calculus iii - florida international...
TRANSCRIPT
Calculus III
Philippe Rukimbira
Department of MathematicsFlorida International University
PR (FIU) MAC 2313 1 / 104
13.3 Partial Derivatives
Example:f (x , y) = (x2 + 2xy)5
The partial derivative of f with respect to x is:
∂f∂x
(x , y) = 5(x2 + 2xy)4(2x + 2y).
What is ∂f∂y ?
The answer is:5(x2 + 2xy)4(2x).
PR (FIU) MAC 2313 2 / 104
13.3 Partial Derivatives
Example:f (x , y) = (x2 + 2xy)5
The partial derivative of f with respect to x is:
∂f∂x
(x , y) = 5(x2 + 2xy)4(2x + 2y).
What is ∂f∂y ?
The answer is:5(x2 + 2xy)4(2x).
PR (FIU) MAC 2313 2 / 104
The second order partial derivative of f with respect to y is:
∂2f∂y2 (x , y) = 40(x2 + 2xy)3(2x)(2x).
∂2f∂x∂y = ∂
∂x ( ∂∂y f )
= ∂∂x (10x(x2 + 2xy)4)
= 10(x2 + 2xy)4 + 40x(x2 + 2xy)3(2x)
= 20(x2 + 2xy)3(5x2 + xy)
PR (FIU) MAC 2313 3 / 104
What is∂2f∂x2 ?
PR (FIU) MAC 2313 4 / 104
13.4 Differentiability, Differentials and local linearity
Denote by ∆f = f (x0 + ∆x , y0 + ∆y)− f9X0,Y0).
f is said to be differentiable at (x0, y0) if:
∆f =∂f∂x
(x0, y0)∆x +∂f∂y
(x0, y0) + ε1∆x + ε2∆y
wherelim
(∆x ,∆y)→(0,0)ε1 = 0 = lim
(∆x ,∆y)→(0,0)ε2.
PR (FIU) MAC 2313 5 / 104
Example
f (x , y) = x2 + y2, at (x0, y0) = (2,3).
∆f = f (2 + ∆x ,3 + ∆y)− f (2,3)
= (2 + ∆x)2 + (3 + ∆y)2 − 4− 9= 4 + 4∆x + (∆x)2 + 9 + 6∆y + (∆y)2 − 4− 9= 4∆x + 6∆y + (∆x)2 + ∆y)2.
PR (FIU) MAC 2313 6 / 104
Let ε1 = ∆x , ε2 = ∆
lim(∆x ,∆y)→(0,0)
ε1 = 0; lim(∆x ,∆y)→(0,0)
ε2 = 0
∂f∂x
(2,3) = 4, and∂f∂y
(2,3) = 6
as expected!
PR (FIU) MAC 2313 7 / 104
If all first order partial derivatives of f exist and are continuous at apoint, then f is differentiable at that point.
Differentials:
∆f ' fx (x0, y0)∆x + fy (x0, y0)∆y
(∆f ' fx ∆x + fy ∆y + fz∆z)
these are approximations for ∆f
PR (FIU) MAC 2313 8 / 104
If z = f (x , y), then the differential dz of f is
dz = fx (x0, y0)dx + fy (x0, y0)dy .
The same way,dw = fxdx + fydy + fzdz.
dz or dw are called the Total differentials of f at (x0, y0).
dz = df is an approximation for ∆z = ∆f .
PR (FIU) MAC 2313 9 / 104
Use dz to approximate ∆z where z = xy2, from its value at (0.5,1.0)
to its value at (0.503,1.004).
Definition (Linear Approximation)
L(x , y) = f (x0, y0) + fx (x0, y0) + fy (x0, y0)dy
Or
L(x , y) = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0).
L(x , y)
is known as the local linear approximation for f (x , y) at the point(x0, y0).
PR (FIU) MAC 2313 10 / 104
Example
f (x , y) =√
x2 + y2
Find L(x , y) at (1,2). The formula is similar for functions of threevariables f (x , y , z).
PR (FIU) MAC 2313 11 / 104
13.5 The Chain Rules: Chain Rule number 1
z = f (x(t), y(t))
dzdt
=∂f∂x
dxdt
+∂f∂y
dydt.
Example:
z = ln(2x2 + y), x(t) =√
t , y(t) = t23 .
dzdt = 1
2x2y 4x dxdt + 1
2x2+ydydt
= 1
2t+t23
4√
t 12√
t+ 1
2t+t23
23 t−
13
= 2
2t+t23
+ 2
3[2t2+t23 ]t
13
PR (FIU) MAC 2313 12 / 104
The Chain Rules: Chain Rule number 2
z = f (x(u, v), y(u, v))
∂z∂u
=∂f∂x
∂x∂u
+∂f∂y
∂y∂u
∂z∂v
=∂f∂x
∂x∂v
+∂f∂y
∂y∂v
Example:z = 8x2y − 2x + 3y , x = uv , y = u − v
∂z∂u = ∂z
∂x∂x∂u + ∂z
∂y∂y∂u
= (16xy − 2)v + (8x2 + 3)(−v)
= (16uv(u − v)− 2)v − v(8u2v2 + 3)
= 16u2v2 − 24uv3 − 5vPR (FIU) MAC 2313 13 / 104
Compute∂z∂v.
PR (FIU) MAC 2313 14 / 104
13.6 Directional derivatives and gradients
Let ~u be a unit vector. There are infinitely many curves C(t) withdCdt (0) = ~u.
