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Unformatted text preview: x = p 2 y1 , 5 8 ≤ y ≤ 1 about the yaxis. Solution: We have the following information about the curve: x = p 2 y1 dx dy = 1 √ 2 y1 Then the area of the surface of revolution, A , is A = Z 1 5 8 2 πx s 1 + ± dx dy ² 2 dy = Z 1 5 8 2 π p 2 y1 r 1 + 1 2 y1 dy = Z 1 5 8 2 π s (2 y1) ± 1 + 1 2 y1 ² dy = Z 1 5 8 2 π p (2 y1) + 1 dy = 2 π Z 1 5 8 p 2 y dy = 4 √ 2 π 3 y 3 2 ³ ³ ³ 1 5 8 = 4 √ 2 π 35 √ 5 π 12 . 2...
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This note was uploaded on 10/08/2010 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.
 Spring '08
 Vershynin
 Calculus

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