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This is a question in which I'm getting the right answer, I just want to make sure what I'm doing its legitimate, as it seems a bit off to me.

The volume enclosed by

[tex]z=16[x^{2}+y^{2}], z=8-16[x^{2}+y^{2}][/tex]

I set my triple integral up as follows:

[tex]\int^{?r}_{?r}\int^{2\pi}_{0}\int^{8-16r^{2}}_{16r^{2}} r dz d\vartheta dr[/tex]

So, my height is the bottom function to the upper, and obviously completing a full circle.

Now here is what I'm unsure about.

I know I need to either do two integrals for r and add them together 0 to 1 and 0 to -1,

or -1 to 1 so that I'm only left with 2/pi (the answer) but what is the logic behind getting the limits for r?

Thanks in advance

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# Homework Help: Volume enclosed by two paraboloids

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