MATH 251
Practice Problems for Examination 2
Spring 2008
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. Let E be the volume in R3 lying between the parabolic cylinders z = y 2 and z = y 2 + 2
and aboveRRR
the region in the xy-plane bounded by the lines x = 0, y = 0, and y = 1 − x.
Compute
xy dV .
E
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 − 32 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
−y
Z
0
Z
2
5 cos(y ) dx dy =
−1
Z
h
ix=−y
2
5x cos(y )
dy =
x=0
−1
0
0
−5y cos(y 2 ) dy
−1
i0
5
5
= − sin(y 2 )
= sin(1).
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 θ2 1 2 i π4
1 π2
= 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 2
−2 2x2 +6
Z 2
Z 2
5 2 2
1
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
Z 1Z y
2
2
(6x + y ) dx dy =
V =
0
y2
1
Z 1
h
ix=y
3
2
2x + y x
(3y 3 − 2y 6 − y 4 ) dy
dy =
x=y 2
0
0
3
2
1 i1
37
= y4 − y7 − y5 =
.
4
7
5 0 140
4. The mass and moments are
Z
Z 5 Z x2
2
x dy dx =
m=
5
1 i5
x4 dx = x5 = 54 ,
5
0
0
0
0
Z 5h
Z 5 Z x2
Z 5
2
i
1 2 2 y=x
1 6
1 7 i5
57
2
x y dy dx =
Mx =
xy
dx =
x dx = x
=
2
14
14
y=0
0
0
0
0
0 2
Z 5 Z x2
Z 5
i
6
5
1
5
x3 dy dx =
x5 dx = x6 = ,
My =
6
6
0
0
0
0
, 125 ).
and so the center of mass is ( 25
6 14
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 0Z 1
Z
−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
h
1
2
dy
xe−y + x2 y
2
x=−y
0
1 −y2 1 4 i1
1
3
1 3
=− + .
− y
− y dy = − e
2
2
8 0
2e 8
−y 2
(e
0
Z
Z
0
1
3
2
h
1 iy=4
32
2y 2 − y 3
dx = .
3 y=0
3
8.
ZZZ
Z
1
Z
1−x
Z
y 2 +2
1
Z
1−x
xy dz dy dx =
xy dV =
0
E
Z
Z
y2
0
1Z
=
0
1−x
1
Z
2xy dy dx =
0
Z
=
0
h
(x − 2x2 + x3 ) dx =
0
1
= .
12
4
z=y 2
0
iy=1−x
dx
y=0
0
1
xy 2
h
iz=y2 +2
xyz
dy dx
h1
2
1 i1
x2 − x3 + x4
2
3
4
0