Anda di halaman 1dari 16

# Section 14.

9
C14S09.001: It is easy to solve the given equations for
u+v u−v
x= , y= .
2 2
Hence
 1 
 1 
∂(x, y)  2 2

 1
= =− .
∂(u, v)  1 1  2
 − 
2 2

## C14S09.002: It is easy to solve the given equations for

u + 2v v − 3u
x= , y= .
7 7
Thus
 1 
 2 
∂(x, y)  7 7
 1

= = .
∂(u, v)  3 1  7
− 
7 7

C14S09.003: When we solve the equations u = xy and v = y/x for x and y, we ﬁnd that there are two
solutions:
 
u √ u √
x= , y = uv and x=− , y = − uv .
v v
It doesn’t matter which we choose; the value of the Jacobian will be the same. (Why?) So we choose the
ﬁrst solution. Then
 
 1 u1/2 
 − 
∂(x, y)  2u1/2 v 1/2 2v 3/2 
= 1 .
=  2v
∂(u, v)  v 1/2 u1/2 
 1/2

2u 2v 1/2

C14S09.004: When we solve the equations u = 2(x2 + y 2 ), v = 2(x2 − y 2 ) for x and y, we get four
solutions—all possible combinations of
√ √
u+v u−v
x=± , y=± .
2 2
We choose the solution for which x and y are both nonnegative. Then
 
 √ 1 1
√ 
 
∂(x, y)  
4 u+v 4 u+v
 1
= =− √ 2 .
∂(u, v)   8 u − v2
 1 1 
 √ − √ 
4 u−v 4 u−v

C14S09.005: When we solve the equations u = x + 2y 2 , v = x − 2y 2 for x and y, we get two solutions:

1

u+v u−v
x= , y=± .
2 2
We choose the solution for which y is nonnegative. Then
 1 1 
 
 
∂(x, y)  
2 2
 1
= =− √ .
∂(u, v)   4 u−v
 1 1 
 √ − √ 
4 u−v 4 u−v

C14S09.006: Given

2x 2y
u= , v=− , (1)
x2 + y2 x2 + y2

## note ﬁrst that

4x2 + 4y 2 4
u2 + v 2 = 2 2 2
= 2 ,
(x + y ) x + y2

so that

4
x2 + y 2 = . (2)
u2 + v 2
Therefore, using the equations in (1), then Eq. (2), we have

1 2u 1 2v
x= u(x2 + y 2 ) = 2 and y = − v(x2 + y 2 ) = − 2 .
2 u + v2 2 u + v2
Then
 
 2(v 2 − u2 ) 4uv 
 2 − 
 (u + v 2 )2 (u2 + v 2 )2 
∂(x, y)  
= 4
=  (u2 + v 2 )2 .
∂(u, v) 
 4uv 2(v 2 − u2 ) 
 2 
(u + v 2 )2 (u2 + v 2 )2

## C14S09.007: First we solve the equations u = x + y and v = 2x − 3y for

3u + v 2u − v
x= , y= .
5 5
Substitution in the equation x + y = 1 then yields

3u + v 2u − v 5u
1= + = = u.
5 5 5
Similarly, x + y = 2 yields u = 2, 2x − 3y = 2 yields v = 2, and 2x − 3y = 5 yields v = 5. Moreover,
 3 
 
1
 
∂(x, y)  5 
5
 1
= =− .
∂(u, v)   5
 2 1 
 − 
5 5
2
Therefore
  5  2
1 1 3
1 dy dx = du dv = 3 · 1 · = .
v=2 u=1 5 5 5
R

Note: Because R is a parallelogram with adjacent sides represented by the two vectors a =  35 , 25 , 0  and
b =  35 , − 35 , 0 , we have the following alternative method of ﬁnding the area A of R:
 
 i j k 

   
 3 2 
 0  = 0, 0, −
3
a×b =  ,
 5 5  5
 
 3 3 
 − 0
5 5
3
and therefore A = |a × b| = .
5
y
C14S09.008: Given u = xy and v = , we have
x
y u x
uv = xy · = y 2 and = xy · = x2 ,
x v y

