Multiple Integrals

**qbkr21** · April 5th 2009, 10:21 AM

Thanks for the help!

**DeMath** · April 5th 2009, 12:48 PM

Originally Posted by qbkr21

Thanks for the help!

$z = 2 + y,{\text{ }}z = 3y.$

$2 + y = 3y \Leftrightarrow y = 1.$

$y = {x^2} = 1 \Leftrightarrow x = \pm 1.$

So $D = \left\{ {\left( {x,y} \right):{\text{ }}{x^2} \leqslant y \leqslant 1,{\text{ }} - 1 \leqslant x \leqslant 1} \right\}.$

$V = \iint\limits_D {\left( {2 + y} \right)dydx} - \iint\limits_D {3ydydx} = 2\iint\limits_D {\left( {1 - y} \right)dydx} =$

$= 2\int\limits_{ - 1}^1 {\int\limits_{{x^2}}^1 {\left( {1 - y} \right)dydx} } = \ldots = \frac{{16}}{{15}}.$

**qbkr21** · April 7th 2009, 05:37 PM

Great I got the same answer. I thank you fine sir for the help!

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