Math 251
Practice Exam 4 Solutions
Cerro Coso Community College
1. Sketch the region of integration and write an equivalent double integral with the order
of integration reversed and evaluate the integral.
3
∫∫
1
0
3
∫∫
0
1
3
e y dydx
x3
y3
e dydx =
x3
1
∫∫
0
3 y2
y3
e dxdy =
0
∫
3 y2
1
0
xe
y3
dy =
∫
1
3
3 y 2 e y dy = e y
3
0
0
0
1
= e −1
2. Find the volume of the solid cut from the first octant by the surface z = 4 – x2 – y.
We are in the first octant which implies that x, y, and z ≥ 0. Set z = 0 and solve for y
in terms of x. y = 4 – x2. A graph of the region follows.
2
∫∫
0
4 − x2
4 − x − y dydx =
2
0
∫
2
0
y2
4 y − x2 y −
2
x4
= 16 − 4 x − 4 x + x − 8 + 4 x − dx =
0
2
∫
2
= 16 −
2
2
4
2
4 − x2
dx =
2
∫ 4(4 − x ) − x (4 − x )
2
2
2
(4 − x )
−
2
2
0
0
∫
2
0
2
x4
4 x3 x5
+
8 − 4 x + dx = 8 x −
2
3 10 0
2
32 32 256 128
+
=
=
.
3 10 30
15
Continued on the next page.
2
dx
3. Find the centroid of the region between the x–axis and arch y = sin x, 0 ≤ x ≤ π.
Finding the centroid implies that δ(x, y) = 1. Draw a picture first.
π
M=
π
∫∫
0
Mx =
My =
x=
sin x
0
∫∫
0
sin x
ydydx =
0
π
∫∫
M
∫
0
sin x
xdydx =
π
2
dx =
y
∫
π
∫
π
∫ sin xdx = − cos x
0
0
0
0
=
sin x
π
0
π
My
dydx =
y2
2
sin x
dx =
0
0
sin x
π
dx =
xy
0
∫
π
∫
π
0
0
π
0
=2
π
sin 2 x
x sin 2 x
π
dx = −
=
2
4
8 0 4
π
x sin xdx = − x cos x + sin x 0 = π
π
; y=
Mx 4 π
= =
M
2 8
4. Find the center of mass and moment of inertia and radius of gyration about the y–axis
of a thin plate bounded by the line y = 1 and the parabola y = x2 if the density is
δ(x, y) = y + 1.
Continued on the next page.
Problem 4 continued.
M=
y
1
∫∫
Mx =
My =
y + 1 dxdy =
− y
0
∫∫
∫ xy + x
0
y
1
1
y 2 + y dxdy =
− y
0
1
y
0
− y
∫∫
∫
1
∫
x ( y + 1) dxdy =
∫
− y
x ( y2 + y )
0
1
0
1
4 5 4 3
32
dy = 2 y + 2 y dy = y 2 + y 2 =
0
5
3
15
0
y
1
3
y
1
2
∫
dy =
− y
1
2
5
1
3
2 y 2 + 2 y 2 dy =
0
y
x2
( y + 1) dy =
2
− y
4 52 4 32
48
y + y
=
7
5
35
0
1
∫ 0dy = 0
0
48
M x 35 9
0
x=
=
= 0; y =
=
= .
32
32 14
M
M
15
15
My
Iy =
1
∫∫
y
x
2
− y
0
( y + 1) dxdy = ∫
1
0
y
x3 2
1
y + y ) dy =
(
3
3
− y
1
1  4 9 2 4 5 2  64
2 y + 2 y dy =  y + y  =
0
3  9
7
 63
0
∫
1
7
5
2
2
5. Change this Cartesian integral into an equivalent polar integral. Then evaluate the
polar integral.
π
2
1− ( x −1)
2 cos θ
2
r 2 ( cos θ + sin θ )
x+ y
dydx
=
drdθ
0
0
0
0
x2 + y 2
r2
2
∫∫
=
∫
π
2
0
∫ ∫
r ( cos θ + sin θ )
2 cos θ
dθ =
∫
π
2
2 cos 2 θ + 2 cos θ sin θ dθ
0
0
π
θ sin 2θ sin 2 θ  2
π 1  π
= 2 +
+
= 2  +  = + 1.

4
2 0
 4 2 2
2
6. Find the area of a region that lies inside the cardioid r = 1 + cos θ and outside the
circle r = 1.
π
Area = 2
∫ ∫
0
=
∫
π
0
2
1+ cos θ
1
rdrdθ = 2
∫
π
0
2
r2
2
1+ cos θ
dθ =
1
∫
π
2
0
1+ cosθ
r
dθ =
2
1
π
2
θ sin 2θ  2
π

2 cos θ + cos θ dθ =  2sin θ + +
= 2+

2
4 0
4

2
Continued on the next page.
∫
π
0
2
1 + 2 cosθ + cos 2 θ − 1 dθ
3− 3 x
1
∫∫ ∫
7. Evaluate
0
3− 3 x
1
∫∫ ∫
0
=
0
1
3−3 x
0
=
∫
0
1
dz dy dx .
0
3− 3 x − y
0
∫∫
0
3− 3 x − y
dz dy dx
y2
3 − 3 x − y dydx = 3 y − 3 xy −
0
2
∫
9 − 9x − 9x + 9x
2
( 3 − 3x )
−
2
0
=
1
2
∫
1
18 − 18 x + 9 x 2 dx =
0
1
2
dx =
3− 3 x
dx =
1
∫ 3 ( 3 − 3x ) − 3x ( 3 − 3x )
0
0
( 3 − 3x )
−
2
2
dx
1 1
18 − 36 x + 18 x 2 − 9 + 18 x − 9 x 2 dx
2 0
∫
1
1
18 x − 9 x 2 + 3 x 3  0 = 6
2
.
1
1
1
−1
−1
−1
∫ ∫ ∫ ( x + y + z ) dy dx dz .
y
1 
1

x + y + z ) dy dx dz =
xy + + yz dx dz =
x + z +  −  − x − z +  dx dz
(

∫∫∫
∫∫ 2
∫ ∫  2 
2
=
∫ ∫ 2x + 2z dx dz = ∫ x + 2 xz dz = ∫ (1 + 2z ) − (1 − 2 z ) dz = ∫ 4zdz = 2z = 0.
8. Evaluate
1
1
1
1
1
−1
−1
−1
−1
−1
1
1
1
2
−1
−1
−1
1
−1
1
2
−1
−1
−1
1
−1
1
1
−1
−1
2 1
−1