx

F x G x H x

f x f x H x

H x 0

Showing that H(x) is constant.

Hence, F(x) – G(x) = A constant

RULE OF INTEGRATION:

i. n + 1n f x

f x f x dx = , n 1n + 1

2

ii. f x

dx = ln f xf x

SOME STANDARD INTEGRALS:

i. 1 1

2 2

dx x xsin or cosa aa x

ii. 1 12 2

dx 1 x xtan or cota + x a a a

iii. 1 1

2 2

dx 1 x 1 xsec or csca a a ax x a

iv. 2 2

1

2 2

x + a x dx xsinh or lna aa + x

v. 2 2

1

2 2

x + x a dx xcos h or lna ax a

vi. 2 2 2

2 2 1x a x a xa x dx sin2 2 a

vii. 2 2 2 22

2 2 x x a x + x a ax a dx ln2 2 a

viii. 2 2 2 22

2 2 x x a x + x a ax a dx ln2 2 a

FOUR STANDARD INTEGRALS:

i. cotx dx = ln sinx

ii. tanx dx = ln cosx or ln sec x

iii. x πsecx dx = ln secx + tanx or ln tan2 4

iv. xcosecx dx = ln cosecx cot x or ln tan2

INTEGRATION BY SUBSTITUTION:

The evaluation of certain integrals becomes easy if we change the variable of integration by some suitable substitution. Suppose, we have to evaluate f x dx ;

3

We make the substitution x = g(z) to change the variable x into z, then dx = g z dz .

Hence, x dx = g z g z dzf f

The substitution x = g(z) is to be such that the transformed integral on the RHS of the

above equation is easier to evaluate than the given integral.

Q.NO.1 Evaluate the following integrals:

i. 1 x dx1 x [Hint: Put x = cosθ or Rationalizing with Nume.]

ii. 1

2

sin x dx1 x

[Hint: Put 1sin x = z ]

iii.

dxx logx.log logx [Hint: Put log (logx) = z]

iv.

dxx logx.log logx .log log log x [Hint: Put log [log{logx} = z]

v. secx cosecx dxlog tanx [Hint: Put log (tanx) = z]

vi.

x

2 x

e 1 + xdx

cos x e [Hint: Put xx e = z]

vii. 2

t

t te dt

e +3e + 2 [Hint: Put te z ]

viii.

52

2 7

x dxa x [Hint: Put

72x = z ]

ix. dxa sinx + b cosx [Hint: Put a = r cosα, b = r sinα ]

x. secx dxa + b tanx [Hint: Put a = r cosα, b = r sinα ]

xi. dx sinx + cosx [Hint: Use the preceding one]

xii. sinx dx

sin x a

4

xiii.

dxsin x a sin x b

SOL – (xii): Put sin x = sin x a a

sin x =sin x a cos a cos x a sin a

sin x cos a +sin a cot x a

sin x a

Thus,

sinx dx

sin x a = cos a +sin a cot x a dx

x cos a sin a ln sin x a

SOL – (xiii): We have

dxsin x a sin x b

=

sin a b1 dxsin a b sin x a sin x b

sin x b x a1 dxsin a b sin x a sin x b

sin x b cos x a cos x b sin x a1 dx

sin a b sin x a sin x b

1= cot x a cot x b dx

sin a b

1= ln sin x a ln sin x bsin a b

sin x a1= lnsin a b sin x b

5

INTEGRATION BY PARTS:

