Solutions - Homework sections 17.7-17.9
17.7
6.
Evaluate
RR
S
xy dS, where S is the triangle with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 2).
The three points - and therefore the triangle between them - are on the plane 2x + y + z = 2, and
their projection onto the xy-plane forms the triangle between the points (1, 0, 0), (0, 2, 0), and (0, 0, 0).
So S is the portion of the plane z = 2−2x−y over the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2−2x}.
Since we have z as a function of x and y on our surface S, we can use formula 4 from this section:
ZZ
ZZ
p
xy dS =
xy · (−2)2 + (−1)2 + 1 dA
S
D
√ Z 1Z
=
6
√ Z
6
xy dydx
0
0
=
2−2x
1
0
xy 2
2
2−2x
dx
0
√ Z 1
(x − 2x2 + x3 ) dx
= 2 6
0
√
1 2 2 3 1 4
x − x + x
2
3
4
√ 1
= 2 6·
2
1
= √ .
6
=
1
2 6
0
9.
RR
Evaluate S yz dS, where S is the surface with parametric equations x = u2 , y = u sin v, z = u cos v, for
0 ≤ u ≤ 1, 0 ≤ v ≤ π/2.
Our parametrization is r(u, v) = hu2 , u sin v, u cos vi, so:
ru
=
h2u, sin v, cos vi;
rv
=
h0, u cos v, −u sin vi;
ru × rv
=
h−u sin2 v − u cos2 v, 2u2 sin v, 2u2 cos vi
= h−u, 2u2 sin v, 2u2 cos vi;
p
|ru × rv | =
u2 + 4u4 sin2 v + 4u2 cos2 v
p
=
u2 + 4u4
p
= u 1 + 4u2 .
So, using formula 2 from this section,
ZZ
ZZ
yz dS =
y(u, v)z(u, v)|ru × rv | dA
S
D
Z 1Z
π/2
=
0
0
Z 1Z
π/2
sin v cos v · u3
=
0
3
=
u
p
0
=
1
2
p
4u2 + 1 dvdu
0
1
Z
p
u cos v · u sin v · u 4u2 + 1 dvdu
Z
1
u3
4u2
1
+1
sin2 v
2
p
4u2 + 1 du
0
1
π/2
du
0
Substitute
:
=
=
=
=
1
1
t = 4u2 + 1, u2 = (t − 1), dt = u du : changing the bounds, we get:
4
8
Z
√ 1
1 51
(t − 1) · t · dt
2 1 4
8
Z 5
1
t3/2 − t1/2 dt
64 1
5
1 2 5/2 2 3/2
t − t
64 5
3
1
1
5√
5+
.
48
240
11.
Evaluate
RR
S
x2 z 2 dS, where S is the part of the cone z 2 = x2 + y 2 between the planes z = 1 and z = 3.
The widest point of S is at the intersection of the cone and the plane z = 3, where x2 + y 2 = 32 = 9;
its thinnest point is where x2 + y 2 = 12 = 1. p
Thus, S is the portion of the surface z = x2 + y 2 over the region D = {(x, y) : 1 ≤ x2 + y 2 ≤ 9}.
So:
v
!2
!2
u
ZZ
ZZ
u
p
y
x
+ p
+ 1 dA
x2 z 2 dS =
x2 ( x2 + y 2 )2 · t p
x2 + y 2
x2 + y 2
S
D
s
ZZ
x2 + y 2
=
x2 (x2 + y 2 )
+ 1 dA
x2 + y 2
D
ZZ
√
=
x2 (x2 + y 2 ) · 2 dA
D
=
√ Z
2
2πZ 3
0
=
√ Z
2
=
√ Z
2
2πZ 3
0
=
=
r5 cos2 θ drdθ
1
2π
0
=
(r cos θ)2 (r2 )r drdθ
1
r6
6
3
dθ
1
Z 2π
364 √
2
cos2 θ dθ
3
0
2π
1
364 √ 1
2 θ + sin 2θ
3
2
4
0
√
364 2
π.
3
R
(Use the Reference Pages in the back of your textbook to evaluate cos2 θ dθ.)
13.
Evaluate
RR
S
y dS, where S is the part of the paraboloid y = x2 + z 2 inside the cylinder x2 + z 2 = 4.
We already have y as a function of the other two variables, and we want to use x and z as parameters.
