This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Name:
You have ﬁfteen minutes to complete the quiz. If anything seems unclear,
please ask.
1. Evaulate the integral
f (x, y )dA,
R where f (x, y ) = 6x2 y + 2y and R = [−1, 1] × [2, 3]. (3 points) 2. Let f (x, y ) be a general function of two variables. Sketch the region of
integration, and switch the order for the integral below. (4 points)
√ 2 x f (x, y )dydx
1 0 3. Set up (do not try to solve) an integral for the volume of the region
where 0 ≤ x ≤ 1 − y , 0 ≤ y and 0 ≤ z ≤ x2 + ex  cos(y ). (3 points) 1 Solutions
If you have any questions and / or think you’ve spotted a mistake / typo, please
get in touch with me.
1. The integral we want to solve is
3 1 6x2 y + 2ydxdy.
2 −1 The inner integral is
1 6x2 y + 2ydx = 2x3 y + 2yx
−1 x=1
x=−1 = 2y + 2y − (−2y − 2y ) = 8y, and the outer integral is
3 8ydy = [4y 2 ]3 = 36 − 16 = 20.
2
2 2. I can’t get the computer to draw the region of integration nicely (sorry),
√
but it’s the region between the lines y = 0 and y = x in the vertical direction, and between the lines x = 1 and x = 2 in the horizontal direction.
The integral with the order switched is
1 √ 2 2 2 f (x, y )dxdy +
0 f (x, y )dxdy.
1 1 y2 3. The desired integral is
1−x 1 x2 + ex  cos(y )dydx
0 0 (drawing a picture of the region you are integrating over might help see
why this is). 2 ...
View
Full Document
 Spring '08
 ALDROUBI
 Region, ex  cos

Click to edit the document details