## and thus we choose


u √
x= and y= uv . (1)
v

Then
 
 1 u1/2 
 − 
 2u1/2 v 1/2 2v 3/2 
∂(x, y)  
= 1 .
=
∂(u, v)   2v

 v 1/2 u1/2 
 
2u1/2 2v 1/2

## Also, if y = x, then substitution of the equations in (1) yields

u1/2 u
(uv)1/2 = ; uv = ; v 2 = 1.
v 1/2 v
So we choose v = 1. (This choice implies that if we have a similar choice with u, we must choose u > 0
because of the equations in (1).) Similarly, y = 2x yields v = 2, xy = 1 yields u = 1, and xy = 2 yields
u = 2. Hence the area of the region of Fig. 14.9.7 is
  2  2  2
1 1 1
A= 1 dx dy = du dv = dv = ln 2 ≈ 0.3465735902799727.
v=1 u=1 2v 1 2v 2
R

## C14S09.009: If u = xy and v = xy 3 , then

v v 1/2
uy 2 = xy 3 = v, so that y2 = ; y= .
u u1/2
Then

3
u u1/2 u3/2
x= = u · 1/2 = 1/2 .
y v v

(We do not need the solution in which x and y are negative.) Then
 
 3u1/2 u3/2 
 − 
 1/2 2v 3/2 
∂(x, y)  2v 
= 3 − 1 = 1 .
=
∂(u, v)   4v 4v
 2v
 v 1/2 1 
 − 3/2 
2u 2u1/2 v 1/2

## We also ﬁnd by substitution that xy = 2 corresponds to u = 2, xy = 4 corresponds to u = 4, xy 3 = 3

corresponds to v = 3, and xy 3 = 6 corresponds to v = 6. Hence the area of the region shown in Fig. 14.9.8
is
  6  4  6
1 1
A= 1 dx dy = du dv = dv = ln 2 ≈ 0.6931471805599453.
v=3 u=2 2v 3 v
D

## C14S09.010: If y = ux2 and x = vy 2 , then

1 1
y = uv 2 y 4 ; y3 = ; y= .
uv 2 u1/3 v 2/3
Then it follows that

v 1
x = vy 2 = = .
u2/3 v 4/3 u2/3 v 1/3
Next,
 
− 2

1 
 
∂(x, y)  3u v 
5/3 1/3 3u2/3 v 4/3
 1
=  = 2 2.
∂(u, v)   3u v
 1 2 
− − 
3u4/3 v 2/3 3u1/3 v 5/3

## Next, y = x2 corresponds to u = 1, y = 2x2 corresponds to u = 2, x = y 2 corresponds to v = 1, and

x = 4y 2 corresponds to v = 4. Therefore the area of the region shown in Fig. 14.9.9 is

  4  2  4  2  4  4
1 1 1 1 1
A= 1 dx dy = du dv = − dv = dv = − = .
v=1 u=1 3u2 v 2 v=1 3uv 2 u=1 1 6v 2 6v 1 8
R

C14S09.011: Given: the region R bounded by the curves y = x3 , y = 2x3 , x = y 3 , and x = 4y 3 . Choose
u and v so that y = ux3 and x = vy 3 . Then

1
y = uv 3 y 9 ; y8 = ;
uv 3
1 v 1
y= ; x = vy 3 = = .
u1/8 v 3/8 u3/8 v 9/8 u3/8 v 1/8

## Then the curve y = x3 can be written as

4
1 1
= 9/8 3/8 ;
u1/8 v 3/8 u v
u1/8 v 3/8 = u9/8 v 3/8 ;

u = 1.

Similarly, the curve y = 2x3 corresponds to u = 2, the curve x = y 3 corresponds to v = 1, and the curve
x = 4y 3 corresponds to v = 4. Next,
 
− 3

1 
 
 8u 11/8 v 1/8 8u3/8 v 9/8 
∂(x, y)   1
=   = .
∂(u, v)   8u v 3/2
3/2
 1 3 
 − − 
8u v 3/8
9/8 8u1/8 v 11/8

## Hence the area of R is

  4  2  4  2  4

1 1 1 1
1 dx dy = du dv = − dv = − √ dv
v=1 u=1 8u3/2 v 3/2 v=1 4u1/2 v 3/2 u=1 1 4v 3/2
4 2 v 3/2
R
√ 4 √
2 −2 2− 2
= = ≈ 0.07322330470336311890.
4v 1/2 1 8

