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Quiz6_sol - Z 1 Z 1 xy p x 2 y 2 dydx = Z 1 x ±Z 1 y p x 2...

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NAME: Spring 2011, MAC 2313, Quiz 6 UFID: Section: 3124 1. Use Lagrange multiplier to find the maximum and minimum values of the function f ( x, y ) = 4 x +6 y subject to the constraint x 2 + y 2 = 13. (2 points) Solution . Let g ( x, y ) = x 2 + y 2 . We have f = h 4 , 6 i , and g = h 2 x, 2 y i . Solveing the system f = λ g and g ( x, y ) = 13 for ( x, y, λ ) we get 2 = λx 3 = λy x 2 + y 2 = 13 (1) and hence 13 = λ 2 ( x 2 + y 2 ) = 13 λ 2 which means λ = ± 1. The solutions to this system are therefore ( x, y, λ ) = ± (2 , 3 , 1). So (2 , 3) has the maximum value 26 and ( - 2 , - 3) has the minimum value - 26. 2. Evaluate the double integral RR R (4 - 2 y ) dA where R = [0 , 1] × [0 , 1] (any method counts). (2 points) Solution . The iterated integral is ZZ R (4 - 2 y ) dA = Z 1 0 Z 1 0 (4 - 2 y ) dydx = Z 1 0 (4 y - y 2 ) | 1 0 dx = Z 1 0 3 dx = 3 . (2) 3. Calculate the iterated integral R 1 0 R 1 0 xy p x 2 + y 2 dydx . (2 points) Solution
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Unformatted text preview: Z 1 Z 1 xy p x 2 + y 2 dydx = Z 1 x ±Z 1 y p x 2 + y 2 dy ² dx = Z 1 x ±Z 1 1 2 p x 2 + y 2 dy 2 ² dx = Z 1 x ³ 1 3 ( x 2 + y 2 ) 3 / 2 ´ 1 dx = Z 1 x 3 h ( x 2 + 1) 3 / 2-x 3 i dx = 1 9 h ( x 2 + 1) 5 / 2-x 5 i 1 = 1 9 (2 5 / 2-1) . (3) 4. Find the the volume of the solid in the first octant bounded by the cylinder z = 16-x 2 and the plane y = 5. (2 points) Solution . The zero of z = 16-x 2 = 0 is x = 4 (in the first octant). The volume of the solid is the integral Z 4 Z 5 (16-x 2 ) dydx = 5 × ± 16 x-x 3 3 ² 4 = 5 × ± 64-64 3 ² = 640 3 . (4)...
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