Let C(t) = (x(t), y(t), z(t))
Given a function f (x , y , z),
DefinitionThe directional derivative of f in the direction of ~u is by definition:
ddt
f (x(t), y(t), z(t))(0).
It is denoted by~uf .
PR (FIU) MAC 2313 15 / 104
DefinitionIf f (x , y , z) is a function, the gradient of f , ∇f , is the vector defined by :
∇f =∂f∂x~i +
∂f∂y~j +
∂f∂z~k .
PR (FIU) MAC 2313 16 / 104
Proposition
~uf = ~u.∇f .
PR (FIU) MAC 2313 17 / 104
Proof
~uf = ddt f (x(t), y(t), z(t))(0)
= ∂f∂x
dxdt + ∂f
∂ydydt + ∂f
∂zdzdt
= ∂f∂x u1 + ∂f
∂y u2 + ∂f∂z u3
= ~u.∇f
PR (FIU) MAC 2313 18 / 104
Property (I)The component of ∇f in any given direction gives the directionalderivative in that direction.
Proof.
Let ~u be a direction (a unit vector). Then
~uf = ~u.∇f = ‖~u‖‖∇f‖ cos θ = ‖∇f‖ cos θ.
PR (FIU) MAC 2313 19 / 104
Property (II)At each point, the vector ∇f points in the direction of the maximum rateof increase of the function f .
Proof.
~uf = ~u.∇f
takes its maximum when ~u and ∇f are parallel, and θ = 0.
PR (FIU) MAC 2313 20 / 104
Property (III)The magnitude of ∇f equals the maximum rate of increase of f per unitdistance.
PR (FIU) MAC 2313 21 / 104
Property (IV)
Through any point (x0, y0, z0) where ∇f 6= ~O, there passes a levelsurface f (x , y , z) = f (x0, y0, z0). The vector ∇f is normal to this surfaceat the point (x0, y0, z0).
PR (FIU) MAC 2313 22 / 104
Proof:
Let C(t) be a curve on the level surface f (x , y , z) = f (x0, y0, z0) suchthat C(0) = (x0, y0, z0) and dC
dt (0) = ~u a unit tangent vector. On onehand,
~uf = ddt f (x(t), y(t), z(t))(0)
= ddt f (x0, y0, z0)(0)
= 0
on the other hand,~uf = ~u.∇f
So, we see that~u.∇f = 0
therefore ∇f is perpendicular to any vector tangent to the level surface;equivalently, ∇f is perpendicular to the level surface.
PR (FIU) MAC 2313 23 / 104
Example
Given f (x , y , z) = x2 + y2 + z2, find the maximum value of the rate ofchange of f at (3,0,4).
Solution
We know the ∇f gives the direction of maximum rate of change.
∇f (3,0,4) = (2x ,2y ,2z)(3,0,4) = (6,0,8).
But also that ‖∇f‖ =√
36 + 64 = 10 gives the maximum rate ofchange.
PR (FIU) MAC 2313 24 / 104
13.7 Tangent Planes and Normal Lines
TheoremLet P0 = (x0, y0, z0) be a point on the surface z = f (x , y). If f (x , y) isdifferentiable at (x0, y0), then the surface has a tangent plane at P0.This plane has equation:
∂f∂x
(x0, y0)(x − x0) +∂f∂y
(x0, y0)(y − y0)− (z − z0) = 0.
PR (FIU) MAC 2313 25 / 104
Proof
To prove the theorem, it is enough to show that < ∂f∂x ,
∂f∂y ,−1 > is a
normal vector to the surface. To that end, let C = (x(t), y(t), z(t)) beany curve on the surface such that C(t0) = P0. We need only to showthat < ∂f
∂x ,∂f∂y ,−1 > is perpendicular to to c, that is, perpendicular to
the tangent vector of C, which is < dxdt ,
dydt ,
dzdt > at t = t0.
Fromf (x , y)− z = 0, we get
0 = ∂f∂x
dxdt + ∂f
∂ydydt −
dzdt
= << ∂f∂x ,
∂f∂y ,−1 > . < dx
dt ,dydt ,
dzdt >
PR (FIU) MAC 2313 26 / 104
Proof
To prove the theorem, it is enough to show that < ∂f∂x ,
∂f∂y ,−1 > is a
normal vector to the surface. To that end, let C = (x(t), y(t), z(t)) beany curve on the surface such that C(t0) = P0. We need only to showthat < ∂f
∂x ,∂f∂y ,−1 > is perpendicular to to c, that is, perpendicular to
the tangent vector of C, which is < dxdt ,
dydt ,
dzdt > at t = t0. From
f (x , y)− z = 0, we get
0 = ∂f∂x
dxdt + ∂f
∂ydydt −
dzdt
= << ∂f∂x ,
∂f∂y ,−1 > . < dx
dt ,dydt ,
dzdt >
PR (FIU) MAC 2313 26 / 104
Take Notice: If f (x , y) is differentiable at x0, y0), then the vector
N =<∂f∂x
(x0, y0),∂f∂y
(x0, y0),−1 >
is a normal vector to the surface z = f (x , y) at P0 = (x0, y0, z0).
The line through P0 parallel to N is called the normal line to the surfaceat P0.
PR (FIU) MAC 2313 27 / 104
Example
Find the equation for the tangent plane and the normal line to thesurface
z = 4x3y2 + 2y
at P = (1.− 2,12).
