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Unformatted text preview: y ( x, y ) dA. The iterated integrals have the same limits as above, but in x the integrand is x = x 2 , and in y the integrand is y = xy. The integrals evaluate to 7 and 1, so dividing by mass, ( x, y ) = 1 10 3 (7 , 1) = (2 . 1 , . 3) . Problem 15.6.x Evaluate the iterated integral Z 1 Z z Z x + z 6 xz dydxdz. Solution. The rst inside integral has the antiderivative Z 6 xz dy = 6 xyz, so the denite integral is Z x + z 6 xz dy = [6 xyz ] y = x + z y =0 = 6 x ( x + z ) . From this point, the problem is a double iterated integral, with value =1. Finally, we covered problems 15.7.9 (used for #35) then 15.8.10 and 15.8.18....
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 Spring '06
 YUKICH
 Calculus, Integrals

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