Z=8−x2−y2 S z=x2+y2 R x2 +y2
Solution to Section 5.4, assigned April 23
(1) (Section 5.4, Problem 10) Evaluate the iterated triple integral
2 4-x2
4-x2-y2
xydzdydx
00
0
Solution:
2
4-x2
4-x2 -y2 xydzdydx =
2
4-x2
[xyz]zz==0 4-x2=y2
dydx
00
0
00
2 4-x2
=
xy
00
4 - x2 - y2dydx =
2 0
-
x 3
(4
-
x2
-
y2)3/2
y=4-x2
dx
0
=
2 0
x 3
(4
-
x2)3/2dx
=
-
1 15
4 - x2
5/2
2 0
=
32 15
(2) (Problem 5.4, Problem 14) Evaluate the volume integral (triple integral) of f (x, y, z) = x2 over S, where S is the solid bounded by the paraboloids z = x2 + y2 and z = 8 - x2 - y2.
Solution:
z=8-x2 -y2
S z=x2+y2
R x 2 +y2 =4
Figure 1. Region S bounded above by paraboloid z = 8-x2 -y2 and below by paraboloid z = x2 +y2. Surfaces intersect on the curve x2 + y2 = 4 = z. So boundary of the projected region R in the x - y plane is x2 + y2 = 4.
Where the two surfaces intersect z = x2 + y2 = 8 - x2 - y2. So, 2x2 + 2y2 = 8 or x2 + y2 = 4 = z, this is the curve at the intersection of the two surfaces. Therefore, the boundary of projected region R in the x - y plane is given by the circle x2 + y2 = 4. So R can be treated as a y simple region in the
1
2
x - y plane, with upper and lower curves y = ? 4 - x2 for -2 x 2. Therefore,
x2dV =
8-x2 -y2
x2dz dA
S
R x2+y2
=
2 -2
4-x2 -4-x2
x2z
z=8-x2-y2 z=x2+y2
dydx
=
2 -2
4-x2
x2 8 - 2x2 - 2y2 dydx
-4-x2
=
2
x2
-2
8y
-
2x2y
-
2 3
y3
y=4-x2
dx
y=-4-x2
=
2
x2
-2
16 4
-
x2
-
4x2 4
-
x2
-
4 3
4 - x2 3/2
dx
Substituting x = 2 sin and noting that dx = 2 cos d we get
x2dV =
2
sin2
256 cos - 256 sin2 cos - 256 cos3
cos d
S
-2
3
=
/2
d
0
512
cos2
sin2
-
512
sin4
cos2
-
512 3
cos4
sin2
Using 512 cos2 sin2 = 128 sin2(2) = 64 (1 - cos[4]),
-512 sin4 cos2 = -128 sin2(2) sin2 = -32 [1 - cos 4] [1 - cos 2] = -32 + 32 cos 2 + 32 cos 4 - 16 cos 2 - 16 cos 6
-
512 3
cos4
sin2
=
-
128 3
sin2(2)
cos2
=
-
32 3
[1
-
cos
4]
[1
+
cos
2]
=
-
32 3
1
+
cos
2
-
cos
4
-
1 2
cos
2
-
1 2
cos
6
Since the integral of cos [2m] for m = 1, 2, 3 is a multiple of sin[2m] which is zero at = /2, it follows that
x2dV =
S
64
-
32
-
32 3
[]02
=
332
(3) (Section 5.4, Problem 18) Find the volume of the indicated solid region S inside the cylinders x2 + y2 = a2 and x2 + z2 = a2.
Solution: Consider only the part of S that lies in the region
x 0, y 0, z 0 From symmetry of the region under the
transformation x -x, y -y and z -z, it follows that
the volume of this region S1 is
V 8
,
where
V
is the volume of S.
We treat S1 as an x-simple region in 3-D.
3
O plot3d({sqrt(9-x^2), sqrt(9-y^2)},x=-3..3,y=-3..3);
Figure 2. Part of the region S bounded by x2 + z2 = a2 and x2 + y2 = a2 for x 0
Note that the projection of region S1 on the y - z plane,
call it R is a a square 0 y a, 0 z a. We break
up R into two region R1 = {(y, z) : a y z 0, } and R2 = {(y, z) : a z > y 0}. In region (y, z) R1, x-ranges from x = 0 to x = a2 - y2 (since this is smaller than a2 - z2. In region (y, z) R1, x-ranges from x = 0 to x = a2 - z2
(since this is smaller than a2 - y2. So, it follows that the
total volume of S1 is
V 8
=
R1
a2 -y2
dx dA +
0
R2
a2 -z 2
dx dA
0
ay
a z
a
=
a2 - y2dzdy +
a2 - z2dydz = 2 y a2 - y2dy
00
00
0
=
-
2 3
(a2
- y2)3/2
a 0
=
2 3
a3
Therefore,
volume
of
S
is
V
=
16 3
a3.