## C14S09.012: The transformation

2x 2y
u= , v=
x2 + y2 x2 + y2

yields

4(x2 + y 2 ) 4 4
u2 + v 2 = 2 2 2
= 2 ; x2 + y 2 = ;
(x + y ) x + y2 u2 + v2
1 2u 1 2v
x= u · (x2 + y 2 ) = 2 ; y= v · (x2 + y 2 ) = 2 .
2 u + v2 2 u + v2

## The circle x2 + y 2 = 2x is thereby transformed into

4 4u
= 2 : u = 1.
u2 + v 2 u + v2
1 1
Similarly, the other three circles are transformed into u = 3, v = 1, and v = 4. The Jacobian of this
transformation is
 
 2(u2 − v 2 ) 4uv 
− 2 − 
 (u + v 2 )2 (u + v 2 )2
2 
∂(x, y)   4
=   = −
 .
∂(u, v)  (u + v 2 )2
2
− 4uv 2(v 2 − u2 ) 
 − 2 
(u + v 2 )2
2 (u + v 2 )2

## Note also that

5
16 1 (u2 + v 2 )2
(x2 + y 2 )2 = , so that = .
(u2 + v 2 )2 (x2 2
+y )2 16

Therefore

  1/4  1/3
1 (u2 + v 2 )2 4
dx dy = · 2 du dv
(x2 + y 2 )2 v=1 u=1 16 (u + v 2 )2
R
 1/4  1/3

1 3 2 1 1
= du dv = − · − · = .
v=1 u=1 4 4 3 4 8

## C14S09.013: The Jacobian of the transformation x = 3r cos θ, y = 2r sin θ is

 
 3 cos θ −3r sin θ 
∂(x, y) 
=   = 6r cos2 θ + 6r sin2 θ = 6r.

∂(r, θ)  2 sin θ 2r cos θ 

x2 y2
The ellipse + = 1 is transformed into
9 4
9r2 cos2 θ 4r2 sin2 θ
+ =1: the circle r = 1.
9 4
The paraboloid has equation

## Therefore the volume of the solid is

 2π  1  2π

27
V = (9r cos θ + 4r sin θ) · 6r dr dθ =
2 2 2 2
cos2 θ + 6 sin2 θ dθ
θ=0 r=0 0 2
 2π
 2π
15 1 + cos 2θ 39 15 39
= 6+ · dθ = θ+ sin 2θ = π ≈ 61.26105674500096815.
0 2 2 4 8 0 2

C14S09.014: The Jacobian of the transformation x = au, y = bv, z = cw (a, b, and c are positive
constants) is
 
a 0 0
 
 
∂(x, y, z)  
= 0 b 0  = abc.
∂(u, v, w)  
 
 
0 0 c

## The ellipsoid with equation

x2 y2 z2
+ + = 1
a2 b2 c2

becomes the sphere S with equation u2 + v 2 + w2 = 1. Let B denote the ball bounded by that sphere.
Then the volume of the ellipsoid is

6
 
4
V = 1 dx dy dz = abc dV = πabc.
3
R B

C14S09.015: We are given the transformation u = xy, v = xz, w = yz. Then uvw = x2 y 2 z 2 . Hence

v 1/2 w1/2
u1/2 v 1/2 w1/2 = xyz = uz : z= ;
u1/2

u1/2 w1/2
u1/2 v 1/2 w1/2 = xyz = vy : y= ;
v 1/2

u1/2 v 1/2
u1/2 v 1/2 w1/2 = xyz = wx : x= .
w1/2

The surface xy = 1 corresponds to the plane u = 1. Similarly, the other surfaces correspond to the planes
u = 4, v = 1, v = 9, w = 4, and w = 9. The Jacobian of the given transformation is
 
 v 1/2 u1/2 u1/2 v 1/2 
 − 
 2u1/2 w1/2 2v 1/2 w1/2 2w3/2 
 
 
 
 
∂(x, y, z)  w1/2 u1/2 w1/2 u1/2  1
=  −  = − 1/2 1/2 1/2 .
∂(u, v, w)  2u1/2 v 1/2 2v 3/2 2v 1/2 w1/2  2u v w
 