The projection of S onto the xz-plane is the disk D = {(x, z) : x2 + z 2 ≤ 4}.
ZZ
s
ZZ
y dS
=
y
S
D
ZZ
=
Z
2
+
∂y
∂z
2
+ 1 dA
p
(x2 + z 2 ) 4x2 + 4z 2 + 1 dA
D
2πZ 2
=
0
∂y
∂x
r2
0
2
p
4r2 + 1r drdθ
1
1
(u − 1), and du = r dr: changing the bounds, this is:
4
8
Z 2πZ 17
√ 1
1
(u − 1) · u · dudθ
4
8
1
0
Z 2πZ 17
1
u3/2 − u1/2 dudθ
32 0 1
17
Z 2π 1
2 5/2 2 3/2
u − u
dθ
32 0
5
3
1
Z 2π
1 2 5/2 2 3/2 2 2
( 17 − 17 − + ) ·
dθ
32 5
3
5 3
0
√
π
(391 17 + 1).
60
Make a substitution: u = 4r2 + 1; r2 =
=
=
=
=
=
15.
Evaluate
RR
S
(x2 z + y 2 z)dS, where S is the hemisphere x2 + y 2 + z 2 = 4, z ≥ 0.
Parametrize the hemisphere in spherical coordinates: S is parametrized by r(φ, θ) = h2 sin φ cos θ, 2 sin φ sin θ, 2 cos φi,
where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
rφ
= h2 cos φ cos θ, 2 cos φ sin θ, −2 sin φi;
rθ
= h−2 sin φ sin θ, 2 sin φ cos θ, 0i;
rφ × rθ
= h4 sin2 φ cos θ, 4 sin2 φ sin θ, 0i;
|rφ × rθ | =
4 sin φ.
So,
ZZ
(x2 z + y 2 z) dS
ZZ
(x2 + y 2 )z dS
=
S
Z
S
π/2
Z
2π
0
π/2
Z
(4 sin2 φ cos2 θ + 4 sin2 φ sin2 θ)(2 cos φ)(4 sin φ) dθdφ
=
0
Z
=
0
Z
=
2π
32
sin3 φ cos φ dθdφ
0
π/2
32 · 2π
sin3 φ cos φ dθdφ
0
1
= 64π sin4 φ
4
= 16π.
π/2
dφ
0
16.
RR
Evaluate S xz dS, where S is the boundary of the region enclosed by the cylinder y 2 + z 2 = 9 and the
planes x = 0 and x + y = 5.
S has three parts: S1 is the part of S that lies on the cylinder, S2 is the part of x + y = 5 within the
cylinder, and S3 is the part of x = 0 within the cylinder.
Since we can’t come up with one parametrization that describes the entire surface, we’re going to
parametrize each of these three parts separately, and then combine them at the end.
• S1 : S1 is parametrized by r(u, v) = hu, 3 cos v, 3 sin vi, where 0 ≤ v ≤ 2π. We want the range of x
to be 0 ≤ x ≤ 5 − y, so we use 0 ≤ u ≤ 5 − 3 cos v.
This gives us ru = h1, 0, 0i and rv = h0, −3 sin v, 3 cos vi, so ru × rv = h0, −3 cos v, −3 sin vi, and
3
finally |ru × rv | = 3. So:
ZZ
2πZ 5−3 cos v
Z
xz dS
u(3 sin v) · 3 dudv
=
S1
0
2π
0
=
=
Z
9
2
0.
(5 − 3 cos v)2 sin v dv
0
• S2 : On S2 , x = 5 − y. Use y and z as parameters:
= h5 − y, y, zi;
r(y, z)
ry
= h−1, 1, 0i;
rz
= h0, 0, 1i;
ry × rz
=
2;
√
|ry × rz | =
2.
The projection of S2 onto the yz-plane is D = {(y, z) : y 2 + z 2 ≤ 9}, so:
ZZ
ZZ
√
xz dS =
(5 − y)z 2 dS
S2
D
√ Z
2
=
2πZ 3
(5 − r cos θ)(r sin θ)r drdθ
0
0
=
√ Z
2
2πZ 3
0
√ Z
=
2
0
2π
0
√ Z
2
(5r2 − r3 cos θ)(sin θ) drdθ
5 3 1 4
r − r cos θ
3
4
3
sin θ dθ
0
2π
81
cos θ) sin θ dθ
4
0
2π
√
1
81
4
=
· (45 −
cos θ)2
2·
81 2
4
0
= 0.