If u and v be two functions of x, then

duu v dx = u v dx v dx dx dx

Evaluate:

i. 1 xsin dx a + x

ii. x sinx dx

iii. 2nx log x dx

iv. 21sin x dx

v. 1 1

1 1

sin x cos x dxsin x cos x

SOL – (i): Put x = 2a tan z , then 2x = 2 a tanz sec dz z

Therefore, we have

1 2

2

2 2

2 2

2 2

2

1 1

1

tansin 2a tanz sec dsec

2a tanz sec d

tan tan2a d2 2

a tan a tan d

a tan a sec 1 d

a tan a tan z + a z

x x x=x tan a a tana a a

xx + a tana

z z zz

z z z

z zz z

z z z z

z z z z

z z

ax

6

SOL – (v): We have 1 1

1 1

sin x cos x dxsin x cos x

1 1 1

1 1

sin x cos x 2cos x dxsin x cos x

1

1 1

π 2cos x π2 dx ; sin x cos xπ 22

14x cos x dx (A)π

Put 1 2cos x x cos x cos dx sin 2 dzz z z z Therefore,

14 4cos x dx sin 2 dx ; integrate by partsπ π

z z

1 1 24 1cos x dx 2 2x 1 cos x + 2 x xπ π

Hence, (A) becomes

= x + 1 21 2 2x 1 cos x + 2 x xπ

A USEFUL SUBSTITUTION (OR MAGIC SUBSTITUTION):

An integral of a rational function of sin x or cos x can be transformed into an integral of a

rational function of z by the substitution

22

x 1 2 dtan , then dz = sec x dx dx =2 2 1

zzz

The value of sin x and cos x, in terms of z, are

2

2

2 22 2

x x2 tan 1 tan2 12 2sin x = and cos x =x x1 11 tan 1 tan2 2

z zz z

7

Evaluate the integrals:

i. 1 dx1 sin x

ii. 1 sin x dx1 cos x

iii. 2 cos x dx2 cos x

iv. cos x dx2 cos x

v. 1 dxtan x sin x

SOL – (i): Put 22

x 1 2 dtan , then dz = sec x dx dx =2 2 1

zzz

And 22

x2 tan 22sin x = x 11 tan2

zz

So, we have

2

2

1 2 dz2 11

1z zz

22

1 d 2 22 dz 2x1 2 11 1 tan2

zz z zz

SOME SPECIAL INTEGRALS:

TYPE – I: When the integrand involves rational function with irreducible quadratic poly.

Evaluate the following integrals:

i. 2

4 2

x + 1 dxx + x + 1

ii. 4

1 dxx + 1

iii. 4

4 2

x dxx + 2x + 1

8

iv. 2

4 2

x + 1 dxx x + 1

SOL – (i): Dividing both the numerator and denominator by 2x , we get

I = 2

22

11+ x dx (1)1x + 1+

x

Put 2 22 2

1 1 1x , then 1 dx = dz and x 2x x x

z z

Substituting into (1), we get

2

1 12

d 1 1 x 1tan tanz + 3 3 3 3 3

z z

TYPE – II:

FORM: dxlinear linear

Evaluate:

i.

dxx + 2 x + 3

ii.

dx2x + 3 x + 5

iii.

dx1 2x 4x + 1

iv.

4x dxx 1 x + 2

SOL – (i): Put 2 2x + 3 or x + 3= z x + 2 1and dx = 2z dzz z

Therefore,

I = 2 2

2 1 x 3 12 2ln 2ln11 1 x 3 1

z dz dz zzz z z

9

TYPE –III:

FORM: f xdx

Quadratic linear

Evaluate:

i. 2

x dxx 2x + 2 x 1

ii. 2

x dxx 4x + 5 x + 2

SOL – (i): Put 2x 1 or x 1= z 1z

dx = 2 z dz

From (1); 22 2 4x = z 1 x 2x + 2 = x 1 1 1z

Therefore,

I =

2 2 2

4 42

2

11 + 1 + z 2z dz 1 + z2 dz 2 dz (2)

11 1z

z z z zz

Put 2 22 2

1 1 1t then 1 dz = dt and z + t 2 in (2)zz z z

, we get

I = 12dt 2 t2 tan

t + 2 2 2

2

1 12 tan2

zz

1 x 1 12 tan2 x 1

1 x 22 tan2x 2

10

TYPE –IV:

FORM: dxlinear Quadratic

Evaluate:

i. 2 2

dxx x + a

ii. 2

dxx 1 x + 1

iii. 2

dxx + 1 x 1

iv.