 
 
 
 v 1/2 w1/2 w1/2 v 1/2 
− 
2u3/2 2u1/2 v 1/2 2u1/2 w1/2

## Therefore the volume bounded by the surfaces is

 9  9  4  9  9  4
1 u1/2
V = du dv dw = dv dw
w=4 v=1 u=1 2u1/2 v 1/2 w1/2 4 1 v 1/2 w1/2 u=1

 9  9  9  9  9  9
1 2v 1/2 4
= dv dw = dw = dw = 8w1/2 = 8.
4 1 v w1/2
1/2
4 w1/2 v=1 4 w1/2 4

C14S09.016: We are given the solid bounded by the paraboloids z = x2 + y 2 and z = 4(x2 + y 2 ) and the
planes z = 1 and z = 4. We are also given the transformation

r r
x= cos θ, y= sin θ, z = r2 .
t t

Under this transformation, the plane z = 1 corresponds to r = 1 and the plane z = 4 corresponds to r = 2.
The paraboloid z = x2 + y 2 corresponds to

r2 r2 2 r2
r2 = cos 2
θ + sin θ = ,
t2 t2 t2

thus to t = 1. The other paraboloid yields t = 2. Finally, to obtain the entire solid, θ varies from 0 to 2π.
The Jacobian of the given transformation is

7
 1 r r 
 − sin θ − cos θ 
 cos θ
 t t t 2

 
  2

∂(x, y, z)   2 3
=  1 r r  = 2r r sin2 θ + r cos2 θ = 2r .
∂(r, θ, t) sin θ cos θ − 2 sin θ  t3 t3 t3
 t t t 
 
 
 
 
2r 0 0

## Hence the volume of the solid is

 2π  2  2  2  2
2r3 r3
V = dt dr dθ = 2π − dr
θ=0 r=1 t=1 t3 1 t2 t=1
 2  2
3 3 3 1 45
= 2π r dr = π r4 = π ≈ 17.67145867644258696635.
1 4 2 4 1 8

## C14S09.017: The substitution x = u + v, y = u − v transforms the rotated ellipse x2 + xy + y 2 = 3 into

the ellipse S in “standard position,” in which its axes lie on the coordinate axes. The resulting equation of
S (in the uv-plane) is 3u2 + v 2 = 3. The Jacobian of this transformation is
 
1 1 
∂(x, y) 
=   = −2.

∂(u, v) 1 −1 

Therefore
 
I= exp(−x2 − xy − y 2 ) dx dy = 2 exp(−3u2 − v 2 ) du dv. (1)
R S

The substitution u = r cos θ, v = r 3 sin θ has Jacobian
 
 cos θ −r sin θ 
∂(u, v)  √ √
=  √ √
 = r 3 (cos2 θ + sin2 θ) = r 3 .

∂(r, θ)  3 sin θ r 3 cos θ 

## 3 = 3u2 + v 2 = 3r2 cos2 θ + 3r2 sin2 θ = 3r2 ,

and thereby transforms it into the circle with polar equation r = 1. Then substitution in the second integral
in Eq. (1) yields

  2π  1 √
I=2 exp(−3u2 − u2 ) du dv = 2 r 3 exp(−3r2 ) dr dθ
S θ=0 r=0

 1
√ 1 2 √
= 4π 3 − exp(−3r2 ) = π 3 1 − e−3 ≈ 3.44699122256300138528.
6 0 3

C14S09.018: Remember that x = x(u, v), y = y(u, v), u = u(x, y), and v = v(x, y). Then

8
   
x xv   ux uy 
∂(x, y) ∂(u, v)  u · 
· =
∂(u, v) ∂(x, y)    
yu yv   vx vy 

 
 ∂x ∂x 
     
 xu ux + xv vx xu uy + xv vy  
 ∂x ∂y  1 0 
  
=  =
 
=
 
 = 1.

 yu ux + yv vx yu uy + yv vy   ∂y ∂y   0 1
 ∂x ∂y 

C14S09.019: Suppose that k is a positive constant. First we need an integration by parts with

u = ρ2 and dv = ρ exp(−kρ2 ) dρ :

1
du = 2ρ dρ and v=− exp(−kρ2 ).
2k

Thus
 
1 1
ρ exp(−kρ ) dρ = − ρ2 exp(−kρ2 ) +
3 2
ρ exp(−kρ2 ) dρ
2k k
1 2 1
=− ρ exp(−kρ2 ) − 2 exp(−kρ2 ) + C.
2k 2k

