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Solve $$\int\limits_{0}^{3}\int\limits_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int\limits_{0}^{9-3\sqrt{x^2+y^2}}dzdxdy.$$

Update: The problem is setting up the integral. What I have tried was $$\int\limits_{0}^{\pi}\int\limits_{-3}^{3}\int\limits_{0}^{9-3r}dzrdrd\theta.$$ It gave me $54\pi$ and I don't think it's right.

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@Sigur are you sure? –  Jack F Nov 6 '12 at 0:31
    
Sorry, I didn't see the $r$ after $dz$. Now $\int _{0}^{\pi }\!\int _{-3}^{3}\!\int _{0}^{9-3\,r}\!1{dz}r{dr}\,{d\theta} =-54\,\pi$ –  Sigur Nov 6 '12 at 0:34
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1 Answer 1

Without computations, one could recognize this integral as the volume of one half of circular cone with height $9$ and base radius $3$. The volume is $$\frac12 \frac{\pi 3^2\cdot 9}{3} = \frac{27\pi}{2} $$

To set up integrals, begin with the inner variable, $z$: $$ \int_{0}^{9-3\sqrt{x^2+y^2}}\,dz = \int_0^{9-3r}\,r\,dz$$ The $x,y$ limits describe the upper half of disk of radius $3$ (because $y\ge 0$). This means $0\le r\le 3$ and $0\le \theta \le \pi$. So, the result is $$\int\limits_{0}^{\pi}\int\limits_{0}^{3}\int\limits_{0}^{9-3r}r\,dz\, dr\,d\theta =\int\limits_{0}^{\pi}\int\limits_{0}^{3} (9r-3r^2)\,dr\,d\theta = \pi (9\cdot 3^2/2-3^3) = \frac{27 \pi}{2} $$

Negative values of $r$ are sometimes useful in drawing polar curves, but they should never be used in integration.

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