MATH 311
Topics in Applied Mathematics I
Lecture 37:
Review for Test 3.
Topics for Test 3
Vector analysis (Leon/Colley 8.1–8.4, 9.1–9.5, 10.1–10.3,
11.1–11.3)
• Gradient, divergence, and curl
• Fubini’s Theorem
• Change of coordinates in a multiple integral
•
•
•
•
Length of a curve
Line integrals
Green’s Theorem
Conservative vector fields
•
•
•
•
Area of a surface
Surface integrals
Gauss’ Theorem
Stokes’ Theorem
Sample problems for Test 3
Problem 1 Find curl(curl(F)), where
F(x, y , z) = (x 2 + y 2)e1 + ze x+y e2 + (x + sin y )e3.
Problem 2 Evaluate a double integral
¨
2x + 3y − cos(πx + 2πy ) dx dy
P
over a parallelogram P with vertices (−1, −1),
(1, 0), (2, 2), and (0, 1).
Sample problems for Test 3
Problem 3 Find the area of a cardioid which
boundary is given by r = 1 − cos φ in polar
coordinates.
Problem 4 Consider a vector field
F(x, y , z) = (yz + 2 cos 2x, xz − e z , xy − ye z ).
(i) Verify that the field F is conservative.
(ii) Find a function f such that F = ∇f .
Sample problems for Test 3
Problem 5 Let C be a solid cylinder bounded by planes
z = 0, z = 2 and a cylindrical surface x 2 + y 2 = 1. Orient
the boundary ∂C with outward normals and evaluate a surface
‹
integral
(x 2 e1 + y 2 e2 + z 2 e3 ) · d S.
∂C
Problem 6 Let D be a region in R3 bounded by a paraboloid
z = x 2 + y 2 and a plane z = 9. Let S denote the part of the
paraboloid that bounds D, oriented by outward normals.
Evaluate a surface integral
¨
curl(F) · d S,
S
where F(x, y , z) = (e
x 2 +z 2
, xy + xz + yz, e xyz ).
Problem 1 Find curl(curl(F)), where
F(x, y , z) = (x 2 + y 2)e1 + ze x+y e2 + (x + sin y )e3.
For any vector field F = (F1 , F2 , F3) we have,
informally,
e e e 1 2 3
∂ ∂ ∂ curl F = ∇× F = ∂x
∂y ∂z F1 F2 F3 or, formally,
∂F
∂F2 ∂F1 ∂F3 ∂F2 ∂F1 3
curl F =
−
,
−
,
−
.
∂y
∂z ∂z
∂x ∂x
∂y
Problem 1 Find curl(curl(F)), where
F(x, y , z) = (x 2 + y 2)e1 + ze x+y e2 + (x + sin y )e3.
Let G = curl F, G = (G1 , G2 , G3 ). We obtain
G1 =
∂
∂
∂F3 ∂F2
−
=
(x + sin y ) − (ze x+y ) = cos y − e x+y ,
∂y
∂z
∂y
∂z
G2 =
∂F1 ∂F3
∂ 2
∂
−
=
(x + y 2) −
(x + sin y ) = −1,
∂z
∂x
∂z
∂x
G3 =
∂
∂ 2
∂F2 ∂F1
−
=
(ze x+y ) −
(x + y 2 ) = ze x+y − 2y .
∂x
∂y
∂x
∂y
Hence G = curl F = (cos y −e x+y , −1, ze x+y −2y ).
Now let H = curl G, H = (H1 , H2 , H3 ). We obtain
H1 =
∂
∂
∂G3 ∂G2
−
=
(ze x+y − 2y ) −
(−1) = ze x+y − 2,
∂y
∂z
∂y
∂z
H2 =
∂G1 ∂G3
∂
∂
−
=
(cos y −e x+y )− (ze x+y −2y ) = −ze x+y ,
∂z ∂x
∂z
∂x
H3 =
∂
∂
∂G2 ∂G1
−
=
(−1)− (cos y −e x+y ) = sin y +e x+y .
∂x
∂y
∂x
∂y
Thus curl(curl(F)) = (ze x+y −2, −ze x+y , sin y +e x+y ).
Problem 2 Evaluate a double integral
¨
2x + 3y − cos(πx + 2πy ) dx dy
P
over a parallelogram P with vertices (−1, −1), (1, 0), (2, 2),
and (0, 1).