Then the improper triple integral given in Problem 19 will be the limit of Ia as a → +∞, where

 2π  π  a
Ia = ρ3 exp(−kρ2 ) sin φ dρ dφ dθ
θ=0 φ=0 ρ=0

 π  a
1 2 1
= 2π − cos φ − ρ exp(−kρ2 ) − 2 exp(−kρ2 )
φ=0 2k 2k ρ=0

 
1 1 1
= 4π − a2 exp(−ka2 ) − 2 exp(−ka2 ) + 2 .
2k 2k 2k

Because k > 0, it is clear that Ia → as a → +∞.
k2

C14S09.020: Given: The solid ellipsoid R with constant density δ and boundary surface with equation

x2 y2 z2
+ + = 1 (1)
a2 b2 c2

## x = aρ sin φ cos θ, y = bρ sin φ sin θ, z = cρ cos φ

has Jacobian

9
 
 a sin φ cos θ aρ cos φ cos θ −aρ sin φ sin θ 
 
∂(x, y, z)  

J= =  b sin φ sin θ bρ cos φ sin θ bρ sin φ cos θ 
∂(ρ, φ, θ)  
 
 
c cos φ −cρ sin φ 0

= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ

= abcρ2 sin φ.

This transformation also transforms the ellipsoidal surface of Eq. (1) into

## ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,

and thereby transforms R into the solid ball B of radius 1 and center at the origin. Therefore the mass of
the ellipsoid is

 2π  π  1  π  π
1 1 4
M= 2
δabcρ sin φ dρ dφ dθ = 2πδabc sin φ dφ = 2πδabc − cos φ = πδabc.
θ=0 φ=0 ρ=0 φ=0 3 3 0 3

C14S09.021: Given: The solid ellipsoid R with constant density δ and boundary surface with equation

x2 y2 z2
+ + = 1 (1)
a2 b2 c2

## x = aρ sin φ cos θ, y = bρ sin φ sin θ, z = cρ cos φ

has Jacobian

 
 a sin φ cos θ aρ cos φ cos θ −aρ sin φ sin θ 
 
∂(x, y, z)  

J= =  b sin φ sin θ bρ cos φ sin θ bρ sin φ cos θ 
∂(ρ, φ, θ)  
 
 
c cos φ −cρ sin φ 0

= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ

= abcρ2 sin φ.

This transformation also transforms the ellipsoidal surface of Eq. (1) into

## ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,

and thereby transforms R into the solid ball B of radius 1 and center at the origin. Therefore the moment
of inertia of the ellipsoid with respect to the z-axis is

10
 2π  π  1
Iz = (ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ)δabcρ2 sin φ dρ dφ dθ
θ=0 φ=0 ρ=0

 2π  π  1
1
= (δabcρ5 sin3 φ)(a2 cos2 θ + b2 sin2 θ) dφ dθ
0 0 5 ρ=0
 2π  π
1
= (δabc sin3 φ)(a2 cos2 θ + b2 cos2 θ) dφ dθ
0 0 5
 2π  π
1
= δabc (cos 3φ − 9 cos φ)(a2 cos2 θ + b2 sin2 θ) dθ
60 0 0
 2π  2π
4 1
= δabc (a2 cos2 θ + b2 sin2 θ) dθ = δabc 2a2 θ + 2b2 θ + a2 sin 2θ − b2 sin 2θ
15 0 15 0

1
= δabc(4πa2 + 4πb2 ).
15
Because the mass of the sphere (found in the solution of Problem 20) is M = 43 πδabc, we see that

