Exam 3B -- Solution

Exam 3B -- Solution - MAC2313 Exam 3 Version B Solution 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MAC2313 Exam 3 Version B – Solution 1) Evaluate the integral. Z 1 0 Z 1 0 y 1 + x 2 y 2 dxdy. Solution: Begin by using substitution. Let u = xy , which implies du = y dx . Now, Z 1 0 Z 1 0 y 1 + x 2 y 2 dxdy = Z 1 0 Z y 0 1 1 + u 2 dudy = Z 1 0 arctan( y ) dy. Proceed with integration by parts: Let u = arctan( y ) and dv = dy which implies du = 1 1+ y 2 dy and v = y . Z 1 0 arctan( y ) dy = y arctan( y ) | 1 0 - Z 1 0 y 1 + y 2 dy = arctan(1) - 1 2 ln(1+ y 2 ) | 1 0 = π 4 - 1 2 ln(2) . 2) Reverse the order of integration, Z 9 0 Z y 0 f ( x,y ) dxdy. Solution: The region of integration is a type II region: { ( x,y ) R 2 | 0 x y , 0 y 9 } . The region can also be viewed as a type I region: { ( x,y ) R 2 | 0 x 3 , x 2 y 1 } . Therefore, Z 9 0 Z y 0 f ( x,y ) dxdy = Z 3 0 Z 1 x 2 f ( x,y ) dy dx. 3) Evaluate R R D y dA , where D is the region in the first quadrant that lies between the circles x 2 + y 2 = 4 and x 2 + y 2 = 2 x . Solution:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/15/2011 for the course MAC 2313 taught by Professor Keeran during the Spring '08 term at University of Florida.

Page1 / 3

Exam 3B -- Solution - MAC2313 Exam 3 Version B Solution 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online