=
(45 −
• S3 : On S3 , x = 0, and so xz = 0.
RR
RR
So S3 xz dS = S3 0 dS = 0.
Put the three parts together:
ZZ
xz dS
ZZ
ZZ
=
ZZ
xz dS +
S
S1
=
0+0+0
=
0.
xz dS +
S2
xz dS
S3
19.
RR
Evaluate S F · dS, for F(x, y, z) = xyi + yzj + zxk, where S is the part of the paraboloid z = 4 − x2 − y 2
above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, with upward orientation.
∂g
∂g
z = g(x, y) = 4 − x2 − y 2 ,
= −2x, and
= −2y.
∂x
∂y
Setting D to be the square [0, 1] × [0, 1], we can use formula 10 from this section:
ZZ
ZZ
F · dS =
(−xy(−2x) − yz(−2y) + zx) dA
S
D
Z 1Z
=
0
Z
=
=
1
(2x2 y + 2y 2 (4 − x2 − y 2 ) + x(4 − x2 − y 2 )) dydx
0
1
1
11
34
( x2 + x − x3 + ) dx
3
3
15
0
713
.
180
4
20.
RR
Evaluate S F · dS, for F(x, y, z) = yi + xj + z 2 k, where S is the helicoid r(u, v) = u cos vi + u sin vj + vk,
0 ≤ u ≤ 1, 0 ≤ v ≤ π, oriented upwards.
ru = hcos v, sin v, 0i, and rv = h−u sin v, u cos v, 1i, so ru × rv = hsin v, − cos v, ui.
Also, F(r(u, v)) = hu sin v, u cos v, v 2 i. Therefore, by formula 9 from this section:
ZZ
ZZ
F · dS =
F · (ru × rv ) dA
S
D
Z πZ
1
(u sin2 v − u cos2 v + uv 2 ) dudv
=
0
0
Z πZ
=
=
−u cos 2v + uv 2 ) dudv
0
π
0
=
1
Z
1
(− cos 2v + v 2 ) dv
2 0
1 3
π .
6
21.
RR
Evaluate S F · dS, for F(x, y, z) = xzey i − xzey j + zk, S the part of the plane x + y + z = 1 in the first
octant, oriented downwards.
z = g(x, y) = 1 − x − y, and the projection of S onto the xy-plane is the triangle D = {(x, y) : 0 ≤
x ≤ 1, 0 ≤ y ≤ 1 − x}.
We can use formula 9, but because S is oriented downwards in this case, we add a minus sign at the
beginning.
ZZ
ZZ
F · dS
(−xzey (−1) − (−xzey )(−1) + z) dA
= −
S
D
ZZ
= −
z dA
Z
D
1Z 1−x
= −
(1 − x − y) dydx
0
Z
= −
0
0
1
1
1
( x2 − x + ) dx
2
2
1
= − .
6
22.
p
RR
Evaluate S F · dS, for F(x, y, z) = xi + yj + z 4 k, where S is the part of the cone z = x2 + y 2 below
the plane z = 1, oriented downwards.
p
∂g
∂g
x
y
z = g(x, y) = x2 + y 2 , so
=p
and
=p
.
2
2
2
∂x
∂y
x +y
x + y2
The projection of S onto the xy-plane is the disk D = {(x, y) : x2 + y 2 ≤ 1}.
Again, we can use formula 9, but because S is oriented downwards, we add a minus sign to the
5
beginning:
ZZ
ZZ
F · dS =
−
(−x
S
p
D
−
D
2πZ 1
Z
=
−
(
0
Z
=
0
2πZ 1
−
2π
= −
0
=
=
p
x2 + y 2
!
+ z 4 ) dA
dA
−r2
+ r4 )r drdθ
r
(r5 − r2 ) drdθ
0
0
Z
−y
x2 + y 2
!
y
p
−(x2 + y 2 )
p
+ ( x2 + y 2 )4
x2 + y 2
ZZ
=
!
x
1 1
( − ) dθ
6 3
1
· 2π
6
π
.
3
17.8
2.
RR
Evaluate S curl F·dS, for F(x, y, z) = 2y cos zi+ex sin zj+xey k, where S is the hemisphere x2 +y 2 +z 2 =
9, z ≥ 0, oriented upward.