Iz 1 1
= (a2 + b2 ), and hence that Iz = M (a2 + b2 ).
M 5 5

y
C14S09.022: Given u = xy and v = , we have
x
y u x
uv = xy · = y 2 and = xy · = x2 ,
x v y
and thus we choose

u √
x= and y= uv . (1)
v
The Jacobian of this transformation is
 
 1 u1/2 
 − 
 2u1/2 v 1/2 2v 3/2 
∂(x, y)  
= 1 .
=
∂(u, v)   2v

 v 1/2 u1/2 
 
2u1/2 2v 1/2
Also, if y = x, then substitution of the equations in (1) yields

u1/2 u
(uv)1/2 = ; uv = ; v 2 = 1.
v 1/2 v
So we choose v = 1. (This choice implies that if we have a similar choice with u, we must choose u > 0
because of the equations in (1).) Similarly, y = 2x yields v = 2, xy = 1 yields u = 1, and xy = 2 yields
u = 2. Hence the area of the region of Fig. 14.9.7 is
  2  2  2
1 1 1
A= 1 dx dy = du dv = dv = ln 2 ≈ 0.3465735902799727.
v=1 u=1 2v 1 2v 2
R

## Its moments with respect to the coordinate axes are

11
    2  √  √ 2 √
2 2
1 u1/2 2
u3/2 2
2 2 −1 2−4 2 5 2 −6
My = · du dv = dv = dv = = ;
v=1 u=1 2v v 1/2 v=1 3v 3/2 u=1 1 3v 3/2 3v 1/2 1 3
    2
2 2
1 √ 2
u3/2
Mx = · uv du dv = 1/2
dv
v=1 u=1 2v v=1 3v u=1
 2 √  √ 2 √
2 2 −1 4 2 − 2 1/2 10 − 6 2 ,
= dv = v = .
1 3v 1/2 3 1 3

## Hence the coordinates of its centroid are

√ √ 
10 2 − 12 20 − 12 2
(x, y) = , ≈ (1.030149480423, 1.456851366485) .
3 ln 2 3 ln 2

## C14S09.023: If u = xy and v = xy 3 , then

v v 1/2
uy 2 = xy 3 = v, so that y2 = ; y= .
u u1/2
Then

u u1/2 u3/2
x= = u · 1/2 = 1/2 .
y v v
(We do not need the solution in which x and y are negative.) Then
 
 3u1/2 u3/2 
 − 3/2 
 1/2
∂(x, y)  2v 2v 
= 3 − 1 = 1 .
=  4v 4v
∂(u, v)   2v
 v 1/2 1 
 − 3/2 
2u 2u1/2 v 1/2

## We also ﬁnd by substitution that xy = 2 corresponds to u = 2, xy = 4 corresponds to u = 4, xy 3 = 3

corresponds to v = 3, and xy 3 = 6 corresponds to v = 6. Hence the area of the region shown in Fig. 14.9.8
is
  6  4  6
1 1
A= 1 dx dy = du dv = dv = ln 2 ≈ 0.6931471805599453.
v=3 u=2 2v 3 v
D

## Its moments with respect to the coordinate axes are

   6  5/2 4  6 √
6
1 u3/24
u 32 − 4 2
My = · 1/2 du dv = 3/2
dv = dv
v=3 u=2 2v v v=3 5v u=2 3 5v 3/2
 √ 6 √ √
8 2 − 64 72 3 − 40 6
= = ;
5v 1/2 3 15
 6  4  6  1/2 4  6 √
1 v 1/2 u 2− 2
Mx = · 1/2 du dv = 1/2
dv = dv
v=3 u=2 2v u v=3 v u=2 3 v 1/2
 6
√  √ √
= 4 − 2 2 v 1/2 = 6 6 − 8 3 .
3

12
Therefore its centroid is located at the point
√ √ √ √ 
72 3 − 40 6 6 6 − 8 3
(x, y) = , ≈ (2.570696785449, 1.212631342551) .
15 ln 2 ln 2

## C14S09.024: If y = ux2 and x = vy 2 , then

1 1
y = uv 2 y 4 ; y3 = ; y= .
uv 2 u1/3 v 2/3
Then it follows that

v 1
x = vy 2 = = 2/3 1/3 .
u2/3 v 4/3 u v
Next,
 
− 2

1 
 
∂(x, y)  3u v 
5/3 1/3 3u2/3 v 4/3
 1
=  = 2 2.
∂(u, v)   3u v
 1 2 
− − 
3u4/3 v 2/3 3u1/3 v 5/3

## Next, y = x2 corresponds to u = 1, y = 2x2 corresponds to u = 2, x = y 2 corresponds to v = 1, and

x = 4y 2 corresponds to v = 4. Therefore the area of the region shown in Fig. 14.9.9 is

