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I have the integral $$\int_{-3}^3 \int_0^\sqrt{9-x^2} (x^2 + y^2)^{3/2} {dy}{dx}$$

I cannot solve this in it's current form so I realize that the limit is a circle ${x^2} + {y^2} = 9$ using this I attempted to convert the integral to polar coordinates. The integral I came up with is $$\int_{0}^{\pi} \int_{-3}^{3} r(r^2)^{3/2} {dr}{d{\theta}}$$

When I integrate that I get $$\frac{(2)9^{5/2}{\pi}}5$$

When I compute the original integral using mathematica I get $$\frac{81{\pi}}{4}$$

Is my conversion to polar coordinates wrong or my integration?

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Indeed, as we see the region of the integration, we have $$\theta|_0^{\pi},~~~r|_0^3$$.

So we have the following integrals instead:

$$\int_0^{\pi}d\theta\times\int_0^3 r^4dr=(243/5)\pi$$

enter image description here

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  • $\begingroup$ even correcting for that my answer would just be half of what it currently is. I take it my integration must be wrong? $\endgroup$ – 0x41414141 Nov 20 '13 at 14:11
  • $\begingroup$ @bh3244: your final answer(s) is not correct. $\endgroup$ – mrs Nov 20 '13 at 14:22
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The lower limit for $r$ is zero, not $-3$. $r$ is the distance so it can never be negative. In this case, it should be $0\le r\le3$.

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