4
(4) (Section 5.4, Problem 24). Find the centroid of the given solid bounded by the paraboloids z = 1 + x2 + y2 and z = 5 - x2 - y2 with density proportional to the distnace from the z = 5 plane. Solution: From the problem statement, density = k|z - 5| = k(5-z) since region is below plane z = 5. The plot of the region S between the two paraboloids is similar to (Secion 5.4, Problem 14 ) we have solved above, whose projection R in the x-y plane is bounded by the curve given by 1 + x2 + y2 = 5 - x2 - y2, or x2 + y2 = 2. So, we have mass
2 2-x2 5-x2-y2
M=
k(5 - z)dzdydx
-2 -2-x2 1+x2+y2
=k
2 -2
2-x2 0
16 - 8x2 - 8y2 dydx = k
2 -2
16 2
-
x2
-
8x2 2
-
x2
-
8 3
(2
-
x2)3/2
dx
=
32 3
k
2
(2-x2)3/2dx
0
=
128 3
k
/2
cos4 d
0
=
32 3
k
/2 0
1
+
1 2
+
1 2
cos(4)
+
2
cos(2)
d
= 8k
Now, from symmetry of the shape, it follows that xc = 0 = yc. So, we only need to calculate
2 2-x2 5-x2-y2
kz(5 - z)dzdydx
-2 -2-x2 1+x2+y2
= 2k
2 -2
2-x2 0
5z2 - z3
z=5-x2-y2
dydx
2
3 z=1+x2+y2
= 2k 2 2-x2 -4x2 - 4y2 - 8x2y2 + 2x4y2 + 2x2y4 + 56 + 2 x6 + 2 y6 - 4x4 - 4y4
-2 0
33 3
= 4k
2 0
1140254 2
-
x2
-
152 35
x2 2
-
x2
-
64 35
x4
2
-
x2
+
32 105
x6
2
-
x2
dx
= 4k
16 3
x(2
-
x2)1/2
+
32 3
arcsin
x 21/2
+
110552 x(2
-
x2)3/2
+
76 315
x3
(2
-
x2)3/2
-
4 105
x5(2
-
x2
)3/2
2 0
=
64 3
k
dydx
So,
zc
=
64k 3(8k)
=
8 3
and
xc
=
0,
0,
8 3
.
5
(5) (Section 5.4, Problem 27) Reverse the order of integration ap-
propriate for a z-simple and x-simple regions.
2
1-z2/4 3 1-x2-z2/4
f (x, y, z)dydxdz
00
0
Solution: Since y ranges from 0 to y = 3 1 - x2 - z4/4, we
have
the
upper
surface
y2 9
+ x2 +
z2 4
=
1,
which
is
an
ellipsoid.
We also note that the projected region R in the x - z plane has
goes between x = 0 and x = 1 - z2/4, the latter being the
boundary of an ellipse, while z ranges from 0 to 2. Therefore,
it is clear that the region S is the first octant of an ellipsoid
bounded
by
x2
+
y2 9
+
z2 4
=
1.
Treating S as a z-simple region, we have lower surface z = 0
and upper-surface z = 2
1 - x2
-
y2 9
.
The
projected
region
in
the
x-y
is
the
the
inside
of
the
ellipse
x2
+
y2 9
=
1
in
the
first
quadrant, which may be described as a y-simple region in the
2-D x - y plane:
(x, y) : 0 y 3 1 - x2, 0 x 1
So, the integral above is the same as
1
31-x2
q
2
2
1-x2-
y 9
f (x, y, z)dzdydx
00
0
Treating S as a x simple region, we have for fixed y - z, x
going from 0 to
1-
y2 9
-
z2 4
.
The
projected
region
in
the
y-z
plane can be described as a z-simple region in the y - z plane
and described by
(y, z) : 0 z 2
1
-
y2 9
,
0
y
3
So, the above integral is the same as
q2
3
2
1-
y 9
q 1-
2 y 9
-
z2 4
f (x, y, z)dxdzdy
00
0
(6) (Section 5.4, Problem 30) Using Theorem 5.4.3, determine whether
the integral
S zdV is positive, negative or 0, where S is the
solid bounded by the paraboloid z = -x2 - y2 and the plane
z = -4.
................
................
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