  4  2  4  2  4  4
1 1 1 1 1
A= 1 dx dy = du dv = − dv = dv = − = .
v=1 u=1 3u2 v 2 v=1 3uv 2 u=1 1 6v 2 6v 1 8
R

## Its moments with respect to the coordinate axes are

 4  2  4  2
1 1 1
My = · du dv = − dv
v=1 u=1 3u2 v 2 u2/3 v 1/3 v=1 5u5/3 v 7/3 u=1
  4
4
4 − 21/3 3 · 21/2 − 12 96 − 36 · 21/3 + 3 · 22/3
= dv = = ;
v=1 20v 7/3 80v 4/3 1 640
 4  2  4  2
1 1 1
Mx = · du dv = − dv
v=1 u=1 3u2 v 2 u1/3 v 2/3 v=1 4u4/3 v 8/3 u=1
  4
4
4 − 22/3 3 · 22/3 − 12 96 + 3 · 21/3 − 30 · 22/3
= dv = = .
1 16v 8/3 80v 5/3 1 640

## 96 − 36 · 21/3 + 3 · 22/3 96 + 3 · 21/3 − 30 · 22/3

(x, y) = , ≈ (0.692563066996, 0.651971644883) .
80 80

C14S09.025: Given: The solid ellipsoid R with constant density δ and boundary surface with equation

x2 y2 z2
+ + = 1 (1)
a2 b2 c2
(where a, b, and c are positive constants). The transformation

13
x = aρ sin φ cos θ, y = bρ sin φ sin θ, z = cρ cos φ

has Jacobian

 
 a sin φ cos θ aρ cos φ cos θ −aρ sin φ sin θ 
 
∂(x, y, z)  

J= =  b sin φ sin θ bρ cos φ sin θ bρ sin φ cos θ 
∂(ρ, φ, θ)  
 
 
c cos φ −cρ sin φ 0

= abcρ2 cos2 φ sin φ cos2 θ + abcρ2 sin3 φ cos2 θ + abcρ2 cos2 φ sin φ sin2 θ + abcρ2 sin3 φ sin2 θ

= abcρ2 sin φ.

This transformation also transforms the ellipsoidal surface of Eq. (1) into

## ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ = ρ2 sin2 φ + ρ2 cos2 φ = ρ2 = 1,

and thereby transforms R into the solid ball B of radius 1 and center at the origin. Note also that

x2 + y 2 = a2 ρ2 sin2 φ cos2 θ + b2 r2 sin2 φ sin2 θ = (ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ).

Assume that the solid R has constant density δ. Then its moment of inertia with respect to the z-axis is

 2π  π  1
Iz = (ρ2 sin2 φ)(a2 cos2 θ + b2 sin2 θ)δabcρ2 sin φ dρ dφ dθ
θ=0 φ=0 ρ=0

 2π  π
1
= (δabc sin3 φ)(a2 cos2 θ + b2 sin2 θ) dφ dθ
0 0 5
 2π
4 4 1
= (δabc)(a2 cos2 θ + b2 sin2 θ) dθ = πδabc(a2 + b2 ) = M (a2 + b2 )
0 15 15 5

## where M is the mass of the ellipsoid. By symmetry,

4 1 4 1
Iy = πδabc(a2 + c2 ) = M (a2 + c2 ) and Ix = πδabc(b2 + c2 ) = M (b2 + c2 ).
15 5 15 5

C14S09.026: Assume that the solid of Problem 16 has constant density δ. By symmetry its centroid lies
on the z-axis. A consequence of the solution of Problem 16 is that the solid has mass M = 45
8 πδ. Its moment
with respect to the xy-plane is

 2π  2  2  r 3  2π  2  2
r5
Mxy = 2 · r2 dt dr dθ = − dr dθ
θ=0 r=1 t=1 t 0 1 t2 t=1
 2  2
3 5 1 63 63
= 2π r dr = 2π r6 = 2π · = π.
1 4 8 1 8 4

Therefore the centroid of the solid is located at the point 0, 0, 14
5 . Next, by symmetry,

