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# HW12 - HW 12 Solutions Math 1920 Spring 2009 Problem 16.5.1...

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HW 12 Solutions Math 1920 Spring 2009 April 28, 2009 Problem 16.5.1 Let f ( x, y, z ) = x 2 + y 2 - z so that f = (2 x, 2 y, - 1) gives the normal to the surface. We have that ZZ S = ZZ R |∇ f | | k · ∇ f | dA. Here R is the planar region | ( x, y ) | ≤ 2. Surface area is then ZZ R p 4 x 2 + 4 y 2 + 1 dx dy Change to polar coordinates: Z 2 π 0 Z 2 0 p 4 r 2 + 1 r dr dθ = Z 2 π 0 13 6 = 13 3 π. Problem 16.5.17 Let f ( x, y, z ) = 2 x + 2 y + z . Then f = (2 , 2 , 1), |∇ f | = 3, and g ( x, y, z ) = x + y + z = x + y + 2 - 2 x - 2 y = 2 - x - y on the surface f ( x, y, z ) = 2. When z = 0, x + y = 1, so surface area is Z 1 0 Z 1 - x 0 3(2 - x - y ) dy dx = 3 Z 1 0 (2 - x )(1 - x ) - 1 2 (1 - x ) 2 dx = 3 Z 1 0 3 2 - 2 x + x 2 2 dx = 2 . Problem 16.5.23 g = (2 x, 2 y, 2 z ), so |∇ g | = p 4 x 2 + 4 y 2 + 4 z 2 = 2 a . Then n = ( x,y,z ) a and = a z dA . F · n = xy a - xy a + z a = z a . Then flux is ZZ R ( z a )( a z ) dA = πa 2 4 Problem 16.5.34 f = (0 , 2 y, 2 z ), so |∇ f | = p 4 y 2 + 4 z 2 = p 4( y 2 + z 2 ) = 6. |∇ f · k | = 2 z so = 6 2 z dA = 3 z dA . M = RR S 1 = Z 3 - 3 Z 3 0 3 z dx dy = Z 3 - 3 Z 3 0 3 p 9 - y 2 dx dy = 9 π.

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HW12 - HW 12 Solutions Math 1920 Spring 2009 Problem 16.5.1...

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