Adjacent edges of the parallelogram P are represented by
vectors v1 = (1, 0) − (−1, −1) = (2, 1) and
v2 = (0, 1) − (−1, −1) = (1, 2).
Consider a transformation L of the plane R2 given by
u
2 1
u
−1
2u + v − 1
L
=
+
=
v
1 2
v
−1
u + 2v − 1
(columns of the matrix are vectors v1 and v2 ). By
construction, L maps the unit square [0, 1]×[0, 1] onto the
parallelogram P. The Jacobian matrix J of L is the same at
2 1
any point: J =
.
1 2
Changing coordinates in the integral from (x, y ) to (u, v ) so
that (x, y ) = L(u, v ) = (2u + v − 1, u + 2v − 1), we obtain
¨
2x + 3y − cos(πx + 2πy ) dx dy
P¨
=
7u + 8v − 5 − cos(4πu + 5πv − 3π) |det J| du dv
L−1 (P)
1
ˆ 1ˆ
3 7u + 8v − 5 + cos(4πu + 5πv ) du dv
0 0
ˆ 1ˆ 1
21
=
+ 12 − 15 +
3 cos(4πu + 5πv ) du dv .
2
0 0
ˆ 1
1
3
sin(4πu + 5πv ) Further,
3 cos(4πu + 5πv ) du =
u=0
4π
0
3
=
sin(4π + 5πv ) − sin(5πv ) = 0 for all v .
4π
¨
15
It follows that
2x + 3y − cos(πx + 2πy ) dx dy = .
2
P
=
Problem 3 Find the area of a cardioid which
boundary is given by r = 1 − cos φ in polar
coordinates.
Let H denote the cardioid. The boundary ∂H is parametrized
2
(in Cartesian coordinates) by a path x : [0,
2π] → R ,
x(t) = (1 − cos t) cos t, (1 − cos t) sin t . The cardioid is on
the left when going along x. By Green’s Theorem,
ˆ
¨
(−y dx + x dy ) =
2 dx dy = 2 area(H).
x
H
Evaluating the line integral, we will find the area of the
cardioid.
Problem 4 Consider a vector field
F(x, y , z) = (yz + 2 cos 2x, xz − e z , xy − ye z ).
(i) Verify that the field F is conservative.
Since F is a smooth vector field on the entire space, it is
conservative if and only if its Jacobian matrix is symmetric
everywhere in R3 . For vector fields on R3 , this is equivalent
to curl(F) = 0. We have to verify three identities.
∂F1
∂F2
∂
∂
=
:
(yz +2 cos 2x) =
(xz −e z ) ⇐⇒ z = z,
∂y
∂x
∂y
∂x
∂F3
∂
∂
∂F1
=
:
(yz +2 cos 2x) =
(xy −ye z ) ⇐⇒ y = y ,
∂z
∂x
∂z
∂x
∂F2
∂F3
∂
∂
=
:
(xz − e z ) =
(xy − ye z )
∂z
∂y
∂z
∂y
⇐⇒ x − e z = x − e z .
Problem 4 Consider a vector field
F(x, y , z) = (yz + 2 cos 2x, xz − e z , xy − ye z ).
(ii) Find a function f such that F = ∇f .
We are looking for a function f : R3 → R such that
∂f
∂f
∂f
= yz + 2 cos 2x,
= xz − e z ,
= xy − ye z .
∂x
∂y
∂z
Integrating the third
ˆ equality by z, we get
f (x, y , z) =
(xy − ye z ) dz = xyz − ye z + g (x, y ).
Substituting this into the other equalities, we obtain that
yz + gx′ = yz + 2 cos 2x and xz − e z + gy′ = xz − e z .
Hence gy′ = 0 so that g does not depend on y . Since
gx′ = 2 cosˆ2x, we obtain that
g (x, y ) = 2 cos 2x dx = sin 2x + c, where c is a constant.
Finally, f (x, y , z) = xyz − ye z + sin 2x + c.
Problem 4 Consider a vector field
F(x, y , z) = (yz + 2 cos 2x, xz − e z , xy − ye z ).
(ii) Find a function f such that F = ∇f .
Alternative solution:ˆ If F = ∇f , then
F · d s = f (B) − f (A)
x
for any points A, B ∈ R3 and any path x joining A to B.
We can use this relation to recover the function f .
For any given point A = (x, y , z) we consider a linear path xA
from the origin to A, xA : [0, 1] → R3 , xA (t) = (tx, ty , tz).
