Challenge Problems 9
Math126C. 05/10/2011
Problem 1. [Midterm 2, Milakis, Winter 2009, 5] Let
D := {(x, y) | 1 ≤ x ≤ 2, ln x ≤ y ≤ ex }.
Compute the area of D.
Problem 2. [Midterm 2, Perkins, Winter 2009, 3a] Evaluate the following integral:
ZZ
xy sin(x2 y)dxdy, R := [0, 1] × [0, π/2];
R
Problem 3. [Midterm 2, Perkins, Winter 2009, 3b] Evaluate the following integral:
ZZ
y 2 exy dxdy, D := {(x, y) | 0 ≤ y ≤ 3, 0 ≤ x ≤ y}.
D
Problem 4. [Final Exam, Spring 2010, 6] Find the volume of the solid that lies under the plane
3x + 2y + z = 12 and above the rectangle R = [0, 1] × [2, 3].
Problem 5. [Final Exam, Spring 2008, 7] Evaluate the integral
Z2 4−x
Z 2
0
xe2y
dydx.
4−y
0
1
Solutions
Problem 1. The area of D is
ZZ
1dxdy.
D
By Fubini’s Theorem, this integral equals


Z2
Z2 Zex
 1dy  dx = (ex − ln x)dx.
1
1
ln x
But
Z2
x=2
= e2 − e.
e dx = e x
x
x=1
1
How to compute the integral of ln x? We need to find
mistake was that
Z2
ln xdx =
1
the antiderivative of this function. One common
x=2
1 .
x x=1
0
But this is not true, since (1/x) 6= ln x! To find the antiderivarive of ln x, we need to integrate by parts:
Z
Z
Z
Z
Z
1
0
0
ln xdx = x ln xdx = x ln x − x(ln x) dx = x ln x − x dx = x ln x − dx = x ln x − x.
x
Hence (x ln x − x)0 = ln x, and by the Fundamental Theorem of Calculus
Z2
ln xdx = (x ln x − x)|x=2
x=1 = (2 ln 2 − 2) − (ln 1 − 1) = 2 ln 2 − 1 = ln 4 − 1.
1
Hence the area of D is
e2 − e − 2 ln 2 − 1.
Problem 2. First, find this one-variable integral:
Z1
xy sin(x2 y)dx =
Zy
sin t
dt
,
2
0
0
2
where we have changed the variables: t = x y, 0 ≤ t ≤ y, dt = 2xydx. And
Zy
0
t=y
dt
1
1
sin t = (− cos t)
= (1 − cos y).
2
2
2
t=0
Hence by Fubini’s Theorem
ZZ
xy sin(x2 y)dxdy =
Zπ/2Z1
0
R
1
2
xy sin(x2 y)dx dy =
0
Zπ/2
1
(1 − cos y)dy =
2
0
y=π/2
Zπ/2
Zπ/2
1
1π 1
π 1
dy −
cos ydy =
− (sin y)
= − .
2
22 2
4 2
y=0
0
0
2
Problem 3. Since (yexy )x = y 2 exy , we have by the Fundamental Theorem of Calculus:
Zy
x=y
2
= yey − y.
y e dx = ye 2 xy
xy x=0
0
Hence by Fubini’s Theorem
ZZ
y 2 exy dxdy =
Z3 Zy
0
D
y 2 exy dx dy =
Z3
0
Z3
y2
(ye − y)dy =
0
y2
Z3
ye dy −
0
ydy.
0
To find the first integral, change variables: t = y 2 , 0 ≤ t ≤ 9, dt = 2ydy.
Z3
Z9
y2
ye dy =
0
0
t=9
1 t e9 − 1
e
= e =
.
2
2 t=0
2
t dt
And the second integral is much easier to compute:
Z3
0
y=3
y 2 9
ydy = = .
2 y=0 2
Thus, the answer is
1 9
9
e9
(e − 1) − =
− 5.
2
2
2
Problem 4. This body is between z = 0 and z = 12 − 3x − 2y and above the rectangle R. So its
volume is
ZZ
Z3 Z1
(12 − 3x − 2y)dA =
(12 − 3x − 2y)dxdy.
2
R
But
Z1
0
0
x=1
3x2
3
21
(12 − 3x − 2y)dx = 12x −
= 12 − − 2y =
− 2xy − 2y.
2
2
2
x=0
Therefore,
Z3 Z1
Z3 (12 − 3x − 2y)dxdy =
2
0
2
y=3
21
21
21
11
2 − 2y dy =
y−y =
−5= .
2
2
2
2
y=2
√
Problem 5. Let us change the order of integration. y √≤ 4 − x2 ⇔ x2 ≤ 4 − y ⇔ x ≤ 4 − y.
Therefore, the domain of integration is 0 ≤ y ≤ 4, 0 ≤ x ≤ 4 − y. And the integral is equal to
√

Z4
Z4−y
Z4 2 x=√4−y 2y
Z4
x e
4 − y e2y

 e2y
xdx
dy =
dy =
dy =

4−y
2 x=0
4−y
2 4−y
0
0
0
1
2
Z4
0
0
y=4
1 e2y e8 − 1
e dy =
=
.
2 2 y=0
4
2y
3