A normal vector at P is
(12x2y2,8x3y + 2,−1)(1,−2) = (48,−14,−1).
So the tangent plane has equation:
48(x − 1)− 14(y + 2)− (z − 12) = 0
and the normal line has parametric equations:
x = 1 + 48ty = −2− 14tz = 12− t
PR (FIU) MAC 2313 28 / 104
Example
Find the equation for the tangent plane and the normal line to thesurface
z = 4x3y2 + 2y
at P = (1.− 2,12).
A normal vector at P is
(12x2y2,8x3y + 2,−1)(1,−2) = (48,−14,−1).
So the tangent plane has equation:
48(x − 1)− 14(y + 2)− (z − 12) = 0
and the normal line has parametric equations:
x = 1 + 48ty = −2− 14tz = 12− t
PR (FIU) MAC 2313 28 / 104
Exercise
Find a point on the surface z = 3x2 − y2 at which the tangent plane isparallel to the plane 6x + 4y − z = 5.
We need to find points where the normal to the surface is parallel to(6,4,−1). That is, we need to solve
(6x ,−2y ,−1) = k(6,4,−1)
Necessarily, k = 1, and y = −2, x = 1.
PR (FIU) MAC 2313 29 / 104
Exercise
Find a point on the surface z = 3x2 − y2 at which the tangent plane isparallel to the plane 6x + 4y − z = 5.
We need to find points where the normal to the surface is parallel to(6,4,−1). That is, we need to solve
(6x ,−2y ,−1) = k(6,4,−1)
Necessarily, k = 1, and y = −2, x = 1.
PR (FIU) MAC 2313 29 / 104
So at (1,−2,−1), the tangent plane is parallel to 6x + 4y − z = 5.
PR (FIU) MAC 2313 30 / 104
13.8 Maxima and Minima of Functions of two variables
TheoremIf f has a relative extremum at a point (x0, y0) and if the first orderpartial derivatives of f exist at this point, then
fx (x0, y0) = 0 and fy (x0, y0) = 0
A point (x0, y0) where fx (x0, y0) = 0 and fy (x0, y0) = 0 is called acritical point.
So, we will remember that local extrema occur at critical points. But notevery critical point leads to a local extremum as the last of the followingexamples shows:
PR (FIU) MAC 2313 31 / 104
Examples
z = f (x , y) = x2 + (y − 1)2 + 5
f (0,1) = 5 is a local minimum. Notice that both partial derivativesof f vanish at (0,1)
z = g(x , y) = −x2 − (y − 1)2 + 2
f (0,1) = 2 is a local maximum. Observe that both partialderivatives of g vanish at (0,1).
z = h(x , y) = x2 − (y − 1)2 + 10
Notice that both partial derivatives of h at (0,1) are equal to zero,but f (0,1) = 10 is neither a local maximum, nor a local minimum.A point like (0,1,h(0,1)) = (0,1,10) is called a saddle point forthe graph of h(x , y).
PR (FIU) MAC 2313 32 / 104
Second Order Partial Derivatives Test for LocalExtrema
TheoremLet f (x , y) be a differentiable function of 2 variables and assume (a,b)
is a critical point for f .Let A = fxx (a,b), B = fxy (a,b), C = fyy (a,b) and D = AC − B2.
1. If D > 0 and A > 0, then f (a,b) is a local minimum.
2. If D > 0 and A < 0, then f (a,b) is a local maximum.
3. If D < 0, then the graph of f has a saddle point at (a,b, f (a,b)).
4. If D = 0, the test is inconclusive!
PR (FIU) MAC 2313 33 / 104
Examples
1. Locate all local maxima, local minima and saddle points if any, for
f (x , y) = x3 − 3xy − y3.
2. Same forf (x , y) = y sin x
PR (FIU) MAC 2313 34 / 104
Absolute Extrema
To find absolute extrema on a closed bounded set R,
1. Find critical points in R
2. Find all boundary points at which the absolute extrema can occur(including corners)
3. Compare values and decide
PR (FIU) MAC 2313 35 / 104
Example
f (x , y) = x2 − 3y2 − 2x + 6y ,
R is the rectangle with vertices (0,0), (2,0), (2,2), and (0,2).
PR (FIU) MAC 2313 36 / 104
13.9 Lagrange Multipliers (Constrained Optimization)
∇f is zero at any local extremum of a continuously differentiablefunction f . Points at which ∇f = ~O are called critical points.
PR (FIU) MAC 2313 37 / 104
Exercise
Find the maximum value of
f (x , y , z) = x4 − 3y2 − 4z4 + 2z2
Sincelim
‖(x ,y ,z)‖→+∞f (x , y , z) = −∞
we know that f has a maximum value.
Also, since a maximum value will occur at a critical point, let us seewhat the critical points are:
PR (FIU) MAC 2313 38 / 104
Exercise
Find the maximum value of
f (x , y , z) = x4 − 3y2 − 4z4 + 2z2
Sincelim
‖(x ,y ,z)‖→+∞f (x , y , z) = −∞
we know that f has a maximum value.
Also, since a maximum value will occur at a critical point, let us seewhat the critical points are:
PR (FIU) MAC 2313 38 / 104
−4x3 = 0 x = 0−4y = 0 y = 0
−16z3 + 4z = 0 z = 0 or z = ±12
PR (FIU) MAC 2313 39 / 104
f (0,0,0) = 0, f (0,0,12
) =14, f (0,0,−1
2) =
14
Thus, the maximum of f is 14 ; it is attained at the points (0,0,−1
2) and(0,0, 1
2).
