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The following integral is given in Spherical Coordinates, which procedure should I follow to express it in Cylindrical Coordinates?

$$\int_{0}^\pi \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \int_{\frac{2}{\sin(\phi)}}^{4} (16-{\rho}^2){\rho}^2 d{\rho}d{\phi}d{\theta}$$

Thank You!

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1 Answer

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Since $\frac{\pi}{6}\le \phi\le \frac{\pi}{2}$, the region lies above the xy-plane and below the cone given by $z=r\sqrt{3}$

$\hspace {.3 in}$since $\tan\phi=\frac{r}{z}$ and $\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$.

Since $r=\rho\sin\phi$, $\frac{2}{\sin\phi}\le\rho\implies r\ge2$ and

since $\rho=\sqrt{r^2+z^2},\;\;$ $\rho\le4\implies r^2+z^2\le16\implies z\le\sqrt{16-r^2}$.

Therefore the projection of the solid in the xy-plane is the semicircular ring defined by

$0\le\theta\le\pi$ and $2\le r\le 4$, and for each point $(r,\theta)$ in this ring, $0\le z\le\sqrt{16-r^2}$.

This gives $\displaystyle\int_0^{\pi}\int_2^4\int_0^{\sqrt{16-r^2}}\big((16-(r^2+z^2)\big)\bigg(\frac{\sqrt{r^2+z^2}}{r}\bigg) \;r \;dz \;dr \;d\theta$

since $\sin\phi=\frac{r}{\rho}$ and $\frac{1}{\sin\phi}=\frac{\rho}{r}$.

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