2

2

x 2x +3 dxx + 2 x 1

SOL – (i): Put 2

1 1x then dx =z z

Therefore,

I 2

22

1dz

1 1 + a

z

z z

2 2

dz1+ a z

2

2

1 dza 1 +

az

11 sinh 1a a

z

11 asinha x

11

TYPE –V:

FORM: dxPure quadratic Pure quadratic

Evaluate:

i. 2 2

dxx 1 x + 1

ii. 2 2

dxx x + 1

iii. 2 2

dx1 x 1 x

iv. 2 2

dx1 2x 1 x

v. 2 2

dx2x 3x +1 3x 2x +1

SOL – (i): Put 2

1 1x then dx =z z

Therefore,

I 2

2 2

1dz

1 11 + 1

z

z z

2 2

(1)1 1

z dzz z

Now put

2

2 2 2 2

1 t in (1)Then 1 + z t 1 z 2 t2 2t dt

z

z dz

Therefore, (1) becomes

12

I 2

t dtt 2 t

2

dt2 t

2

2

2

2

2

2

1 2 tln2 2 2 t

2 11 ln2 2 2 1

12 11 xln2 2 12 1

x

2 x x 11 ln2 2 2 x x 1

zz

TYPE – VI: When the integrand involve surds of x.

Evaluate:

i. 3

dxx x

ii. 3

x dx1 x

iii. 3

dxx 2 x

SOL – (i): Put 1

6 56x or z x then dx = 6zz dz

So, we have

I = 5 5 3

3 236 6

6 d 6 d 6 d1

z z z zz zz z zz z

2 11 d1

z z zz

13

3 2

1 1 113 6 32

2 3 6 6ln 1

2x 3x 6x 6ln x 1

z z z z

INTEGRATION OF TRIGONOMETRIC FUNCTIONS:

To evaluate the integrals n nsin x dx and cos x dx where “n” is a positive integer.

CASE – I: When n is odd;

Put cos x = z in nsin x dx and sin x = z in ncos x dx

Example: Evaluate: I = 5cos x dx

SOL: Put sin x = z so that cos x dx = dz and

I = 2 24 2 2cos x cosx dx = 1 sin x cos x dx 1 dz z

=

2 4

53

3 5

1 2 d

23 5

2 1sin x sin x sin x3 5

z z z

zz z

CASE – II: When “n” is even, then we find the integrals of n nsin x dx and cos x dx ;

i. Let I = n n 1sin x dx sin x sin x dx , then integrating by parts, we have

I

n 1 n 2

n 1 n 2 2

n 1 n 2 2

n 1 n 2 n

n 1 n 2

n 1

sin x cosx n 1 sin x cosx cosx dx

cosx sin x n 1 sin x cos x dx

cosx sin x n 1 sin x 1 sin x dx

cosx sin x n 1 sin x sin x dx

cosx sin x n 1 sin x dx n 1 I

n 1 1 I = cosx sin x

n 2

n 1n 2

n 1 sin x dx

n 1cosx sin xI = sin x dxn n

14

The above integral is named as reduction formula for sine function.

ii. Let I = n n 1cos x dx cos x cos x dx , then integrating by parts, we have

I

n 1 n 2

n 1 n 2 2

n 1 n 2 2

n 1 n 2 n

n 1 n 2

cos x sinx n 1 cos x sinx sinxdx

sinx cos x n 1 cos x sin x dx

sin x cos x n 1 cos x 1 cos x dx

sin x cos x n 1 cos x cos x dx

sin x cos x n 1 cos x dx n 1 I

n 1 n 2

n 1n n 2

n 1 1 I = sin x cos x n 1 cos x dx

n 1sin x cos xcos x dx = cos x dxn n

The above integral is named as reduction formula for cosine function. Similarly, we can find

out the reduction formulae for the remaining.