PR (FIU) MAC 2313 40 / 104
Suppose g(x , y , z) is a continuously differentiable function and thesurface g(x , y , z) = 0 divides all space into two regions:
g(x , y , z) < 0 and g(x , y , z) > 0.
Problem: Maximize f (x , y , z) subject to the constraint g(x , y , z) = 0.
PR (FIU) MAC 2313 41 / 104
Suppose (x0, y0, z0) is a point on the surface g(x , y , z) = 0 andf (x0, y0, z0) is maximum among all points on the surface g(x , y , z) = 0.If ∇f were not perpendicular to the surface, then it would have anonzero tangential component along the surface, that would be adirection of increase for f along the surface, contradicting themaximality of f (x0, y0, z0). So the only possibility is that ∇f isproportional to ∇g, that is, for some constant λ,
∇f + λ∇g = 0.
This is a necessary condition for a constrained maximum (minimum).
PR (FIU) MAC 2313 42 / 104
The parameter λ is called the Lagrange Multiplier.
Example; Find the maximum of f (x , y , z)) = −x4 − 2y2 − 4z4 + 2z2
subject to the constraint g(x , y , z) = x + y − 2 = 0
∇f = (−4x3,−4y ,4z − 16z3)
∇g = (1,1,0)
λ(1,1,0) + (−4x3,−4y ,4z − 16z3) = 0
PR (FIU) MAC 2313 43 / 104
The parameter λ is called the Lagrange Multiplier.
Example; Find the maximum of f (x , y , z)) = −x4 − 2y2 − 4z4 + 2z2
subject to the constraint g(x , y , z) = x + y − 2 = 0
∇f = (−4x3,−4y ,4z − 16z3)
∇g = (1,1,0)
λ(1,1,0) + (−4x3,−4y ,4z − 16z3) = 0
PR (FIU) MAC 2313 43 / 104
y = x3, z = 0 or z = ±12
and x + y = 0
x + x3 − 2 = 0, implies (x − 1)(x2 − x + 2) = 0
x = 1
PR (FIU) MAC 2313 44 / 104
So the constrained extrema can be found only at
(1,1,12
), (1,1,−12
) and (1,1,0)
f (1,1,±12 = −11
4 and f (1,1,0) = −3, so the maximum is−11
4 = f (1,1,±12).
PR (FIU) MAC 2313 45 / 104
Generalized Lagrange Multipliers
Maximize f (x , y , z) = z, subject to x2 + y2 = 1 and 2x + 2y + z = 0.
∇f + λ1∇g1 + λ2∇g2 = 0
Complete the solution!
PR (FIU) MAC 2313 46 / 104
Generalized Lagrange Multipliers
Maximize f (x , y , z) = z, subject to x2 + y2 = 1 and 2x + 2y + z = 0.
∇f + λ1∇g1 + λ2∇g2 = 0
Complete the solution!
PR (FIU) MAC 2313 46 / 104
14.1 Double Integrals
Motivated by the volume problem: Given the graph of a function f (x , y)
of two variables, find the volume under the graph and over someregion R in the X -Y plane.
Riemann sums are here again!
The region R is divided into subregions Rij with dimensions ∆xi by ∆yj
and area∆Aij = ∆xi ×∆yj
PR (FIU) MAC 2313 47 / 104
Pick (ui , vj) ∈ Rij and write the Riemann sum:
Sn =l∑
i=1
mi∑j=1
f (ui , vj)∆Aij ,
where m1 + m2 + ...+ ml = n.
The volume, if defined, should be
V = limmax(∆xi ,∆yj )→0 Sn
= limn→∞∑l
i=1∑mi
j=1 f (ui , vj)∆Aij
= liml→∞∑l
i=1 limmi→∞∑mi
j=1 f (ui , vj)∆yj∆xi
PR (FIU) MAC 2313 48 / 104
When the above limit exists, it is called the double integral of f over Rand denoted by ∫ ∫
Rf (x , y)dA
wheredA = dxdy or dydx
A geometric interpretation of the double integral is that for f > 0,∫ ∫R fdA represents the volume under the graph of f , above the region
R in X ,Y plane,
ObservationIf f = 1, then the volume
∫ ∫R dA is numerically equal to the surface
area of R.
Example: Compute the double integral∫ ∫
R(30− xy)dA where R is therectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.
∫ 31
∫ 42 (30− xy)dydx =
∫ 31 (30y − x y2
2 )∣∣42dx
=∫ 3
1 (60− 6x)dx= (60x − 3x2)
∣∣31
= 96
PR (FIU) MAC 2313 49 / 104
When the above limit exists, it is called the double integral of f over Rand denoted by ∫ ∫
Rf (x , y)dA
wheredA = dxdy or dydx
A geometric interpretation of the double integral is that for f > 0,∫ ∫R fdA represents the volume under the graph of f , above the region
R in X ,Y plane,
ObservationIf f = 1, then the volume
∫ ∫R dA is numerically equal to the surface
area of R.
Example: Compute the double integral∫ ∫
R(30− xy)dA where R is therectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 4.