The boundary curve C of S is the circle x2 + y 2 = 9, z = 0, oriented counterclockwise looking down
from above. We can parametrize C as r(u, v) = h3 cos t, 3 sin t, 0i.
This gives us F(r(t)) = h6 sin t, 0, 3 cos t · e3 sin t i, and r0 (t) = h−3 sin t, 3 cos t, 0i.
Via Stokes’s Theorem:
ZZ
Z
curl F · dS =
F · dr
S
C
2π
Z
F(r(t)) · r0 (t) dt
=
0
Z
=
2π
−18 sin2 t dt
0
= −18π.
(To evaluate
R
2
sin t dt, see the Reference Pages in the back of the textbook.)
3.
RR
Evaluate S curl F · dS, for F(x, y, z) = x2 z 2 i + y 2 z 2 j + xyzk, where S is the part of the paraboloid
z = x2 + y 2 inside the cylinder x2 + y 2 = 4, oriented upward.
The boundary curve C of S is the intersection of the paraboloid and the cylinder, that is, z =
x2 + y 2 = 4: the circle of radius 2 at the height z = 4. To match the orientation of S, C should be
oriented counterclockwise as seen from above.
So, we can parametrize C as r(t) = h2 cos t, 2 sin t, 4i; this gives us r0 (t) = h−2 sin t, 2 cos t, 0i, and
F(r(t)) = h64 cos2 t, 64 sin2 t, 16 sin t cos ti. So, by Stokes’s Theorem:
ZZ
Z
curl F · dS =
F · dr
S
C
2π
Z
F(r(t)) · r0 (t) dt
=
0
Z
=
2π
(−128 cos2 t sin t + 128 sin2 t cos t) dt
0
1
1
cos3 t + sin3 t
= 128
3
3
= 0.
6
2π
0
5.
RR
Evaluate S curl F · dS, for F(x, y, z) = xyzi + xyj + x2 yzk, where S is the top and four sides, but not
the bottom, of the cube with vertices (±1, ±1, ±1), oriented outward.
EquationRR3 in this section RR
says that, if two surfaces S1 and S2 have the same oriented boundary
curve, then S1 curl F · dS1 = S2 curl F · dS2 .
So, instead of working the integral over S, which we would have to do in five parts, or over the
boundary curve C, which is a square, and would have to be done in four parts, we’re going to pick
another surface with the same boundary curve, and work the integral over that surface.
Let S 0 be the missing bottom face of the cube: it’s a solid square in the plane z = −1. Since the top
of the cube is oriented outward, we orient S 0 upwards, such that n = k, so that both surfaces orient the
boundary C the same way.
curl F = hx2 z, xy − 2xyz, y − xzi.
Since n = k = h0, 0, 1i on S 0 , curl F · n = y − xz = x + y, since z = −1 on S 0 .
So, using equation 3:
ZZ
ZZ
curl F · dS =
curl F · dS0
S
S0
ZZ
=
curl F · n dS 0
Z
S0
1Z 1
(x + y) dxdy
=
−1 −1
Z 1
=
2y dy
−1
=
0.
8.
R
Evaluate C F · dr, where F(x, y, z) = e−x i + ex j + ez k, and C is the boundary of the part of the plane
2x + y + 2z = 2 in the first octant.
1
Let S be the portion of the plane bounded by C. Then S is the part of the graph of z = 1 − x − y
2
located over the triangle D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 − 2x}.
curl F = h0, 0, ex i.
So, using Stokes’s Theorem and formula 10 from section 17.7,
Z
ZZ
F · dr =
curl F · dS
C
Z ZS
=
(0 + 0 + ex ) dA
D
Z 1Z
2−2x
=
0
Z
=
ex dydx
0
1
(2 − 2x)ex dx
0
=
(integrate by parts, u = 2 − 2x, dv = ex )
=
[(2 − 2x)ex + 2ex ]0
=
2e − 4.
1
11.a)
R
Evaluate C F · dr, where F(x, y, z) = x2 zi + xy 2 j + z 2 k, and C is the curve of intersection of the plane
x + y + z = 1 and the cylinder x2 + y 2 = 9, oriented counterclockwise from above.
The intersection C is an ellipse in the plane x + y + z = 1; thus, it has unit normal vector n =
1
√ (i + j + k).
