q5-sol-1042-sp08

# q5-sol-1042-sp08 - Calculus II—1042 Solution of Quiz 5 1...

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Unformatted text preview: Calculus II—1042 Solution of Quiz # 5 1. Find the volume of the solid generated by rotating the region bounded by y = x 2 , y = 4 , and y = axis (a) about y- axis . Solution: No hole in the solid and cross section is perpendicular to y- axis so limits will be y- values, ≤ y ≤ 4 and integral for volume will be set up in y s (Curve y = x 2 OR x = √ y and the radius is from y- axis to the curve x = √ y ) r = x right- x left = √ y- 0 = √ y Area of the cross section is A ( y ) = πr 2 = π ( √ y ) 2 = π y V = Z 4 A ( y ) dy = π Z 4 y dy = π y 2 2 ¶ | 4 = 8 π (b) about y = 4 Solution: No hole in the solid and cross section is perpendicular to x- axis so limits will be x- values, ≤ x ≤ 2 and integral for volume will be set up in x s (The radius is from y = 4 to the curve y = x 2 ) r = y upper- y lower = 4- x 2 , and r 2 = (4- x 2 ) 2 = 16- 8 x 2 + x 4 Area of the cross section is A ( x ) = πr 2 = π (16- 8 x 2 + x 4 ) V = Z 2 A ( x ) dx = π Z 2 (16- 8 x 2 + x 4 ) dx...
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## This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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q5-sol-1042-sp08 - Calculus II—1042 Solution of Quiz 5 1...

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