Calculus III
Practice Problems 9: Answers
xy that lies inside the cylinder x 2
1. Find the surface area of the part of the hyperbolic paraboloid z
Answer. Let R be the disk x2 y2
R. Now zx y zy x, so that
Sur f ace Area
4. We want the surface area of the part of the surface z
dS
R
R
z
2
x
1
z2y dxdy
R
1
y
2
x2 dA
y
2
4.
xy lying over
It is convenient to switch to polar coordinates. We get
2π
Sur f ace Area
2
0
0
1
2π 3
5
3
r2rdrd θ
y2
2. Find the surface area of the part of the hyperbolic paraboloid z
x2 y2 1 and x2 y2 4.
2π
0
2
1
1
4r2rdrd θ
2π
x2 that lies between the cylinders
Answer. We use the formula dS
1
z
x
12 18 23 1 4r2
2 3 x
3. Find the surface area of the part of the surface z
quadrant bounded by the line x y 1 .
2
x
z2y dxdy:
∂z
∂x
x 3
2
2
1
3
1
17 2
12
2y, dS
y
5
1
4x2
1
1 x
y that lies above the triangle in the first
3 2
3 2
y
∂z
∂y
1
0
1
3 2 1 x
0
0
3 2
x 1 0 y 1 x 0 z 1 x ydz dy dx Mass ρ dV The innermost integral is y 1 x y y 1 x y . Then
y 1 x y dy y 1 x y 1 x 2
3
6
Answer. The solid can be represented by the inequalities, 0
0
1 x
0
1 x y
0
2
1 x
2
3
3
0
and the mass is
Mass
1
6
1
x
1
0
3
dx
1 x
0
1
24
x,
0
4. Find the mass and the x-coordinate of the center of mass of the solid bounded by the planes x
0 z 0 x y z 1 with the density function ρ x y z
y
1
4y2dA.
3
2
1 x ydxdy. The region is the type 1 domain given by the inequalities 0 x 1 0 y 1 1 x ydy dx 2 1 x y dx 2 2 dx 2 8 S 3
3
3
0
2
4 . Since z 2x z
Answer. The surface lies over the region R
1 x 2 y2
Now, switching to polar coordinates, the surface area is
so dS
so
1 3
0 y
y. Thus
xydz dy dx The innermost integral is xy 1 x y x y 1 x y , and
x y 1 x y dy x y 1 x y x 1 x 2
3
6
Now
1
Momx
0
1 x
0
1 x y
0
0
2
1 x
2
3
3
0
Finally,
Momx
using the substitution u
1
1
6
0
1
x dx 1 u u du
6
1 120 1 24 1 5.
x. Thus x̄
1
3
x1
0
3
1 x
0
3
1
120
4
0
5. Find the center of mass of the piece of the solid parabolic shell z
16
x
2
y lying above the xy-plane.
2
Answer. Since this is a solid of revolution about the z-axis, the center of mass lies on the z axis, so is of the
form 0 0 z . To find z we must calculate the volume and Momz 0 for the region. For these calculations we
switch to cylindrical coordinates.
r
r dr 2π 8r 4 Mom
zdz rdrd θ π 16 r rdrdθ We conclude the computation with the change of variable u 16 r du 2rdr:
2 π
Mom
π
u du
6
2π
Volume
0
4
r rdrdθ
0
2π
z 0
0
2π
2
16
4
3
16r
2
0
16 r2
4
0
4
4
4
0
27 π
2 2
0
0
2
16
z 0
Thus
z
12
2
0
1 212 π
6 27 π
32
6
5 333
Answer. R can be described as the set of x y z 6. Find the average value of f x y z
x y z over the region R in the first octant (the region where all the
coordinates are positive) under the plane x y z 1.
satisfying 0
description tells us how to calculate by iterated integrals:
x 1 0 y 1 x 0 z 1 x y. This
dz dy dx x y z dz dy dx Total f dV and the average is Total Volume. The computations are tedious and involve only integrations of polynomials.
The final integrations (with respect to x) are
Volume 1 2 x dx 16 Volume
1
1 x
1 x y
dV
R
0
1
R
0
1 x
0
0
0
1 x y
0
2
1
0
Total
1
0
1
3
x
2
x3
dx
6
1
8
Thus the average is (1/8)/(1/6) = 3/4.
7. The curve z
x 1 2 0 z 1 is rotated about the z-axis, enclosing, together with the xy-plane, a
3-dimensional region R. R is filled with a substance whose density is inversely proportional to the distance
from the z-axis. Find the total mass of this object.
r 1
r 1
2π k 3
2
Answer. The region is that under the curve (using polar coordinates) z
r 1 on the xy-plane. The density is δ k r. Thus, the mass is
R
zδ drd θ
2π
0
y
2
z2
1
2k
r
0
8. Evaluate
where R is the ball x2
r 1
R
x
2
R
π
2π
0
2π k
3
1
0
2
Answer. Switch to spherical coordinates. R is given by ρ
ρ 2 dV
0
y z dxdydz
2
4.
3
rdr d θ
above the disc R
0
2
0
2, and the integral is
ρ 4 sin φ d ρ d θ d φ
4π
25
5
9. Find the centroid of the region R described in example 24, Chapter 17.
60. We used the change of variables
Answer. There we found Volume
u
x
x
y z v y z u v y v w w
z
z
w
to do the computation. We had the Jacobian equal to 1, so it is easy to do the computation in the u v w
variables:
Mom
x 0
0 3
5
xdV
R
0
4
v dudvdw 4 u v dv du 5
3
u
0
0
0
The inner integral is
v 3u 9 2
2
5
9
Mom
4
3u du 4 15 9 60 2
2
ydV 0 v w dudvdw 5 v w dv dw Mom
9
5
30 Mom
5
3w dw
9w 3w 2
2
zdV 0 wdudvdw 15 wdw 120 Mom
2
3
0
uv
Thus
5
x 0
0
3
5
y 0
R
4
0
4
0
0
4
y 0
Finally
2
0
3
5
z 0
R
0
0
4
0
4
0
3
4
0
Thus the centroid is at (1,-1/2,2). For completeness, here is Example 24. Find the volume of the region R
given by the inequalities
0
z 4
0
y z 3 0
x y z 5 This region is a parallelipiped, so by the appropriate change of coordinates, can be made to correspond to a
rectangular parallelipiped. That is, we make the change of variables
u
x
y z v
y
z
w
z
u 5 0 v 3 0 w ∂ x y z dudvdw Volume dxdydz ∂ u v w Now, to calculate the Jacobian, we solve for x y z in terms of u v w:
x u v y v w z w so that R corresponds to the region S given by the inequalities 0
R
so that
∂ xyz
∂ uvw
Thus
S
det
5
Volume
0
1 0 0
1 1 0
0 0 1
3
0
4
dudvdw
0
1
60
4. Thus