MATH 251.504
Practice Problems for Examination 2
Fall 2006
1. Find the absolute minimum and maximum values of the function f (x, y) = xy − 6x2
on the closed region bounded by the parabola y = 9x2 − 1 and the x-axis.
2. Calculate the double integral.
RR
2
(a) R 3xex dA, where R = [−1, 0] × [0, 1].
RR
(b) R 5 cos(y 2 ) dA, where R is the region bounded by the lines y = −x, x = 0, and
y = −1.
RR
2
2
(c) R ex +y tan−1 ( xy ) dA, where R is the region described in polar coordinates by
{(r, θ) : 0 ≤ θ ≤ π4 , 0 ≤ r ≤ θ}.
RR
(d) R (5x2 y +2xy 5 ) dA, where R is the region between the two parabolas y = x2 +10
and y = 2x2 + 6.
3. Find the volume of the solid which lies between the paraboloid z = 6x2 + y 2 and the
xy-plane and above the region {(x, y) : 0 ≤ y ≤ 1, y 2 ≤ x ≤ y}.
4. Find the center of mass of a lamina occupying the region {(x, y) : 0 ≤ x ≤ 5, 0 ≤ y ≤
x2 } with density function ρ(x, y) = x2 .
RR
2
5. Determine whether the following is true or false: −2 ≤ R x sin(exy + x5 ) dA ≤ 2,
where R = [0, 1] × [−2, 0].
R0 R1
2
6. Compute −1 −x (e−y + xy) dy dx.
7. Find the volume of the solid bounded by the parabolic cylinder z = y 2 and the planes
z = 4y, x = 1, and x = 2.
8. Consider a solid object which occupies the space between the parabolic cylinders z = y 2
and z = y 2 + 2 and over the region bounded by the lines x = 0, y = 0, and y = 1 − x.
Suppose the object has density function ρ(x, y, z) = xyz. Find the moments of inertia
of the object about the x- and y-axes.
1
Solutions
1. We have fx (x, y) = y − 12x and fy = x, and both of these are zero at the point
(x, y) = (0, 0), which is on the boundary of the region. For the part of the boundary of
the region that lies on the x-axis we have the function f (x, 0) = −6x2 for − 31 ≤ x ≤ 31 ,
which has maximum value 0 at 0 and minimum value − 23 at 13 and − 31 . On the other
part of the boundary we have the function g(x) = f (x, 9x2 − 1) = 9x3 − √6x2 − x for
− 31 ≤ x ≤ 13 . Then g 0 (x) = 27x2 − 12x − 1, which is zero at x0 = 12−54 252 . Since
g 00 (x0 ) < 0 the value g(x0 ) ≈ 0.03754 is maximum for g, while g( 13 ) = g(− 13 ) = − 23
is the minimum. Thus for f the absolute minimum value is − 23 while the absolute
maximum value is approximately 0.03754.
2. (a)
Z
1
Z
0
1
Z
x2
3xe dx dy =
−1
0
h3
2
0
x2
e
3
dy = (1 − e).
2
x=−1
ix=0
(b)
Z
0
Z
−y
Z
2
0
5 cos(y ) dy =
−1
h
Z
ix=−y
5x cos(y )
dy =
x=0
−1
0
0
2
−5y cos(y 2 ) dy
−1
i0
5
5
= sin(1).
= − sin(y 2 )
2
2
−1
(c)
Z
π
4
Z
θ
π
4
Z
r2
θre dr dθ =
0
0
h1
2
0
θe
r2
ir=θ
Z
dθ =
r=0
1 π2
1 θ2 1 2 i π4
= e − θ
= e 16
4
4 0
4
(d)
Z
2
x2 +10
2
π
4
1 θ2
θ(e − 1) dθ
0 2
π2 1
−
− .
64 4
iy=x2 +10
1
x2 y 2 + xy 6
dx
3
y=2x2 +6
−2 2x2 +6
−2 2
Z 2
Z 2
1
5 2 2
2
2
2
2
6
2
6
=
x ((x + 10) − (2x + 6) ) dx +
x((x + 10) − (2x + 6) ) dx.
−2 2
−2 3
Z
2
5
(5x y + 2xy ) dy dx =
Z
h5
Expand to evaluate the first integral, and for the second use substitution.
2
3. The volume is given by
Z 1Z y
Z
2
2
V =
(6x + y ) dx dy =
0
y2
1
h
3
2
2x + y x
0
ix=y
x=y 2
Z
dy =
1
(3y 3 − 2y 6 − y 4 ) dy
0
2
1 i1
37
3
.
= y4 − y7 − y5 =
4
7
5 0 140
4. The mass and moments are
Z 5 Z x2
Z
2
m=
x dy dx =
5
1 i5
x4 dx = x5 = 54 ,
5
0
0
0
0
Z 5
Z 5 Z x2
Z 5h
2
i
1 4
1 5 i5 54
1 2 2 y=x
2
xy
dx =
x dx = x
=
Mx =
x y dy dx =
2
10
2
y=0
0
0 2
0
0
0
Z 5 Z x2
Z 5
i
6
5
1
5
My =
x3 dy dx =
x5 dx = x6 = ,
6
6
0
0
0
0
and so the center of mass is ( 25
, 1 ).
6 2
5. The statement is true. On R the values of the function are bounded
between −1 and
RR
2
1, and the area of R is 2. Therefore −2 = −1 × area(R) ≤ R x sin(exy + x5 ) dA ≤
1 × area(R) = 2.
6. Using Fubini’s theorem,
Z
Z 0Z 1
−y 2
(e
+ xy) dy dx =
−1
−x
1
=
0
−y
1
ye−y
2
0
7. The volume is given by
Z 2Z 4
2
Z
+ xy) dx dy =
Z
(4y − y ) dy dx =
1
1
ix=0
1
2
dy
xe−y + x2 y
2
x=−y
0
1 3
1 −y2 1 4 i1
1
3
− y dy = − e
=− + .
− y
2
2
8 0
2e 8
−y 2
(e
0
Z
Z
0
1
3
2
h
h
1 iy=4
32
2y 2 − y 3
dx = .
3 y=0
3
8. The moment of inertia about the y-axis is given by
ZZZ
Iy =
2
2
1
Z
Z
1−x
Z
y 2 +2
(x + z )xyz dV =
E
Z
1
Z
0
1−x
0
(x3 yz + xyz 3 ) dz dy dx
y2
iz=y2 +2
1
dy dx
x3 yz 2 + xyz 4
2
4
z=y 2
0
0
Z 1 Z 1−x 1
1 3 5 1 9
1 3 2
2
2
4
x y(y + 2) + xy(y + 2) − x y − xy dy dx
=
2
4
2
4
0
0
Z 1h
iy=1−x
1
1
1
1 3 2
x (y + 2)3 + x(y 2 + 2)5 − x3 y 6 − xy 10
dx
=
12
40
12
40
y=0
0
Z 1
1 3
1
=
x ((1 − x)2 + 2)3 + x((1 − x)2 + 2)5
12
40
0
1
1 3
2 3 4 6
10
− x (1 − x) − x(1 − x) − x − x dx.
12
40
3
5
=
h1
Now expand to evaluate. The moment of inertia Ix about the x-axis can be similarly
R 1 R 1−x R y2 +2 2
computed as 0 0
(y + z 2 )xyz dz dy dx.
y2
4