Math 212 Spring 2008: Solutions: HW #8
Instructor: S. Cautis
1. section 5.2 #2a
We have
Z 1 m+1
Z 1Z 1
Z 1
x
1
y n+1 1
1
1
xm y n dxdy =
[
]10 y n dy =
y n dy =
[
]0 =
m+1 n+1
(m + 1)(n + 1)
0
0
0 m+1
0 m+1
2. section 5.2, #5
Note that g(y) does not depend on x so
!
Z
Z d Z b
Z d
g(y)
[f (x)g(y)]dx dy =
c
a
!
b
f (x)dx dy =
a
c
where the last equality follows since
can pull out to the front.
Rb
a
Z
a
b
! Z
f (x)dx)
f (x)dx is just a number which we
3. section 5.2, #6
Notice that sin(y) ≥ 0 if y ∈ [0, π/2] so we don’t have to worry about
calculating a negative volume anywhere. We get
Z π/2 Z 1
Z π/2
π/2
sin(y)dxdy =
sin(y)dy = [− cos(y)]0 = 1
0
0
0
4. section 5.3, #8
We get
Z
0
1
2
[x2 y + xy 2 /2 − y 3 /3]x0 dx
=
=
=
Z
1
0
5
x4 + x5 /2 − x6 /3dx
[x /5 + x6 /12 − x7 /21]10
1/5 + 1/12 − 1/21
This is an integral over the region D which lies below the graph y = x2
and above the x-axis between 0 ≤ x ≤ 1.
5. section 5.3, #12
For this problem it is helpful to draw a picture (I won’t because it is hard
to typeset).
The easiest is to place the tip of the cone at the origin and the cone
facing upwards in the z-direction. Then the side of the cone has equation
z 2 = a(x2 + y 2 ) where the constant a is yet to be determined.
p
When the cone has height h (i.e. z = h) we want the radius (i.e. x2 + y 2 )
to be r. Thus we get
h2 = a(r2 )
c
d
g(y)dy
!
p
so a = h2 /r2 . Now the volume under the graph of z = h/r x2 + y 2 over
the region D = {(x, y) : x2 + y 2 ≤ r2 } (which is a disk of radius r centered
at the origin) is precisely the volume of a cylinder of radius r and height
h minus the volume of the cone we want.
Since the volume of such a cylinder is πr2 h the volume of the cone is
Z Z
p
h/r x2 + y 2 dxdy
πr2 h −
D
where D is the disk defined above.
6. section 5.3, #14
Integrating in the y direction first we get
b
Z
a
Z
φ2 (x)
f (x)g(y)dydx
φ1 (x)
Now we can pull out the factor of f (x) as before to get
Z
b
f (x)
a
but this is not the same as
Z b
Z
φ2 (x)
g(y)dydx
φ1 (x)
f (x)dx
a
Z
φ2 (b)
g(y)dy
φ1 (a)
R φ (x)
in general. The point is that φ12(x) g(y)dy depends on x and so we cannot
pull it out to the front as we did in the past.
7. section 5.4, #2c
Again, drawing a picture helps. You can see that reversing the order of
the integral gives
Z 2
Z 2 Z 2x
2
2
2
2xex dx = [ex ]20 = e4 − 1
ex dydx =
0
0
0
8. section 5.4, #8
Again, drawing a picture helps. √
y = x2 and y√= 10 − x2 intersect when
2
2x = 10 which means at points ( 5, 5) and (− 5, 5). Since we also insist
that x > 0 the integral we want is
Z
0
√
5
Z
10−x2
√
y xdydx
2
=
Z
=
Z
x2
√
5
0
0
√
5
√
2
dx
[ xy 3 /3]10−x
x2
√
√
x(10 − x2 )3 /3 − xx6 /3dx
At this point it is clear that if we expand
all out we can integrate this
√
expression since each term looks like xxk for some integer k.
You can try reversing the order of integration. You find that the integral
is made up of two pieces. The first piece is
Z
0
5
Z
√
y
√
y 2 xdxdy
=
Z
5
0
0
=
=
=
=
2
3
√
y
[y 2 · 2x3/2 /3]0 dy
Z
0
5
y 2 · y 3/4 dy
Z
2 5 11/4
y
dy
3 0
2 4 15/4 5
[ y
]0
3 15
2 4
·
· 515/4
3 15
So the integral is easier to calculate this time and gives
15
8
·5 4
45
The second piece is
Z
10
5
Z
√
10−y
√
y 2 xdxdy
0
which can be computed similarly. Summing the two gives the answer.
9. section 5.5 #6
The elliptic cylinder has its main axis along the y-axis so it makes sense
to first integrate along the y-direction. This gives
Z Z
D
√
4−x2 −z 2
√
− 4−x2 −z 2
1dydA
where D is the region in the x−z plane bounded by the ellipse 2x2 +z 2 = 1.
The integral over D can also be broken down just like the integral over a
disk.
10. section 5.5 #24
The region W is the region inside the unit ball lying between the planes
z = 1 and z = 1/2. The plane z = 1 intersects the ball only at the point
(0, 0, 1). The plane z = 1/2 intersects it over x2 + y 2 ≤ 3/4. Thus the
integral is
Z Z √1−x2 −y2
f (x, y, z)dzdA
D
1/2
where D is the disk x2 + y 2 ≤ 3/4. Thus γ1 (x, y) = 1/2 and γ2 (x, y) =
p
1 − x2 − y 2 .
Next we integrate in the y direction so the second integrand is
Z √
2
3/4−x
−
Thus φ1 (x) = −
√
stuf f dy
3/4−x2
p
p
3/4 − x2 and φ2 (x) = 3/4 − x2 .
Notice that if the region would have been 1/2 ≤ z ≤ t where t is some
constant less than 1 then the expression for γ2 would have been more
complicated (if you drew a picture this should be clear).