Example: Evaluate: 6sin x dx

SOL: Using reduction formula, we have

56 4cosx sin x 5sin x dx = sin x dx (1)

6 6

34 2cosx sin x 3sin x dx sin x dx

4 4

2 0cosx sin x 1sin x dx sin x dx

2 2

cosx sin x x

2 2

Therefore,

34

3

cosx sin x 3 cosx sin x xsin x dx 4 4 2 2

cosx sin x 3 3 cosx sin x x4 8 8

15

Hence, (1) becomes

5 36

53

cosx sin x 5 cosx sin x 3 3sin x dx = cosx sin x x6 6 4 8 8

cosx sin x 5 5 5= cosx sin x cosx sin x x6 24 16 16

16

THE DEFINITE INTEGRAL

Definite Integral as limit of a Sum :

DEF: Let f be a continuous real – valued function defined on a finite closed interval

[a, b]. A partition (or subdivision/ subinterval) P of [a, b] is a finite set of points;

0 1 2 n 1 nP x , x , x ,..., x , x

Such that

0 1 2 n 1 na x x x . . . x x b

It subdivides [a, b] into n closed subintervals:

0 1 1 2 2 3x , x , x , x , x , x , ..., 1n nx , x

The rth subinterval [xr – 1, xr] and its length x r – x r – 1 will both be denoted by ∆xr. The norm

(or mesh) of P, written ∥P∥, is defined as

∥P∥ = 1 r nmax ∆xr

Let cr be any point of [xr – 1, xr], r = 1, 2, 3, . . . , n. The expression

n n

1 0 1 2 1 2 1 1=1 =1

x x c x x c ... x x c x x c cf f f f f n n n r r r r rr r

Δx

is called the Riemann Sum of f corresponding to the partition P of [a, b]. We denote this sum by

S(P, f). The limit of S(P, f), if it exists, as the number of subintervals associated with the

x

nx / b 1x r x r 2x 1x 0

0x / a

17

partition P tends to infinity and ∥P∥→ 0, is called the definite integral of f over [a, b] and is

symbolically written as

b b

a ax dx or f f

In this case f is said to be Integrable over [a, b]. The numbers “a” and “b”are called lower and

upper limits of integration.

Example – 1: Evaluate 3

02x 1 dx by definition.

SOL: We consider a partition P of [0, 3] into n subintervals of equal length i.e.,

r3 0 3 ; r = 1, 2, 3, . . . , n

n nx

Then, 3 n 13 6 9 3n0, , , , . . . , , 3

n n n n nP

The subintervals are

3 n 13 3 6 6 9 3n0, , , , , , . . . , ,

n n n n n n n

The number cr is any element of the rth subinterval [xr – 1, xr]. We take the right – end point xr = cr, r = 1, 2, 3, . . ., n. Then

S(P, f) = S(P, 2x – 1)

n

=1 cf r r

rΔx

18

3 3 3 6 3 9 3 3n= . . .n n n n n n n n3 3 3 6 3 9 3 3n= 2 1 2 1 2 1 . . . 2 1n n n n n n n n

3 32 1 2 3 . . . n nn n

n n +13 32 nn n 2

f f f f

9 n +13

n

Proceeding to limits as n → ∞, we get

3

n0

2x 1 dx lim S P, f

nlim S P,2x 1

n

9 n 1lim 3 9 3 6

n

Example – 2: Compute x

b

ae dx by definition.

SOL: We divide [a, b] into n subintervals each of length

b a

nΔx

Then

S(P, f) = S(P, e x)

2 n

a 2 n

a a a a

e

= e + e e . . . e

= 1+ e e . . . e

Δx

Δx

Δx Δx Δx

Δx Δx Δx

na 1 ee

1 eΔx

Δx

Δx

19

a n

a b a

e 1 e1 e

e 1 e ; n b a1 e

Δx

Δx

ΔxΔx

Δx Δx

Taking limit as n → ∞ and ∆x → 0, we have

x

0

0

ba b a

a

a b a

a b b a

e dx lim e 1 e1 e

e 1 e lim1 e

e e 1 e e

Δx

Δx

Δx

Δx

Δx

Δx

Example – 3: show that

n 1

n n 2n 1sin x cos xcos x dx = cos x dxn n

Hence evaluate,

π2

nn

0

I cos x dx , where n is a positive integer.

