Calculus III
Practice Exam 3, Answers
1. Find the volume under the plane z
x
2y 1 over the triangle bounded by the lines y
Answer. We view this as the region lying over the triangle T and under the plane z
type 1 domain 0 x 1 0 y 2x. Thus
x
0 x
2y
1 y
2x.
1. T is the
x 2y 1 dy
dx x 2y 1 dy xy y y 6x 2x 6
Volume 6x 2x dx
1 3
3
Volume
The inner integral is
1
2x
zdxdy
T
0
0
2x
2
2x
0
2. Let R be the region in the plane bounded by the curves x
y 2, x
0
Thus
1
2
2
0
I
y x dx 3
2y2 . Calculate
2
R
Answer. This is the region between two parabolas with axis the x-axis. Integrate first with respect to x along
curves y = constant from y 2 to 3 2y2 . To find the range in y find the point of intersection of the parabolas:
y2 3 2y2 ; solutions y
1. Thus
I y x dx dy 2y 6y 9 x
y x
2
2
3 y2
1
The inner integral is
2
2
Integrating over y, we obtain
1
I
2y4
6y2
1
2
y2
1
3 y2
y2
4
2
92 dy 2 25 2 92 3. Let R be the region in the first quadrant bounded by the curves y
of its centroid?
x and y
29
5
x 3 . What are the coordinates
x 1 x y x. Thus
1 1 1
Area dydx x x dx 2 4 4 1 1
2
Mom xdydx x x dx
3
5
15
ydydx y dx 1 x x dx 1 1 1 Mom 2
2 3 7
2
3
Answer. R is the type 1 region 0
1
x
x3
0
1
x 0
1
y 0
0
The centroid is at (8/15,8/21).
1
0
x
1
x3
2
4
0
x
x3
3
0
1
0
2
x
x3
1
0
2
6
2
21
4. What is the mass of the lamina bounded by the curves y
is δ x y
xy?
3x and y
6x
x 2 where the density function
x 3 3x y 6x x . Thus
xydy
dx Mass δ dA 2
Answer. This is the type 1 region 0
6x x2
3
R
The inner integral is
6x x2
x
Mass
1
2
3
0
x5
12x4
5. As u v runs through the region u 2
3x
x
x 3x x 2
27x dx 9 81 1 4 3 2 6 5 4
x
6x
2
ydy
3x
Thus
0
2 2
3
2
2
12x
v2
729 7
2 60
3
27 42 525
1, the vector function
u v I u v J
2
2
Xuv
2
2
uvK
describes a surface S in three space. Write down the double integral which must be calculated to find the
surface area of S.
2uI
Xu
u
2uJ
2
v
u
vK
Xv
2
2vI
2
2vJ
2
2
v
uK X X
2v 2u I 2v 2u J 4uvK so, when the smoke clears we find
X X
2 2 u v Thus
S 2 2 u v dudv where R is the unit disc. We now switch to polar coordinates: u v
r cos θ sin θ cos 2θ r drdθ 2 cos 2θ dθ 4 2 cos 2θ dθ
S 2 2 2
1 2
4 2 2
Answer.
2
2
2
R
2
2
2π
0
1
2
2π
3
0
2
2
r2 cos2θ , so
π 4
0
0
2 2
4x 2
6. Find the volume of the region bounded below by the surface z
25y2, and above by the plane z
Volume Answer. This can be viewed as the volume of the region between the surfaces z
which lies above the ellipse E : 4x 2 25y2 100:
E
100 4x2
25y2 dxdy
Let u 2x v 5y. Then E corresponds to the disc D : u 2 v2 100 and 100 4x 2
Since x u 2 y v 5, the Jacobian of the transformation is 1 10.This gives us
Volume
∂ x y
!
dudv
∂
u v
f xuv yuv
D
1
10
"
25y2
100 u v dudv 2
D
100 and z
2
4x 2
100
100.
25y2
u
2
v2 .
Now, moving to polar coordinates in u v space, this becomes
1
10
2π
10
0
0
r rdrdθ
π
5
2
100
7. Find the centroid of the region under the cone z 2
x2
100 r dr
100
500π
2
0
2π
Volume
3
r
0
0
rdzdrd θ
0
y2 lying over the disc x 2
Answer. In cylindrical coordinates, this region is describe by the inequalities r
2π
3
y2
9.
z
3, 0
18π
r2 dr
0
r. Thus,
Because of the symmetry about the z-axis, the centroid lies on the z-axis. To find z̄, we calculate the moment
2π
Mom
Thus z̄
81π 4# 18π z 0
3
0
r
0
9 8.
8. Find the volume inside the hyperboloid x 2
rzdzdrd θ
0
y
2
z2
2π
3 r3
0
z
1, for 0
2.
Answer. In cylindrical coordinates, the region is described by the inequalities 0
Thus
$
2π
Volume
0
0
The inner integral is
$
1 z2
1
rdr
z2
2
0
0
' 1
1
&% 1 z x2
2
1
z2
2
0
14π
3
dz
2, 0
r 1
z2 .
y2 lying between the planes z
z2y dxdy. Since zx
2
x
z
rdrdzd θ
2π
so Volume
9. Find the surface area of the piece of the paraboloid z
Answer. We start with the equation dS
1 z2
2
81π
2
dr
2
2x zy
0 z
2.
2y, this gives us dS
4x2 4y2dxdy
4r 2 rdrd θ , where we have switched to polar coordinates because of the circular
symmetry. Then the area is the integral of dS over the disc r
2:
(
2π
Sur f ace Area
using the substitution u
1
0
r
2
2 1 2
1
0
r2.
rdrd θ
2π
3
1) 2π
3 3
3
u1 2 du
1
Answer. The mass is the triple integral over R of δ dV . Switching to spherical coordinates, R is described by
the inequalities 0 θ π 2 0 φ π 2 0 ρ 1, and δ ρ cos φ sin φ cos θ sin θ .
ρ cos φ sin φ cos θ sin θ ρ sin φ dφ dθ Mass 10. The part R of the sphere of radius 1 centered at the origin which lies in the first octant is filled with a
material whose density function is δ x y z
z 2 xy. Find the mass of this object.
2
π 2
0
π 2
0
1
2
2
2
2
2
2
0
Now, the integral with respect to ρ is 1/5. To calculate the integral with respect to φ , we use
π 2
0
cos2 φ sin φ d φ
Thus
Mass
1
15
1
π 2
0
1
3
π 2
0
sin2 φ d φ
2 cos θ sin θ d θ
2
3
1 1 π
15 2