Hw3+Solutions

# Hw3+Solutions - = 3 √ 34 8.2.22 t = π 2 8.2.38 cos 5 π...

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Solutions to Homework 3 7.6.2 Z 1 - 1 Z 1 - 1 xy dx ! dy = Z 1 - 1 " x 2 y 2 # 1 - 1 dy = Z 1 - 1 0 dy = 0 . 7.6.8 Z 1 0 Z x 0 e x + y dy ! dx = Z 1 0 " e x + y # x 0 dx = Z 1 0 ( e 2 x - e x ) dx = " e 2 x 2 - e x # 1 0 = e 2 2 - e + 1 2 . 7.6.12 Z 2 0 Z 3 2 e y - x dy dx = Z 2 0 ( e 3 - x - e 2 - x ) dx = e 3 - e 2 - e + 1 . 7.6.14 Z 1 0 Z 3 x 0 ( x 2 + y 2 ) dy dx = Z 1 0 x 2 · x 1 / 3 + x 3 dx = 7 15 8.1.2 π 10 , 2 π 5 , 5 π 6 . 8.1.5 4 π . 8.1.8 9 π/ 2 8.1.18 2 π/ 3 , - π/ 6 , 7 π/ 2 8.2.2 sin t = 2 / 3 , cos t = 5 / 3 8.2.6 hypotenuse= 34. So sin t = 5 / 34. cos t
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Unformatted text preview: = 3 / √ 34. 8.2.22 t = π/ 2. 8.2.38 cos 5 π = cos π =-1 cos-2 π = cos 0 = 1 cos 17 π/ 2 = cos π/ 2 = 0 cos-13 π/ 2 = cos-π/ 2 = 0 1...
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## This note was uploaded on 04/29/2011 for the course MATH 16B taught by Professor Sarason during the Spring '06 term at Berkeley.

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