Calculus III
Practice Problems 7: Answers
1. What is the volume of the region under the surface z
vertices (1,0), (1,5), (3,0), (3,5)?
ln x
y and over the rectangle in the xy-plane with
Answer. Let R represent the rectangle, so that R is given by the inequalities 1
then is given by dV zdA, so is
The inner integral is
3
5
ln x
R
1
0
2
5
0
V
3
1
5 ln x
5. The volume
y dA ln x y dy dx ln x y dy y ln x y 5 lnx 12 5 2
V
Thus
x 3 0 y 12 5 dx 5 x ln x x 12 5x 3
1
3
1
5
0
15 ln3
15
37 5 5 12 5
15 ln3
5
2. What is the volume of the solid bounded by the surfaces z x 3 and z x2 2y2 lying directly over the
rectangle 0 x 1 0 y 3?
Answer. Note that for 0 x 1 we must have x 3 x2 x2 2y2 . Thus we shall integrate over the rectangle
0 x 1 0 y 3, with the height ranging from x 3 to x2 2y2. Thus
x 1
The inner integral is 3x 2
V
18
3
0
2
x dy dx 2y2
3
0
3x3 , and the volume is 18.25 .
y2
3. What is the volume of the solid bounded above by the surface z
T : 0 y 2 y x y?
x2 lying directly over the triangle
Answer. This is a type 2 regular domain, so
Volume
The inner integral is
Then
y
y
y x dx dy 2
zdA
T
y
0
2
3
2
3
3
3
0
ex
4. What is the volume of the region under the surface z
vertices (0,0), (1,0), (0,2).
Answer. This is the volume under the graph of z
x 0, y 0 and y 2 2x. Thus
y
y
3
V
2
x
4
x dx
y x
y 3
3
4
16
Volume
y dy
y2
2
2
y
ex
y
T
0
y
and over the triangle in the xy-plane with
and over a triangle T . T is bounded by the lines
1
zdxdy
2 2x
0
ex y dy dx
The inner integral is
2 2x
ex y dy
0
ex
y 2 2x
0
e2
ex
x
Now we integrate with respect to x:
1
0
e dx e e e2
x
2 x
x
x 1
0
y 2, x
5. Let R be the region in the plane bounded by the curves x
I
2e 1 e2
3
y x dxdy 2y2 . Calculate
2
R
Answer. Draw the region described by the curves:
x
y2
3
x
2y2
PSfrag replacements
Integrate first with respect to x along curves y = constant from y 2 to 3
point of intersection of the parabolas: y 2 3 2y2 ; the solutions are y
y x dx dy 3 y2
1
I
x 2 29 5.
y2 x
Integrating over y, we obtain I
2
2
. To find the range in y find the
1. Thus
2
y2
1
The inner integral is
2y
3 y2
y2
2y4
6. What is the mass of the lamina bounded by the curves y
density function is δ x y
x2?
Answer. Draw the region described by the curves:
6y2
1
92 x and y
1
x 3 and the x-axis, where the
y
1
1
x3
x
PSfrag replacements
x
x y2
3 2y2
y
If we insist on viewing this as a type 1 domain, we must split the region into two parts: 1 x 0 0 y
1 x; 0 x 1 0 y 1 x 3 , and integrate dydx. As a type 2 domain, it is easier: the region is given as
1 y 1 3 and integrate dxdy. In the second case, the mass is given by
0 y 1 y 1 x
1 y1 3
1
Mass
The integral with respect to dx is
1
x2 dx dy
y 1
0
y y 1 3
3
3
Now, to complete the integration, make the change of variable u
1
3
1
! y" y 1 dy
0
1
3
3
1
1
1
y:
u du
3
u
0
1
4
7. A lamina filled with a homogeneous material (the density is identicaly equal to 1) is in the shape of the
region R bounded by the curves y 1 and y x 2 . What is its center of mass?
Answer. This is the region
x̄ 0. Now
1 x 1 x y 1. Call the center of mass x̄ ȳ . Because of symmetry,
4
Mass dydx 2
1
1
3
1 x2
Momy
0 3 5$
#
0
1
1
1
x2
ydydx
4
5
Thus x̄ ȳ
8. Find the mass of the solid bounded by the surface z
y 2, where the density is δ x y z
x.
Answer. Draw the region bounded by the surfaces:
&% x 2
y, the coordinate planes and the planes x
1,
z
PSfrag replacements
y
y2
x
x
3 2y2
y 1 x
y 1 x3
x
y
1
2
x
δ dV
Now, over an infinitesimal rectangle of side lengths dx dy, the mass is dM
Mass
R
zδ dxdy
1
2
2
0
ydydx
0
% x ydy 2 x y 2 x 2 x ' 3
3 ! 2
2 1
x 2
x 2 3 1(
x x 2 x dx
3
3 5
5
15 2
2
2
0
1
Mass
x% x δ zdxdy, so
0
2
3 2
3
3 2 2
0
2
2
3 2
3 2
5
3
1
0
5 2