3
1
curl F = h0, x2 , y 2 i, so curl F · n = √ (x2 + y 2 ).
3
7
Finally, if S is the part of the plane contained inside the ellipse C, the projection of S onto the
xy-plane is the disk D = {(x, y) : x2 + y 2 ≤ 9}. Over D, S is given by z = g(x, y) = 1 − x − y, so
∂g
∂g
=
= −1.
∂x
∂y
So, by Stokes’s Theorem,
Z
ZZ
F · dr =
curl F · n dS
C
Z ZS
1
√ (x2 + y 2 ) dS
=
3
Z ZS
p
1 2
√ (x + y 2 ) (−1)2 + (−1)2 + 1 dA
=
3
Z ZD
2
=
(x + y 2 ) dA
Z
D
2πZ 3
=
0
0
Z
r3 drdθ
2π
=
0
81
dθ
4
81π
.
2
=
13.
Verify Stokes’s Theorem for F(x, y, z) = y 2 i + xj + z 2 k, S the part of the paraboloid z = x2 + y 2 below
the plane z = 1, oriented upward.
The projection of S onto the xy-plane is D = {(x, y) : x2 + y 2 ≤ 1}.
curl F = h0, 0, 1 − 2yi.
So, by equation 10 from section 17.7,
ZZ
ZZ
curl F · dS =
(0 + 0 + 1 − 2y) dA
S
D
ZZ
=
(1 − 2y) dA
Z
D
2πZ 1
(1 − 2r sin θ) · r drdθ
=
0
Z
0
2π
=
0
1 2
( − sin θ) dθ
2 3
= π.
Also, the boundary curve C of S is the circle x2 + y 2 = 1, z = 1, oriented counterclockwise from
above. So a parametrization is r(t) = hcos t, sin t, 1i, 0 ≤ t ≤ 2π.
This gives us r0 (r) = h− sin t, cos t, 0i, and F(r(t)) = hsin2 t, cos t, 1i.
So,
Z
Z 2π
F · dr =
F(r(t)) · r0 (t) dt
C
0
Z
2π
(cos2 t − sin3 t) dt
=
0
2π
Z 2π
1
=
(1 + cos 2t) dt −
(1 − cost ) sin t dt
2
0
0
2π 2π
1
1
1
=
t + sin 2t
− − cos t + cos3 t
2
2
3
0
0
= π.
Z
Therefore,
R
C
F · dr =
RR
S
curl F · dS, and Stokes’s Theorem holds.
8
17.
A particle moves along line segments from the origin to (1, 0, 0) to (1, 2, 1) to (0, 2, 1) and back to the
origin, under the force field F(x, y, z) = z 2 i + 2xyj + 4y 2 k.
Find the work done.
R
If C is the path the particle follows, then the work done is C F · dr.
1
All four points are in the plane z = y. So, if S is the flat surface with boundary C, S is the portion
2
1
of the plane z = y over the rectangle D = [0, 1] × [0, 2], oriented upward (to match the direction in
2
which the particle moves around C).
curl F = h8y, 2z, 2yi.
So, using equation 10 from section 17.7, the work done is
Z
ZZ
F · dr =
curl F · dS
C
Z ZS
1
=
(−8y(0) − 2z( ) + 2y) dA
2
Z ZD
=
(2y − z) dA
D
ZZ
1
=
(2y − y) dA
2
Z ZD
3
y dA
=
D 2
Z 1Z 2
3
=
y dydx
0 0 2
2
Z 1
3 2
=
y
dx
4
0
0
Z 1
=
3 dx
0
=
3.
18.
Evaluate
R
C
(y+sin x)dx+(z 2 +cos y)dy+x3 dz, where C is the curve r(t) = hsin t, cos t, sin 2ti, 0 ≤ t ≤ 2π.
R
This is C F · dr, for F(x, y, z) = hy + sin x, z 2 + cos y, x3 i, giving us curl F = h−2z, −3x2 , −1i.
C is on the surface z = 2xy; let S be the part of this surface bounded by C. The projection of S
onto the xy-plane is D = {(x, y) : x2 + y 2 ≤ 1}.
Looking down from above, C is oriented counterclockwise; to match this, our surface S needs to be
oriented downwards.
