53 Math Homework 12 Solutions
16.5 # 1: a) curl F =
b) div F =
F =
k
/x /y /z
xyz
0
x2 y
= (x2 0) (2xy xy ) + (0 xz )k = x2 + 3xy xz k.
F=
(xyz ) +
(0) +
(x2 y ) = yz + 0 + 0 = yz .
x
y
z
16.5 # 9: If we write F = P + Q, then P = 0, while Q is independent of x (and z ), and
decreases as y increases, so Q/y < 0.
P
Q
Q
a) div F =
+
=0+
< 0.
x
y
y
b) since we are considering a 2D vector eld, the x and y components of the curl vanish,
Q P
and curl F =
k = (0 0)k = 0.
x
y
(This illustrates the interpretation of divergence for velocity elds: the eld F corresponds
to a ow that compresses areas, hence div< 0. On the other hand, there is no spinning
motion, hence curl is zero).
16.5 # 11: If we write F = P + Q, then Q = 0, while P is independent of x (and z ), and
increases as y increases, so P/y > 0.
P
Q
a) div F =
+
= 0 + 0 = 0.
x
y
b) since we are considering a 2D vector eld, the x and y components of the curl vanish,
Q P
P
and curl F =
k=
k points in the negative z direction.
x
y
y
(This illustrates the interpretation of curl for velocity elds in terms of rotation, see also
exercise # 37. The ow described by F is neither compressing nor expanding, hence divergence is zero; however the shearing motion causes a particle placed in this velocity eld
to spin clockwise).
k
16.5 # 15:
F = /x
/y
/z = (2y 2y ) (0 0) + (2x 2x)k = 0.
2 + 2yz
2
2xy x
y
Since F is dened everywhere and curl F = 0, we conclude that F is conservative.
f (x, y, z ) is a potential function if it satises: fx = 2xy , fy = x2 + 2yz , fz = y 2 .
Integrating fx = 2xy , we get that f (x, y, z ) = x2 y + g (y, z ) for some function g (y, z ). Thus
fy = x2 + gy = x2 + 2yz , which implies that gy = 2yz , hence g (y, z ) = y 2 z + h(z ), and
f (x, y, z ) = x2 y + y 2 z + h(z ). Finally, fz = y 2 + h (z ) = y 2 , so h (z ) = 0, so h(z ) = c for
some constant c. Hence f (x, y, z ) = x2 y + y 2 z + c.
(f P ) (f Q) (f R)
+
+
=
x
y
z
= (f Px + fx P ) + (f Qy + fy Q) + (f Rz + fz R)
16.5 # 25: if F = P, Q, R then div(f F ) =
= f (Px + Qy + Rz ) + P, Q, R fx , fy , fz = f div(F ) + F
1
f.
16.5 # 33: Using Greens theorem in normal form, and the result of 16.5 # 25,
C f ( g ) n ds =
D div(f g ) dA =
D (f div( g ) + g f ) dA.
2 g (by denition, see p. 1066). Hence
However, div( g ) =
2 g dA =
Df
C f ( g ) n ds
D g f dA.
16.5 # 34: By the previous exercise, D f 2 g dA = C f ( g ) n ds D f g dA, and
similarly (switching f and g ), D g 2 f dA = C g ( f ) n ds D g f dA. Subtracting
these from each other, we get:
D (f
2g
g
2 f ) dA
=
C (f
=
C (f
g g f ) n ds
g g f ) n ds.
D(
f
g
g
f ) dA
16.5 # 36: Greens rst identity (the result of 16.5 # 33) for g = f reads:
2 f dA =
2
Df
C f f n ds
D | f | dA.
If we assume that f is harmonic in D, i.e. 2 f = 0, then the left hand side is zero. If
moreover we assume that f is zero at every point of C , then C f f n ds = C 0 ds = 0.
Therefore, we conclude that
2
D | f | dA = 0.
(Since we assume continuity of the partial derivatives of f , this implies that | f |2 = 0
everywhere in D, hence f = 0, hence f is constant in D; however, because f = 0 at
the boundary of D, this implies that f is zero everywhere in D; an important result about
harmonic functions.)
16.5 # 37: a) We know that the speed v at the given point P (at distance d from the z axis)
is related to the angular speed by = v/d, i.e. v = d. However, d = |r| sin where
r = OP and is as in the gure; so v = |r| sin = |w r|. Moreover v is perpendicular
to both w and r, and the right-hand rule holds (cf. gure), so v = w r.
k
b) From (a), v = w r = 0 0 = y, x, 0 .
xyz
k
c) curl v = v = /x /y /z
y
x
0
= (0 0) (0 0) +
x (x)
y (y )
k = ( ( ))k = 2 k = 2w.
16.6 # 13: The parametric equations for the surface are: x = u cos v , y = u sin v , z = v .
We look at the grid lines (the curves obtained by xing one of u and v and varying the other).
If we keep v constant and vary u, then x and y move along a straight line in the horizontal
plane z = v , intersecting the z -axis at (0, 0, v ) (for u = 0) and directed along the vector
cos v, sin v, 0 (a vector horizontal whose direction rotates with the height z = v ).