14
 2π  2
2   
 2 
r r 3
Iy = Ix = δ sin θ + r4 · 2 dt dr dθ
θ=0 r=1 t=1 t t
 2π  2  5 2
r (1 + 4r2 t2 − cos 2θ)
=δ − dr dθ
θ=0 r=1 4t4 t=1
 2π  2  2π  2
3 5 3 8 5 2 2
=δ r (5 + 16r2 − 5 cos 2θ) dr dθ = δ r + r sin θ dθ
θ=0 r=1 64 θ=0 32 64 r=1
 2π
 2π
765 315 45 3375
=δ + sin2 θ dθ = δ (7 sin 2θ − 150θ) = πδ.
0 32 64 64 0 64

## Thus Ix = Iy ≈ (165.6699250916492528)δ. Finally, to compute Iz , note that

r 2 r 2 r2
x2 + y 2 = cos θ + sin θ = .
t t t2
Therefore the moment of inertia of the solid with respect to the z-axis is

 2π  2  2  2π  2  2
2r5 r5
Iz = δ dt dr dθ = δ − 4 dr dθ
θ=0 r=1 t=1 t5 θ=0 r=1 2t t=1
 2π  2  2
15 5 5 6 315
=δ r dr = 2πδ r = πδ ≈ (30.9250526837745272)δ.
θ=0 r=1 32 64 1 32

C14S09.027: The average distance of points of the ellipsoid from its center at (0, 0, 0) is
 2π  π  1 
1
d= (abcρ2 sin φ) (aρ sin φ cos θ)2 + (bρ sin φ sin θ)2 + (cρ cos φ)2 dρ dφ dθ
V 0 0 0

where V = 43 πabc is the volume of the ellipsoid. In particular, if a = 4, b = 3, and c = 2, we ﬁnd (using
the NIntegrate command in Mathematica 3.0) that d ≈ 2.300268522983.

## C14S09.028: Following the instructions in Problem 28, we have

    ∞
 ∞ 


1 1
1 1 1   1 1
dx dy = 1+ n n
x y dx dy = 1 + xn dx · y n dy
0 0 1 − xy 0 0 n=1 n=1 0 0



1 1 1 1 1
=1+ · = 1 + 2 + 2 + 2 + · · · = ζ(2). —C.H.E.
n=1
n+1 n+1 2 3 4

## C14S09.029: Part (a): First note that

 1  1
 1  1
1 1 2xy
− dx dy = dx dy.
0 0 1 − xy 1 + xy 0 0 1 − x2 y 2

## The Jacobian of the substitution u = x2 , v = y 2 is

 
 2x 0 
∂(u, v) 
=   = 4xy,
∂(x, y)  0 2y 

15
so

 1  1  1  1  1  1
2xy 1 1 1 1 1
dx dy = · 4xy dx dy = du dv = ζ(2).
0 0 1 − x2 y 2 2 0 0 1 − x2 y 2 2 0 0 1 − uv 2

Part (b): Addition as indicated in Problem 29, and cancellation of the integrals involving 1/(1 + xy), yields
the equation
 1  1  1  1
1 1 1
2 dx dy = ζ(2) + 2 dx dy,
0 0 1 − xy 2 0 0 1 − x2 y 2

## which we readily solve for

 1  1  1  1
1 1 1 3
dx dy = dx dy − ζ(2) = ζ(2).
0 0 1 − x2 y 2 0 0 1 − xy 4 4

Part (c): The Jacobian of the transformation T : R2uv → R2xy that we deﬁne by x = (sin v)/(cos u),
y = (sin u)/(cos v) is
 
 cos u sin u sin v 
 − 
 cos v cos2 v 
  2 2
JT =   = 1 − sin u sin v = 1 − tan2 u tan2 v.
 cos2 u cos2 v
 sin u sin v cos v 
− 
 cos2 u cos u 

Reading the limits for the transformed integral from Fig. 14.9.10(a) in the text, we therefore ﬁnd that

 1  1
4 1
ζ(2) = dx dy
3 0 0 1 − x2 y 2

 π/2  (π/2)−v
−1  π/2  (π/2)−v
4 sin2 u sin2 v 4
= 1− · (1 − tan u tan v) du dv =
2 2
1 du dv
3 0 0 cos2 u cos2 v 3 0 0

  π/2
4 π/2 π 4 π 
1 2 4 π2 π2
= − v dv = v − v = · = . —C.H.E.
3 0 2 3 2 2 3 8 6
0

16
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