Math 221
Quiz 6
Spring 2013
Several Variable Calculus
1
1
2
4𝑦 𝑒𝑥 d𝑥 d𝑦.
∫0 ∫𝑦2
Hint: Reverse the order of integration.
1. Evaluate the iterated integral
Solution. The indicated bounds on 𝑥 are the parabola 𝑥 = 𝑦2 (opening
sideways) and the vertical line 𝑥 = 1. Here is a picture of the integration
region:
1
1
Reversing the limits leads to the following integral:
1
√
𝑥
1
𝑥2
∫0 ∫0
4𝑦 𝑒 d𝑦 d𝑥 =
∫0
[
2𝑦
2
]√𝑥
0
2
𝑒𝑥 d𝑥
1
=
2
2𝑥 𝑒𝑥 d𝑥
∫0
[ 2 ]1
= 𝑒𝑥
0
= 𝑒 − 1.
2. Evaluate the double integral
quadrant where 𝑥2 + 𝑦2 ≤ 1.
February 28, 2013
∬𝐷
𝑥2 d𝐴, where 𝐷 is the region in the first
Page 1 of 2
Dr. Boas
Math 221
Quiz 6
Spring 2013
Several Variable Calculus
Solution. Converting to polar coordinates is the simplest approach:
1
𝜋∕2
2
∬𝐷
𝑥 d𝐴 =
∫0
(𝑟 cos 𝜃)2 𝑟 d𝑟 d𝜃
∫0
𝜋∕2
1
𝑟3 cos2 𝜃 d𝑟 d𝜃
∫0
]1
𝜋∕2 [
1 4
=
𝑟 cos2 𝜃 d𝜃
∫0
4 0
=
∫0
1
=
4 ∫0
𝜋∕2
cos2 𝜃 d𝜃
𝜋∕2
1
(1 + cos 2𝜃) d𝜃
8 ∫0
[
]𝜋∕2
1
1
=
𝜃 + sin 2𝜃
8
2
0
[(
) (
)]
1
1
𝜋 1
+ sin 𝜋 − 0 + sin 0
=
8
2 2
2
𝜋
= .
16
=
3. The polar equation 𝑟 = 2 + sin 𝜃 corresponds to which one of the following
graphs? Explain how you know.
(A)
(B)
(C)
Solution. Observe that 1 ≤ 2 + sin 𝜃 ≤ 3 (since −1 ≤ sin 𝜃 ≤ 1), so 𝑟 is
never equal to 0. Graph (B) is the only one that does not pass through the
origin.
February 28, 2013
Page 2 of 2
Dr. Boas