Then
ˆ
ˆ 1
f (A) − f (0) =
F · ds =
F(xA (t)) · x′A (t) dt.
xA
0
f (A) − f (0) =
ˆ
F · ds =
xA
1
=
ˆ
1
=
ˆ
1
=
ˆ
ˆ
1
F(xA (t)) · x′A (t) dt
0
(t 2 yz + 2 cos 2tx, t 2 xz − e tz , t 2 xy − tye tz ) · (x, y , z) dt
0
0
(t 2 yz + 2 cos 2tx)x + (t 2 xz − e tz )y + (t 2 xy − tye tz )z dt
(3t 2 xyz + 2x cos 2tx − ye tz − tyze tz ) dt
0
1
= t 3 xyz t=0
1
+ sin 2tx t=0
1
− yte tz t=0
= xyz + sin 2x − ye z .
Thus f (x, y , z) = xyz + sin 2x − ye z + c, where c = f (0) is
a constant.
Problem 5 Let C be a solid cylinder bounded by planes
z = 0, z = 2 and a cylindrical surface x 2 + y 2 = 1. Orient
the boundary ∂C with outward normals and evaluate a surface
‹
integral
(x 2 e1 + y 2 e2 + z 2 e3 ) · d S.
∂C
By Gauss’ Theorem,
‹
˚
2
2
2
(x e1 + y e2 + z e3 )·d S =
∇· (x 2 e1 + y 2e2 + z 2 e3 ) dV
∂C
C
=
˚ =
˚
C
C
∂ 2
∂ 2
∂ 2
(x ) +
(y ) +
(z ) dx dy dz
∂x
∂y
∂z
2(x + y + z) dx dy dz.
Consider an invertible linear transformation L : R3 → R3
given by L(x, y , z) = (−x, −y , z). The matrix of L (relative
to the standard basis) is 

−1
0 0
M =  0 −1 0 .
0
0 1
It is also the Jacobian matrix of L at every point. Changing
coordinates from (x, y , z) to (u, v , w ) so that
(x, y , z) = L(u, v , w ), we obtain
˚
˚
2(x + y ) dx dy dz =
2(−u − v ) |det M| du dv dw
C
L−1 (C )
˚
=−
2(u + v ) du dv dw .
C
It follows that
˚
2(x + y ) dx dy dz = 0.
C
By linearity of the integral,
‹
˚
2
2
2
(x e1 + y e2 + z e3 )·d S =
2(x + y + z) dx dy dz
∂C
C
˚
˚
˚
=
2(x +y ) dx dy dz +
2z dx dy dz =
2z dx dy dz.
C
C
C
The cylinder C can be represented as C = U × [0, 2], where
U is the unit disc in the plane,
U = {(x, y ) ∈ R2 : x 2 + y 2 ≤ 1}.
By Fubini’s Theorem,
˚
¨ ˆ 2
2z dx dy dz =
2z dz dx dy
C
U
=
¨
U
0
4 dx dy = 4 area(U) = 4π.
Problem 6 Let D be a region in R3 bounded by a paraboloid
z = x 2 + y 2 and a plane z = 9. Let S denote the part of the
paraboloid that bounds D, oriented by outward normals.
Evaluate a surface integral
¨
curl(F) · d S,
S
where F(x, y , z) = (e
We have curl F =
x 2 +z 2
∂F
3
∂y
= (xze xyz − x − y , 2ze x
, xy + xz + yz, e xyz ).
−
2 +z 2
∂F2 ∂F1 ∂F3 ∂F2 ∂F1 ,
−
,
−
∂z
∂z
∂x ∂x
∂y
− yze xyz , y + z).
Direct evaluation of the surface integral seems problematic.
By Stokes’ Theorem, the surface integral equals the integral of
the field F along the circle ∂S. However evaluation of this
line integral seems problematic as well.
By the corollary of Stokes’ Theorem,
¨
curl(F)·d S = 0.
∂D
It follows that
¨
curl(F)·d S = −
S
¨
curl(F)·d S.
∂D\S
We observe that ∂D \ S is a horizontal disc Q ×{9}, where
Q = {(x, y ) ∈ R2 : x 2 + y 2 ≤ 9}. It is oriented by the upward
normal vector n = (0, 0, 1). Now
¨
¨
curl(F)·d S =
curl(F)·n dS
∂D\S
∂D\S
¨
¨
=
(y + 9) dx dy =
9 dx dy = 9 area(Q) = 81π.
Q
Thus
¨
curl(F)·d S = −81π.
S
Q