∫ 31
∫ 42 (30− xy)dydx =
∫ 31 (30y − x y2
2 )∣∣42dx
=∫ 3
1 (60− 6x)dx= (60x − 3x2)
∣∣31
= 96
PR (FIU) MAC 2313 49 / 104
14.2 Double Integrals over non-rectangular regions
For double integration, we identify two types of region to integrate over!The X regions and the Y regions. Some regions are of both types,looking at them from one type to the other is related to reversing theorder of integration.
PR (FIU) MAC 2313 50 / 104
X-Regions
An X -region R is one that can be described by:
a ≤ x ≤ b
g(x) ≤ y ≤ h(x)
for some functions g and h and constants a and b.
A double integral over such a region can be written as :
∫ ∫R
fdA =
∫ b
a
∫ h(x)
g(x)f (x , y)dydx
Example: Compute the integral∫ ∫R
(x2 − y2)dA
where R is the region bounded by y = x and y = x2.PR (FIU) MAC 2313 51 / 104
The region R is described by:
0 ≤ x ≤ 1
x2 ≤ y ≤ x
The integral is given by∫ ∫R(x2 − y2)dA =
∫ 10
∫ xx2(x2 − y2)dydx
=∫ 1
0 x2y − y3
3
∣∣y=xy=x3dx
=∫ 1
023x3 − x4 + x6
3 !dx= x4
6 −x5
5 + x7
21
∣∣10
= 16 −
15 + 1
21
PR (FIU) MAC 2313 52 / 104
Y-Regions
A Y -region R is one that can be described by:
c ≤ y ≤ d
h(y) ≤ x ≤ g(y)
for some constants c and d and function h and g.
An integral over such a region can be written as:
∫ ∫R
fdA =
∫ d
c
∫ g(y)
h(y)f (x , y)dxdy
PR (FIU) MAC 2313 53 / 104
Reversing the order of integration
Example: Evaluate ∫ 12
0
∫ 1
2xey2
dydx
Notice that the first integration is not easy!
The region R over which it is being integrated is described by:
0 ≤ x ≤ 12
2x ≤ y ≤ 1
PR (FIU) MAC 2313 54 / 104
After sketching the region R, we observe that an alternate descriptionof R is
0 ≤ y ≤ 1
0 ≤ x ≤ y2
Using the alternate description of R, we see that the same integral canbe written as:
∫ 1
0
∫ y2
0ey2
dxdy
PR (FIU) MAC 2313 55 / 104
Now we are able to start the integration process! We have reversedthe order of integration. Finish the double integration!
PR (FIU) MAC 2313 56 / 104
14.3 Double Integrals in Polar Coordinates
Consider the region R described in polar coordinates by:
α ≤ θ ≤ β
a ≤ r ≤ b
Divide the region r into subregions Rij , where each Rij is described as
θi−1 ≤ θ ≤ θi
rj−1 ≤ r ≤ rj
PropositionDenoting by ∆ri = ri − ri−1 and ∆θj = θj − θj−1, the area ∆Aij of Rij isgiven by:
∆Aij =12
(ri + ri−1)∆ri∆θj
PR (FIU) MAC 2313 57 / 104
Proof
∆Aij is the difference between areas of sectors with angle ∆θj andradii ri and ri−1. Therefore,
∆Aij =r2i2 ∆θj −
r2i−12 ∆θj = 1
2(r2i − r2
i−1)∆θj
= 12(ri + ri−1)(ri − ri−1)∆θj
= 12(ri + ri−1)∆ri∆θj
PR (FIU) MAC 2313 58 / 104
Choose r̄i = 12(ri + ri−1) and θ̄j in Rij , then
∆Aij = r̄i∆ri∆θj
and the Riemann sum associated to a function f (r , θ) and a partition Pover the region R will be
S(P, f ) =∑
j
∑i
f (r̄i , θ̄j)∆Aij =∑
i
∑j
f (r̄i , θ̄j)r̄i∆ri∆θj
By definition, the double integral∫ ∫
R fdA of f over the region R isgiven by
lim‖P‖→0
∑j
∑i
f (r̄i , θ̄j)r̄i∆ri∆θj =
∫ β
α
∫ b(θ)
a(θ)f (r , θ)rdrdθ
PR (FIU) MAC 2313 59 / 104
Example
Evaluate the integral
∫ 1
0
∫ √4−x2
0(x2 + y2)dydx
using polar coordinates.
From the sketch, the region can be described as
0 ≤ r ≤ 2
0 ≤ θ ≤ π
2Recall: In polar coordinates dA = rdrdθ.
PR (FIU) MAC 2313 60 / 104
So ∫ 20
∫√4−x2
0 (x2 + y2)dydx =∫ π
20
∫ 20 r2rdrdθ
=∫ π
20
r4
4
∣∣20dθ
=∫ π
20 4dθ
= 2π
PR (FIU) MAC 2313 61 / 104
Example 2
The cardioid is given by
r = a(1− cos θ)
Find its enclosed surface area.
Sketch the cardioid on the X Y plane.
Its description is0 ≤ θ ≤ 2π0 ≤ r ≤ a(1− cos θ)
The surface area is given by
A = 2∫ π
0
∫ a(1−cos θ)
0rdrdθ = 3π
a2
2.
Compute the integral in details!
PR (FIU) MAC 2313 62 / 104
Example 2
The cardioid is given by
r = a(1− cos θ)
Find its enclosed surface area.
Sketch the cardioid on the X Y plane.
Its description is0 ≤ θ ≤ 2π0 ≤ r ≤ a(1− cos θ)
The surface area is given by
A = 2∫ π
0
∫ a(1−cos θ)
0rdrdθ = 3π
a2
2.
