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Exam 3A -- Solution

# Exam 3A -- Solution - MAC2313 Exam 3 Version A Solution 1...

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MAC2313 Exam 3 Version A – Solution 1) Evaluate the integral. Z 2 0 Z 1 0 y 1 + xy dx dy. Solution: Begin by using substitution. Let u = 1 + xy , which implies du = y dx . Now, Z 2 0 Z 1 0 y 1 + xy dx dy = Z 2 0 Z 1+ y 1 1 u du dy = Z 2 0 ln(1 + y ) dy. Proceed with integration by parts: Let u = ln(1+ y ) and dv = dy which implies du = 1 1+ y dy and v = y . Z 2 0 ln(1+ y ) dy = y ln(1+ y ) | 2 0 - Z 2 0 y 1 + y dy = 2 ln(3) - (ln(1+ y ) - (1+ y )) | 2 0 = 3 ln(3) - 2 . 2) Reverse the order of integration, Z 9 0 Z 3 y f ( x, y ) dx dy. Solution: The region of integration is a type II region: { ( x, y ) R 2 | y x 3 , 0 y 9 } . The region can also be viewed as a type I region: { ( x, y ) R 2 | 0 x 3 , 0 y x 2 } . Therefore, Z 9 0 Z 3 y f ( x, y ) dx dy = Z 3 0 Z x 2 0 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: The region D can be described as a polar region: { ( r, θ ) R 2 | 0 θ π 2 , 2 cos( θ ) r 2 } .

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

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