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September 22, 2014

September 22, 2014

Posted by **Anonymous** on Friday, July 19, 2013 at 9:29am.

- Math -
**MathMate**, Friday, July 19, 2013 at 9:52amThe volume to be integrated is bounded by the planes

0≤x≤2

0≤y≤2

z=0

and the surface

z=16-x²-2y².

Thus the volume is given by the triple integral:

V=∫∫∫z dz dx dy

Where z is evaluated from 0 to 16-x²-2y²,

x and y are both evaluated from 0 to 2.

Thus

V=∫∫(16-x²-2y²)dx dy

=∫(88-12y²)/3 dy

=48

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