Compute the integral in details!PR (FIU) MAC 2313 62 / 104
14.4 Parametric Surfaces; Surface area
A parameterized curve C is a vector valued function
r : R→ R3 : t 7→ r(t) = (x(t), y(t), z(t)).
A parameterized surface S is a vector valued function
r : R2 → R3 : (u, v) 7→ r(u, v) = (x(u, v), y(u, v), z(u, v)).
Example:r(u, v) = (u, v ,4− u2 − v2)
Eliminating u, v from the parametric equations shows that
z = 4− x2 − v2
The surface S is a piece of a paraboloid of revolution (Circularparaboloid).
PR (FIU) MAC 2313 63 / 104
Tangent Plane to Parameterized Surfaces
Let r(u, v) be a parameterized surface. Its tangent plane at any point isparallel to ∂r
∂u and ∂r∂v , that is, parallel to
(∂x∂u,∂y∂u,∂z∂u
) and (∂x∂v,∂y∂v,∂z∂v
).
Therefore,~N = (
∂r∂u
)× (∂r∂v
)
is a normal vector to the surface.
Definition
~n =~N‖~N‖
is called the principal normal vector.
PR (FIU) MAC 2313 64 / 104
~n =∂r∂u ×
∂r∂v
‖ ∂r∂u ×
∂r∂v ‖
Example:x = uv , y = u, z = v2.
Find an equation of the tangent plane at the point where u = 2 andv = −1.
PR (FIU) MAC 2313 65 / 104
Surface Area of a Parameterized Surface
Let Rij be a rectangle in the subdivision od the domain ofparametrization D. Its edges are (ui , vj), ui + ∆ui , vj), (ui , vj + ∆vj) and(ui + ∆ui , vj + ∆vj). The image of Rij is denoted by σij . Its surface area∆Sij is approximated by the area of the parallelogram spanned byr(ui + ∆ui − r(ui , vj) and r(ui , vj + ∆vj)− r(ui , vj).
PR (FIU) MAC 2313 66 / 104
In turn, from the Mean Value Theorem,
r(ui , vj + ∆vj − r(ui , vj) '∂r∂v
∆vj
r(ui + ∆ui , vj)− r(ui , vj) '∂r∂u
∆ui
PR (FIU) MAC 2313 67 / 104
Therefore,
∆Sij ' ‖∂r∂u
∆ui ×∂r∂v
∆vj‖ = ‖ ∂r∂u× ∂r∂v‖∆ui∆vj
The surface area of S is approximated by the Riemann sum
Sn =∑i,j
‖ ∂r∂u× ∂r∂v‖∆Aij
The surface area S is given by
S = limn→∞
Sij =
∫ ∫D‖ ∂r∂u× ∂r∂v‖dA.
PR (FIU) MAC 2313 68 / 104
In the special case where the surface is the graph of
z = f (x , y)
then a parametrization is
r(x , y) = (x , y , f (x , y))
and the surface area S is given by
S =
∫ ∫D
√(∂f∂x
)2 + (∂f∂y
)2 + 1dA.
Example: 1. r(u, v) = (u, v ,4− u2 − v2), D = {u2 + v2 ≤ 4}.
Compute the surface area.
2. Find the surface area of the portion of the sphere x2 + y2 + z2 = 8that is cut out by the cone z =
√x2 + y2.
3. Find the surface area of the paraboloid portion z = 9− x2 − y2 thatlies between the planes z + 0 and z = 8.
PR (FIU) MAC 2313 69 / 104
1.
2. The domain of parametrization is a disc of radius 2.
−2 ≤ x ≤ 2
−√
4− x2 ≤ y ≤√
4− x2
The surface area is given by
S =
∫ 2
−2
∫ √4−x2
−√
4−x2
√x2
8− x2 − y2 +y2
8− x2 − y2 + 1dydx .
We can use polar coordinates to compute this integral.
PR (FIU) MAC 2313 70 / 104
∫ 2
0
∫ 2π
0
√r2
8− r2 + 1rdrdθ = (16− 8√
2)π
PR (FIU) MAC 2313 71 / 104
For example 3, the domain of parametrization is an annulus 1 ≤ r ≤ 3.
S =
∫ 3
1
∫ 2π
0
√4r2 + 1rdrdθ
PR (FIU) MAC 2313 72 / 104
Think of the surface area S as
S =
∫S
ds
One defines the surface integral of any function f on the surface as
∫S
fdS =
∫ ∫D
f (x(u, v), y(u, v))‖ ∂r∂u× ∂r∂v‖dA
wheredA = dudv or dvdu.
Examples will be provided later!
PR (FIU) MAC 2313 73 / 104
14.5 Triple Integrals
An example of a 3-dimensional region R:
The region bounded by
x = 0, y = 0, z = 0
andx + 2y + 3z = 6.
After sketching the region, it can be described as follows:
0 ≤ x ≤ 60 ≤ y ≤ 6−x
20 ≤ z ≤ 6−x−2y
3
PR (FIU) MAC 2313 74 / 104
There are up to 5 more different descriptions of R. Any triple integral∫ ∫ ∫R fdV could be written as an iterated integral
∫ 6
0
∫ 6−x2
0
∫ 6−x−2y3
0f (x , y , z)dzdydx .