SOL:

Let I = n n 1cos x dx cos x cos x dx , then integrating by parts, we have

I

n 1 n 2

n 1 n 2 2

n 1 n 2 2

n 1 n 2 n

n 1 n 2

cos x sinx n 1 cos x sinx sinxdx

sinx cos x n 1 cos x sin x dx

sin x cos x n 1 cos x 1 cos x dx

sin x cos x n 1 cos x cos x dx

sin x cos x n 1 cos x dx n 1 I

n 1 n 2

n 1n n 2

n 1 1 I = sin x cos x n 1 cos x dx

n 1sin x cos xcos x dx = cos x dxn n

20

Now

ππ π22 2

n n 2n

0 00

n 1 n 1sin x cos xI cos x dx = cos x dxnn

π

2n 2

0

n 1cos x dx

n

Therefore, we have

n

n 2

3 1

2 0

1

n

0

n 2

n 4

n 1I Inn 3I In 2

2I I , if n is odd31I I , if n iseven2

n 1 n 3 n 5 2. . . . . I when n is oddn n 2 n 4 3Hence, I

n 1 n 3 n 5 1. . . . . I when n is evenn n 2 n 4 2

π2 π

21 0

0

π2 π0 2

0 00

Now I cos x dx = sin x 1

πand I cos x dx = x2

π

2n

0

Thus,n 1 n 3 n 5 2. . . . . when n is odd

n n 2 n 4 3cos x dx =n 1 n 3 n 5 1 π. . . . . . when n is even

n n 2 n 4 2 2

This is known as Walli’s Cosine Formula.

21

Similarly, we have

n 1n n 2

π2

n

0

n 1cos x sin xsin x dx = sin x dxn n

andn 1 n 3 n 5 2. . . . . when n is odd

n n 2 n 4 3sin x dx =n 1 n 3 n 5 1 π. . . . . . when n is even

n n 2 n 4 2 2

This is known as Walli’s Sine Formula.

Practice Problem: Prove that:

nm + 1

n n 1m mx lnx nx lnx dx x lnx dxm 1 m 1

Hence calculate;

1

3 nm m

0

x lnx dx x lnx dxi ii

NUMERICAL INTEGRATON

In this section, we will discuss two methods:

(i) Trapezoidal Rule

(ii) Simpson’s Rule

Trapezoidal Rule: Let a function f be continuous on [a, b] and let [a, b] be partitioned

into n equal subintervals [xo, x1], [x1, x2], . . . , [xn – 1, xn] each of length b a

n

, then

0 1 2 nb a 1 1x dx x x x . . . x

n 2 2f f f f f

b

a

Question – 1: Use the trapezoidal rule with n = 4 to approximate

42

0

x 1dxI .

SOL: The interval [0, 4] is partitioned into subintervals by the points

0 1 2 3 4x 0, x 1, x 2, x 3, x 4

22

Length of each subinterval = 4 0 1

4

Now the function values corresponding to each point are given below:

n xn f (xn) 0 0 0 1 1.00000 1 1 1 1 1.41421 2 2 4 1 2.23607 3 3 9 1 3.16228 4 4 16 1 4.12311

Substituting into the trapezoidal rule, we get

1 11 1.41421 2.23607 3.16228 4.12311 9.374112 2

I

Question – 2: Use the trapezoidal rule with n = 6 to estimate;

2

1

dxln 2x

SOL: The interval [1, 2] is partitioned into subintervals by the points

0 1 2 3 4 5 67 8 9 10 11x 1, x , x , x , x , x , x 26 6 6 6 6

Length of each subinterval = 2 1 1

6 6

Now the function values corresponding to each point are given below:

n xn f (xn) 0 1 1 1 7

6 0.85714

2 86 0.75000

3 96 0.66666

4 106 0.60000

5 116 0.54545

6 2 0.50000

23

Substituting into the trapezoidal rule, we get

1ln 2 0.50000 0.85714 0.75000 0.66666 0.60000 0.54545 0.25000 0.694876

Question – 3: Use trapezoidal rule to approximate;

