Download 212 Calculus 3 Review 10 1) Find the centroid of the uniform solid

212 Calculus 3
1)
2)
Review 10
Find the centroid of the uniform solid eighth-sphere x 2 + y 2 + z 2
x 0, y 0, z 0.
R,
Find the moment of inertia around the z-axis, using cylindrical
coordinates, of a sphere of radius a assuming (x, y, z) = x 2 + y 2 .
3)
Find, by a triple integral in spherical coordinates, the volume of the solid
that is bounded above by the sphere
= 4 and below by the cone
=
3
.
4)
Find the volume of the region in the first octant that is bounded by the
hyperbolic cylinders xy =1, xy = 9, xz = 1, xz = 4, yz = 1, and yz = 16. Use
the transformation u = xy, v = xz, w = yz. Note that uvw = x 2 y 2 z 2 .
5)
Compute the value of the triple integral for f(x, y, z) = x + 2y where the
solid is the tetrahedron bounded by the coordinate planes and the first
octant part of the plane with equation 2x + y + 4z = 8.
1
SOLUTIONS
Problem 1. Without the loss of generality we can assume that the density is
identically 1. The mass of the region then equals to its volume. The volume of
the region is one eighth of the volume of a sphere of radius R.
4 R 3  R 3

.
m V 
38
6
Because of symmetry x  y  z . To find x we compute


M yz   xdV   x   dydz dx.


0  Ex
D

R
where Ex  {( y, z ) | y  0, z  0, y  z  R  x } . Notice that the area of
2
Ex is

4
2
2
2
( R 2  x 2 ) and therefore
M yz 
Finally, x  y  z 
M yz
m


R
4 0
x( R  x )dx 
2
2
 R4
16
3R
.
8
Problem 2.
I z   ( x 2  y 2 ) ( x, y, z )dV   ( x 2  y 2 )3 2 dV .
E
E
Changing to cylindrical coordinates x  r cos , y  r sin  , z  z we get
a 2
Iz 
a2  z2
 
a 0
4 a 6

5
2
r rdrd dz 
5
3
0
 2
 cos
6
tdt 
0
2
8
a
 (a
2
 z 2 )5 2 dz 
a
a6
Problem 3.
 3 2 4
V

0
64






 2 
sin
d
d
d
0 0
3
2
Problem 4. To compute the Jacobian
 3

sin  d 
0
64
3
 ( x, y , z )
we notice that it is the
 (u, v, w)
reciprocal to
u
x
 (u , v, w) v

 ( x, y, z ) x
w
x
u
y
v
y
w
y
u
z
y x 0
v
 z 0 x  2 xyz  2 uvw.
z
0 z y
w
z
The volume is
9 4 16
1
1 1 1
V  

dudvdw  24 .
2111 u v w
Problem 5. The projection of the tetrahedron on the
with the vertices (0,0),(0,8),
xy plane is the triangle
and (4,0) . The triple integral now can be set
as
4 8 2 x 2  x /2  y /4

0

0
( x  2 y )dzdydx 
0
4 8 2 x

0
4

(2 x  (1 / 2) x 2  (5 / 4) xy  4 y  (1 / 2) y 2 )dydx 
0
  (128 / 3  24 x  4 x 2  (1 / 6) x 3 ) dx 
0
160
3