If we x u and vary v , then the projection to the xy -plane is a circle (x = u cos v , y = u sin v ,
so x2 + y 2 = u2 ) while z = v increases at a constant rate, so we obtain a helix.
Thus we obtain the surface shown on graph IV (called a helicoid).
16.6 # 23: The cone intersects the sphere in the circle x2 + y 2 = 2, z = 2, and we want
the portion of the sphere above this. Using x and y as parameters, we can parametrize the
surface as x = x, y = y , z = 4 x2 y 2 , where x2 + y 2 2.
2
Alternatively, in cylindrical coordinates, using r and as parameters, we can parametrize
the surface by x = r cos , y = r sin , z = 4 r2 , where 0 r 2, 0 2 .
Or, using spherical coordinates, where the sphere is = 2 and the cone is = /4: using
and as parameters, the parametrization becomes x = 2 sin cos , y = 2 sin sin ,
z = 2 cos , where 0 /4 and 0 2 .
16.6 # 25: This is a cylinder of radius 4 centered on the x-axis; its projection to the yz plane is the circle of radius 4, easiest to parametrize by y = 4 cos , z = 4 sin . Meanwhile,
x varies between 0 and 5. Hence: x = x, y = 4 cos , z = 4 sin , 0 x 5, 0 2 .
16.6 # 41: z = f (x, y ) = xy over the unit disk D : x2 + y 2 1. Since fx = y and fy = x,
2 1
A(S ) = D 1 + (fx )2 + (fy )2 dA = D 1 + y 2 + x2 dA = 0 0 1 + r2 r dr d.
1
1
1
So A(S ) = 2 0 r2 + 1 r dr = 2 3 (r2 + 1)3/2 0 = 23 (2 2 1).
16.6 # 43: z = f (x, y ) = y 2 x2 over the region D : 1 x2 + y 2 4. Then
2 2
A(S ) = D 1 + (fx )2 + (fy )2 dA = D 1 + 4x2 + 4y 2 dA = 0 1 1 + 4r2 r dr d.
2
2
1
So A(S ) = 2 1 1 + 4r2 r dr = 2 12 (1 + 4r2 )3/2 1 = (17 17 5 5).
6
Problem 1:
a) The vector F (x, y ) = x2 + xy = x(x + y) is parallel to the position vector x + y of
the point (x, y ), so it always points in the radial direction: straight away from the origin if
x > 0, straight towards the origin if x < 0, and it vanishes if x = 0.
In particular, F points outwards of the unit circle C for x > 0, and inwards for x < 0. So
the normal component F n, which is the integrand of the line integral for ux, is positive
for x > 0 (the right half of the unit circle), and negative for x < 0 (the left half).
b) The normal vector to C at (x, y ) is n = x + y. Therefore F n = x2 , xy x, y =
3 + xy 2 = x(x2 + y 2 ) = x. (Recall x2 + y 2 = 1 on the unit circle.) This is indeed positive
x
for x > 0 and negative for x < 0.
We parametrize the unit circle by the polar angle : so x = cos , y = sin , and the length
2
element is ds = d. Therefore C F n ds = C x ds = 0 cos d = 0.
We get zero because, by symmetry about the y -axis, the (negative) ux through the left
half-circle exactly compensates the (positive) ux through the right half-circle.
c) div F = x (x2 ) + y (xy ) = 3x. So by Greens theorem, we have
C
where R is the unit disk, and
tation in polar coordinates
is seen to be zero either by symmetry or by compu2
1
r cos r dr d =
x dA =
R
R
R
R x dA
2
x dA = 0
3x dA = 3
F n ds =
0
0
0
1
cos d = 0.
3
Problem 2:
(x2 + y 2 ) 2x2 (x2 + y 2 ) 2y 2
x
y
div F =
+
=
+
= 0.
x x2 + y 2
y x2 + y 2
(x2 + y 2 )2
(x2 + y 2 )2
So, for a < 1, the ux out of the circle Ca around (1, 0) of radius a is zero, because
Ca F n ds =
Ra div F dA = 0, where Ra is the disk enclosed by Ca .
3
Next, consider a > 1. The previous argument does not work because Ra contains the origin.
To avoid the origin, we remove it: consider the region R(a, b) = Ra Db , the disk Ra with
the disk Db of radius b around the origin removed. Let Tb be the circle of radius b around
the origin, oriented counterclockwise. For b suciently small, Tb is inside of Ra and the
oriented boundary of R(a, b) is Ca Tb (see gure below). By Greens theorem,
Ca
F n ds
Tb
F n ds =
div F dA = 0
R(a,b)
(One can apply Greens theorem because F and div F are dened in R(a, b).) Moreover,
for every b > 0,
1
1
F n ds =
ds = (2b) = 2
b
Tb
Tb b
because on Tb , n = x, y /b and F n = x/b2 , y/b2 x, y /b = (x2 + y 2 )/b3 = 1/b. Hence,
for a > 1,
Ca
F n ds =
Tb
F n ds = 2.
R(a, b)
Ra
Tb
Ca (a < 1)
Ca (a > 1)
4