1

0

dx1+ x

Simpson’s Rule: Let a function f be continuous on [a, b] and let [a, b] be partitioned into n

even number of equal by the points a = xo, x1, x2, . . . , xn – 1, xn = b then

0 1 2 1 2 nb ax dx x 4 x 2 x 4 x 2 x . . . x3n

f f f f f f f

b

a

Question – 1: Use Simpson’s rule with n = 4 approximate

1

20

dx1+ x

SOL: Points of subdivision of [0, 1] into subintervals are

0 1 2 3 41 1 3x 1, x , x , x , x 14 2 4

Function values are as follows:

n 0 1 2 3 4

xn 0 14

12

34 1

f (xn) 1 1617

45

1625

12

Substituting into the Simpson’s rule, we get

1

20

dx 1 64 8 64 1 11 9.4247059 0.78539 0.78541 x 12 17 5 25 2 12

Question – 2: Use Simpson’s rule to approximate

1

0

dx1+ x

24

EVALUATION OF DOUBLE INTEGRAL:

Double integral over a region R may be evaluated by two successive integrations as given below:

i. If R can be described by inequalities of the form 1 2x y x and a x bf f

x

x

x, y dx dy = x, y dy dx (A)f

f

f f

2

1R

b

a

The inner integral

2

1

x

x

x, y dyf

f

f is to be integrated first with respect to y between the

limits which are functions of x.

Thus we obtain, as a result of this integration a function say F(x). Then we evaluate the integral;

x dxFb

a

Over the constant limits a to b and get the value of the double integral given in (A).

ii. If R can be described by inequalities of the form 1 2φ y x φ y and c y d

φ y

φ y

x, y dx dy = x, y dx dy (B)f f

2

1R

d

c

The inner integral

φ y

φ y

x, y dxf2

1is to be integrated first with respect to x between the

limits which are functions of y.

Thus we obtain, as a result of this integration a function say φ(y). Then we evaluate the integral;

φ y dy Cd

c

25

Which is the value of the double integral in (C).

NOTE: If however, the limits for both the variables are constant, the integration

can be performed by taking any one of the variables as the first variable and substituting

its limits in the result obtained from the integrations.

Example – 1: Evaluate the double integral 2 1

2 2

0 0

x y dydx

SOL: 21 2 1

2 2 2

00 0 0

y3x y dy dx x y dx3

112 3

00

8 2 8 2 8 10= 2x dx = x x3 3 3 3 3 3

Example – 2: Evaluate:

i. 1 x

2

0 0

x 4xy dydx

ii.

4 25

0 y

ycos x dx dy

iii.

112

0 2 x

e dy dx 2y

iv. Let the region D be bounded by the parabola x2 + 9y = 36 and the straight line 2x

+ 3y = 12. Evaluate

D D

a x ydy dx b x ydx dy 2 2

SOL – (iv) – (a): We have,

2

236 x

36 x6 699

12 2x12 2x 3D 0 0

3

1x ydy dx x y dydx x y dx2

2 2 2 2

26

63 3

0

1 16 4 1x x + x dx2 3 3 81

32 274

35 3581

×

SOL – (iv) – (b): We have,

3 4 y4

3D 08 2y

x ydydx x ydx dy

2 2

4 3 4 y

38 2 y0

31 x y dy3

4 43

2 3 42

0 0

99 4 4 y 4 y dy 64y 48y 12y y dy8

4 4 43 52 3 42 2

0 0 0

44 4 55 72 3 42 2

0 0 0

936 4 y dy 9 4 y dy 64y 48y 12y y dy8

72 18 9 y= 4 y 4 y 32y 16y 3y5 5 8 5274

35

TRIPLE INTEGRAL:

A triple integral may be defined in the same way as a double integral, that is, as the limit

of the sum over a region in space.