PR (FIU) MAC 2313 75 / 104
Example
Find the volume integral of f (x , y , z) = x + y + z over the box boundedby x = 1, y = 2, z = 1 + x and the coordinate planes.
After sketching the region, it can be described as follows:
0 ≤ x ≤ 10 ≤ y ≤ 20 ≤ z ≤ 1 + x
PR (FIU) MAC 2313 76 / 104
Example
Find the volume integral of f (x , y , z) = x + y + z over the box boundedby x = 1, y = 2, z = 1 + x and the coordinate planes.
After sketching the region, it can be described as follows:
0 ≤ x ≤ 10 ≤ y ≤ 20 ≤ z ≤ 1 + x
PR (FIU) MAC 2313 76 / 104
The integral is written as follows:∫ 10
∫ 20
∫ 1+x0 (x + y + z)dzdydx =
∫ 10
∫ 20 (x + y)z + z2
2
∣∣1+x0 dydx
=∫ 1
0
∫ 20 (1 + x)(x + y) + (1+x)2
2 dydx=
∫ 10 (1 + x)(xy + y2
2 )∣∣20 + (1 + x)2dx
= 3∫ 1
0 (1 + x)2dx= (1 + x)3|10 = 7
PR (FIU) MAC 2313 77 / 104
Use a triple integral to find the volume of the solid bounded by thesurface y = x2, the plane y + z = 4 and z = 0.
After sketching, the region is described as follows:
−2 ≤ x ≤ 2x2 ≤ y ≤ 40 ≤ z ≤ 4− y
PR (FIU) MAC 2313 78 / 104
Use a triple integral to find the volume of the solid bounded by thesurface y = x2, the plane y + z = 4 and z = 0.
After sketching, the region is described as follows:
−2 ≤ x ≤ 2x2 ≤ y ≤ 40 ≤ z ≤ 4− y
PR (FIU) MAC 2313 78 / 104
The volume is given by∫ 2−2
∫ 4x2
∫ 4−y0 dzdydx =
∫ 2−2
∫ 4x2(4− y)dydx
=∫ 2−2[4y − y2
2 ]∣∣4x2dx∫ 2
−2(8− 4x2 + x4
2 )dx= 16− 32
3 + 3210 + 16− 32
3 + 3210
PR (FIU) MAC 2313 79 / 104
14.7 Change of variables in multiple integrals:Jacobians
Motivation: Recall the u substitution technique of integration.
PropositionFrom ∫ b
af (x)dx ,
make a substitution x = g(u), dx = g′(u)du.The integral becomes:∫ g−1(b)
g−1(a)f (g(u)g′(u)du =
∫ β
αf (g(u)|g′(u)|du, α < β
PR (FIU) MAC 2313 80 / 104
Proof
If g−1(a) < g−1(b), then g′(u) > 0 and thus, g′(u) = |g′(u)|.∫ b
af (x)dx =
∫ g−1(b)
g−1(a)f (g(u))|g′(u)|du.
If g−1(b) < g−1(a), then g′(u) < 0, thus g′(u) = −|g′(u)|.∫ b
af (x)dx = −
∫ g−1(b)
g−1(a)f (g(u))|g′(u)|du =
∫ g−1(a)
g−1(b)f (g(u))|g′(u)|du.
In each case, the integral is given by∫ β
αf (g(u))|g′(u)|du, α < β.
PR (FIU) MAC 2313 81 / 104
Generalization to higher dimensions
Under a reparameterization (x , y) = r(u, v), from a domain S into adomain R, a small rectangle with vertices(u0, v0), (u0 + ∆u, v0), (u0, v0 + ∆v), (u0 + ∆u, v0 + ∆v) transformsinto R, approximated by a rectangle based on r(u0, v0) and edges
a ' r(u0 + ∆u, v0)− r(u0, v0)
∆u∆u ' ∂r
∂u∆u
and similarly,
b ' ∂r∂v
∆v
The elementary surface ∆A in the x , y parameters is approximately
∆A = ∆x∆y ' ‖ ∂r∂u ∆u × ∂r
∂v ∆v‖ = ‖ ∂r∂u ×
∂r∂v ‖∆u∆v
= |det
(∂x∂u
∂y∂u
∂x∂v
∂y∂v
)∆u∆v
PR (FIU) MAC 2313 82 / 104
The change of variable formula in double integrals is therefore: when(x , y) = r(u, v),∫ ∫
Rf (x , y)dAx ,y =
∫ ∫S
f (u, v)|det(∂(x , y)
∂(u, v))|dAu,v .
The determinant det(∂(x ,y)∂(u,v) ) is called the Jacobian of the coordinate
change.
PR (FIU) MAC 2313 83 / 104
Example
Compute the integral ∫ ∫R
exydA
where R is determined by
y =12
x , y = x , y =1x, y =
2x
using the transformation:
u =yx, v = xy
PR (FIU) MAC 2313 84 / 104
The domain S is described as
u =12, u = 1, v = 1, v = 2
We need the coordinates changes as
x =
√vu, y =
√uv
Then the Jacobian is∂(x , y)
∂(u, v)= − 1
2u.∫ ∫
RexydAx ,y =
∫ ∫S
ev 1|2u|
dAu,v =12
∫ 2
1
∫ 1
12
1u
ev dudv .
PR (FIU) MAC 2313 85 / 104
In triple integrals, the Jacobian is a 3 by 3 determinant
det∂(x , y , z)
∂(u, v ,w).