Since z = g(x, y) = 2xy, we can use equation 10 from the last section, adding a minus sign on the
front because of the downward orientation:
Z
ZZ
F · dr = −
curl F · dS
C
S
ZZ
= −
(−(−4xy)(2y) − (−3x2 )(2x) − 1) dA
D
ZZ
= −
(8xy 2 + 6x3 − 1) dA
Z
D
2πZ 1
= −
0
Z
(8r3 cos θ sin2 θ + 6r3 cos3 θ − 1) · r drdθ
0
2π
8
6
1
( cos θ sin2 θ + cos3 θ − ) dθ
5
5
2
0
2π
8
6
1
1
= −
sin3 θ + (sin θ − sin3 θ) − θ
15
5
3
2 0
= π.
= −
9
19.
If S is a sphere and F satisfies the hypotheses of Stokes’s Theorem, show that
RR
S
curl F · dS = 0.
On its own, a sphere S doesn’t have a boundary curve to which we can apply Stokes’s Theorem. So,
cut the sphere in half: let S1 be the upper half of the sphere, and S2 the lower half, such that S = S1 ∪S2 .
Let C1 be the boundary curve of S1 , and let C2 be the boundary curve of S2 .
Then:
ZZ
ZZ
ZZ
curl F · dS =
curl F · dS1 +
curl F · dS2
S
S
S2
Z 1
Z
=
F · dr +
F · dr
C1
C2
But C1 and C2 are the same circle, oriented in opposite directions: C1 is oriented counterclockwise from
above, to match the orientation ofRRthe top half of Rthe sphere, Rwhile C2 is oriented
clockwise,
to match
R
R
the bottom half of the sphere. So S curl F · dS = C1 F · dr + C2 F · dr = C1 F · dr + − C1 F · dr = 0.
17.9
Calculate
RR
S
F · dS:
5:
F(x, y, z) = ex sin yi + ex cos yj + yz 2 k;
S is the surface of the box bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 2.
∂Q ∂R
∂P
+
+
= (ex sin y) + (−ex sin y) + (2yz) = 2yz.
∂x
∂y
∂z
Setting E to be the box of which S is the boundary, we can use the Divergence Theorem:
div F =
ZZ
ZZZ
F · dS =
div F dV
S
Z
E
1Z 1Z 2
=
2yz dzdydx
0
0
Z 1Z
0
1
=
0
0
Z 1Z
=
2 2
yz 0 dydx
1
4y dydx
0
Z
0
1
=
0
Z
2 1
2y 0 dx
1
=
2 dx
0
=
2.
7.
F(x, y, z) = 3xy 2 i + xez j + z 3 k;
S the surface of the solid bounded by y 2 + z 2 = 1, x = −1, and x = 2.
div F = 3y 2 + 0 + 3z 2 = 3y 2 + 3z 2 .
Let E be the solid bounded by S, and use the Divergence Theorem:
10
ZZ
ZZZ
F · dS
=
S
div F dV
Z Z ZE
3(y 2 + z 2 ) dV
=
E
2
Z
2π Z 1
Z
=
−1
2
Z
0
0
2π
Z
=
−1
=
=
=
=
3
4
Z
0
2
3r2 · r drdθdx
Z
3 4
r
4
dθdx
2π
1 dθdx
−1
0
Z 2
3
· 2π
1 dx
4
−1
3
· 2π · 3
4
9π
.
2
9.
F(x, y, z) = xy sin zi + cos(xz)j + y cos zk;
S is x2 /a2 + y 2 /b2 + z 2 /c2 = 1.
div F = y sin z + 0 − y sin z = 0.
So, if E is the volume inside the ellipsoid,
ZZ
ZZZ
F · dS
=
div F dV
S
Z Z ZE
=
0 dV
E
=
0.
11.
F(x, y, z) = (cos z + xy 2 )i + xe−z j + (sin y + x2 z)k;
S is the surface of the solid bounded by z = x2 + y 2 and z = 4.
div F = y 2 + 0 + x2 = x2 + y 2 .
If E is the solid bounded by S, then we want to think about E in cylindrical coordinates: the
paraboloid and the plane intersect where x2 + y 2 = z = 4. Inside E, z values range from x2 + y 2 = r2
to 4.
So:
ZZ
ZZZ
F · dS =
x2 + y 2 dV
S
Z
E
2π Z 2 Z 4
=
0
Z
0
2π Z 2
=
0
Z
2π Z 2
0
Z
4
r2
drdθ
4r3 − r5 drdθ
2π
2
r6
r4 −
dθ
6 0
2π
32
dθ
3
0
=
0
=
r3 z
0
=
Z
0
=
r2 · r dzdrdθ
r2
32π
.