Example: Evaluate:

i. 3 2 1

0 0 0

x y z dz dy dx

ii.

4 x 4 x y4

0 0 0

dz dy dx

iii. 2 2

S

3 x y + y z dv , where S is bounded by the planes x = 1, x = 3, y = –1 , y = 1,

z = 2, z = 4.

27

SOL – (iii): We have

43 1 4 3 1 2 2

2 2 2 2 2

S 1 1 2 1 1 2

y z3 x y + y z dv 3 x y + y z dz dy dx 3 x y z + dy dx2

3 1

2 2

1 1

3 2x y + 6y dy dx

3 312 2 3

11 1

3 x y + 2y dx 12dx

3112x 24

VOLUME OF SOLID OF REVOLTION:

DEF: If a plane area is revolved about a straight line in the plane the resulting solid a

Solid of revolution and the solid is said to be generated by the plane area. The line about which

the plane area is revolved is called axis of revolution.

Here we discuss some methods of finding volumes of such solids:

i. Disc Method

ii. Washer Method

iii. Shell Method

i. Disc Method:

(a) About x – axis: The volume of the solid generated by revolving about the x – axis

the region between the x – axis and the graph of the continuous function y = R(x),

a x b is

2b

aV π R x dx

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(b) About y – axis: 2

d

cV π R y dy

Example – 1: The region bounded by the curve y = x , 0 x 4 and the x – axis is revolved

about the x – axis to generate a solid. Find its volume.

Example – 2: The region bounded by the curve y = x and the lines y 1, y 4 is revolved

about the line y = 1 to generate a solid. Find its volume.

Example – 3: Find the volume of the solid generated by revolving the region between the

parabola 2y + 1 = x and the line x 3 about the line x 3 .

SOL: The volume is

2

2

2

222

2

24 2

2

253

2

V π R y dy

π 2 y dy

π 4 y 4y dy

y 4 64 2ππ 4y y5 3 15

ii. Washer Method: If we revolve to generate the solid does not border on or cross the

axis of revolution, the solid has a hole in it. The cross sections perpendiculars to the

axis of revolution are washers instead of disc. The dimensions of typical washer are

Outer radius: R(x) and Inner radius: r(x)

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The washer’s area is 2 2 2 2A x π R x π r x π R x r x

The washer formula for finding volume is 2 2b

aV π R x r x dx .

Example – 1: The region bounded by the curve 2x + 1 = y and the line y = – x + 3 is revolved

about the x – axis to generate the solid. Find the volume of the solid.

Example – 2: Find the volume of the solid generated by revolving the area bounded by the

parabola 2y = 8x and its latus rectum x = 2 about the y - axis.

SOL: Here we divide the area by horizontal strips. When an approximating rectangle is

revolved about the y – axis, it generates a washer whose volume is

Volume of ECDF – Volume of EPBF

= 2 2 2π 2 y π x y π 4 x y

Thus, the required volume

24

4= π 4 x dy

2

0

4= 2π 4 x dy

0

4 4y 128π= 2π 4 dy =64 5

30

iii. Shell Method:

(a) About the y – axis: The volume of the solid generated by revolving the region

between the x – axis and the graph of a continuous function y = f (x), a x b ,

about y – axis is

a a

b bV = 2π shell radius shell height dx 2π x x dxf

(b) About the x – axis:

d d

c cV = 2π shell radius shell height dy 2π y y dyf

Example – 1: The region bounded by the curve y = x , the x – axis and the line x = 4 is

revolved about the y – axis to generate a solid. Find the volume of the solid.

SOL: Required volume = 44 4 3 5

2 2

0 0 0

2 128π2πx x dx = 2 π x dx = 2 π x5 5

Example – 2: The region bounded by the curve y = x , the x – axis and the line x = 4 is

revolved about the x – axis to generate a solid. Find the volume of the solid.

SOL: Required volume = 22 2 4

2 3 2

0 0 0

y2πy 4 y dy = 2 π 4y y dy = 2 π 2y 8π4