∫ ∫ ∫R
f (x , y , z)dVx ,y ,z =
∫ ∫ ∫S
f (u, v ,w)| ∂(x , y , z)
∂(u, v ,w)dVu,v ,w .
Example: Find the volume of
x2 +y2
4+
z2
9= 1
using the transformation
x = u, y = 2v , z = 3w
PR (FIU) MAC 2313 86 / 104
14.6 Triple Integrals in Cylindrical and SphericalCoordinates
Example: Evaluate the integral using spherical coordinates:∫ 2
0
∫ √4−y2
0
∫ √4−x2−y2
0
2z√x2 + y2
dzdxdy .
PR (FIU) MAC 2313 87 / 104
15.1 Vector fields
A vector field F is a vector valued function.
F : R3 → R3.
F (x , y , z) = (F1(x , y , z),F2(x , y , z),F3(x , y , z)).
PR (FIU) MAC 2313 88 / 104
Examples
Given a differentiable function
f : R3 → R,
∇f is a vector field, the gradient vector field of f .
∇f : R3 → R3
(x , y , z) 7→ ∇f (x , y , z) = ( ∂f∂x ,
∂f∂y ,
∂f∂z )
PR (FIU) MAC 2313 89 / 104
DefinitionA curve C(t) = (x(t), y(t), z(t)) such that c′(t) = F (C(t)) is called aflow line for the vector field F .
PR (FIU) MAC 2313 90 / 104
Examples
Find the equation of flow lines for F (x , y) = (−y , x).
By definition, C(t)− (x(t), y(t)) is a flow line if
dxdt = −ydydt = x
PR (FIU) MAC 2313 91 / 104
Examples
Find the equation of flow lines for F (x , y) = (−y , x).
By definition, C(t)− (x(t), y(t)) is a flow line if
dxdt = −ydydt = x
PR (FIU) MAC 2313 91 / 104
Solving the system leads to:
d2xdt2 = −dy
dt= −x
That is:
d2xdt2 + x = 0.
The solution is :x(t) = a cos t + b sin t
PR (FIU) MAC 2313 92 / 104
From dydt = x , one obtains
y(t) = a sin t − b cos t + C
with C = 0 since dxdt = −y .
PR (FIU) MAC 2313 93 / 104
C(t) = (a cos t + b sin t ,a sin t − b cos t)
Eliminating t shows that the flow lines are circles of radius√
a2 + b2,centered at (0,0).
PR (FIU) MAC 2313 94 / 104
Example 2
F (x , y , z) = (x
x2 + y2 ,y
x2 + y2 ).
C(t) = (x(t), y(t))
dxdt = x
x2+y2
dydt = y
x2+y2
PR (FIU) MAC 2313 95 / 104
1x2 + y2 =
1x
dxdt
=1y
dydt
1x
dxdt− 1
ydydt
= 0
ddt
(lnxy
) = 0
xy
= C
orx = Cy
The flow lines are straight lines through the origin, minus the originitself.
PR (FIU) MAC 2313 96 / 104
15.2 Line Integrals
PR (FIU) MAC 2313 97 / 104
15.7 The divergence Theorem
Definition of divergence as flux density
The divergence of a vector field is the measure of "flux out" of a closedsurface. The divergence or flux density of a vector field F is defined by
div ~F (x , y , z) = limvolume(σ)→0
∫σ~F . ~dS
(volume enclosed by σ)
In rectangular coordinates:
div ~F =∂F1
∂x+∂F2
∂y+∂F3
∂z
Observation:div curl ~F = 0
This can be shown directly (do the calculation!)PR (FIU) MAC 2313 98 / 104
On a doubly connected region, one has the converse: If div ~G = 0,then ~G = curl ~F for some ~F .
Examples of divergence free vector fields, or solenoidal vector fields,are the magnetic fields ~B. Maxwell’s equation states exactly that:
div ~B = 0.
Another example: For a magnetic dipole (or a current loop), withconstant dipole moment ~µ,
~B = − ~µ
‖~r‖3+ 3
(~µ.~r)~r‖~r‖5
, ~r 6= ~0
Show that div ~B = 0.
PR (FIU) MAC 2313 99 / 104
Alternate notation:With ~∇ =< ∂∂x ,
∂∂y ,
∂∂z >
div ~F = ~∇.~F
PR (FIU) MAC 2313 100 / 104
The divergence Theorem expresses the total flux as the integral of theflux density.
∫S
~F .~n dS =
∫W
div ~FdV
where S is the boundary of the region W , oriented by outer unitnormals.
PR (FIU) MAC 2313 101 / 104
Example: Use the divergence Theorem to find the flux ofF (x , y , z) = (x2y ,−xy2, z + 2) across the surface σ of the solidbounded above by the plane z = 2x and below by the paraboloidz = x2 + y2.
PR (FIU) MAC 2313 102 / 104
16.8: Stokes’ Theorem
Stokes’ Theorem expresses the total circulation of ~F as the integral ofthe circulation density:
∫σ
curl ~F .~n dS =
∫∂σ
~F . ~dr
A special case of Stokes Theorem is Green’s Theorem:
~F =< f ,g,0 >
PR (FIU) MAC 2313 103 / 104
Example: Verify Stokes’ Theorem for
~F (x , y , z) =< x , y , z >
where σ the upper hemisphere z =√
a2 − x2 − y2 with upwardorientation.
PR (FIU) MAC 2313 104 / 104