3
11
13.
F(x, y, z) = 4x3 zi + 4y 3 zj + 3z 4 k;
S is the sphere of radius R centered at the origin.
div F = 12x2 z + 12y 2 z + 12z 3 = 12z(x2 + y 2 + z 2 ).
Letting E be the volume inside the sphere:
ZZ
ZZZ
F · dS
12z(x2 + y 2 + z 2 ) dV
=
S
E
Z
2π Z π Z R
=
0
Z
=
12ρ cos φ · ρ2 · ρ2 sin φ dρdφdθ
0
0
2π
12 ·
Z
π
Z
dθ ·
0
0
2π
[θ]0
1
·
sin2 φ
2
12 ·
=
12 · 2π · 0 ·
=
0.
ρ5 dρ
0
=
R
cos φ sin φ dφ ·
π R
1 6
· ρ
6
0
0
R6
6
17.
1
F · dS, where F(x, y, z) = z 2 xi + ( y 3 + tan x)j + (x2 z + y 2 )k
3
and S is the top half of the sphere x2 + y 2 + z 2 = 1.
Use the Divergence Theorem to evaluate
RR
S
S by itself is not the boundary of a simple solid region - the bottom of S is open. So, we’re going to
add to S so that it does enclose a simple solid region, and then use the Divergence Theorem on that.
To fill in the bottom of S, we need the “missing” disk. Let S1 be the disk x2 + y 2 ≤ 1, z = 0.
Since S is part of the unit sphere, S is oriented outwards; to match that, we make S1 be oriented
downwards.
RR
RR
RR
Then if we set S2 = S ∪ S1 , S2 is a closed surface, and S F · dS = S2 F · dS − S1 F · dS.
Since S1 is oriented downward, the normal vector on S1 points straight downward: n = −k. So
F · n = F · −k = −x2 z − y 2 = −y 2 , since z = 0 on S1 .
Let D be the unit disk. Then:
ZZ
ZZ
F · dS =
F · n dS
S1
S
ZZ 1
=
(−y 2 )dA
D
2π Z 1
Z
=
0
Z
−(r sin θ)2 · r drdθ
0
2π Z 1
= −
0
Z
1
4 0
1
− π
4
= −
=
r3 sin2 θ drdθ
0
2π
sin2 θ dθ
S2 is a closed surface, so we can use the Divergence Theorem on it.
div F = z 2 + y 2 + x2 .
12
Let E be the volume inside the half-sphere S2 encloses. Then:
ZZ
ZZZ
F · dS =
div FdV
S2
E
ZZZ
=
(x2 + y 2 + z 2 )dV
E
πZ
2πZ
2
Z
=
1
5
=
=
RR
S
π
2 sin φ dρdθ
2πZ
0
0
Z
1 2π
dθ
5 0
2
π.
5
=
So, putting the two parts together,
Z
ρ2 · ρ2 sin φ dρdφdθ
0
0
0
1
F · dS =
RR
S2
F · dS −
RR
S1
F · dS =
1
13
2
π − (− π) =
π.
5
4
20
24.
(2x + 2y + z 2 ) dS, where S is the sphere x2 + y 2 + z 2 = 1:
RR
RR
To get this in the right form, we need to find a vector field F such that S F·n dS = S (2x+2y +z 2 ) dS,
so we want F · n = 2x + 2y + z 2 .
xi + yj + zk
(you can show this using the φ and θ
In general, on a sphere, the normal vector is p
x2 + y 2 + z 2
xi + yj + zk
parametrization of the sphere). For S, the sphere of radius 1, then, n = p
= xi + yj + zk.
x2 + y 2 + z 2
So, if F · n = 2x + 2y + z 2 , F = 2i + 2j + zk.
∴ div F = 0 + 0 + 1 = 1.
Use the Divergence Theorem to evaluate
RR
S
So, let E be the volume inside S. Then:
ZZ
(2x + 2y + z 2 ) dS
ZZ
F · n dS
=
S
Z ZS
F · dS
=
Z ZSZ
=
1 dV
E
=
since
4
π is the volume of the ball of radius 1.
3
13
4
π,
3