# Convert from Spherical to Cylindrical Coordinates

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!

• it is easy to solve the integral, what will you do if you change the coordinates? Integration domain is suitable for spherical coordinates. However, the relation between the spherical and cylindrical coordinates is \begin{align} r&=\rho \sin\theta\\ \phi &=\phi\\ z&=\rho\cos\theta. \end{align} Commented Nov 23, 2014 at 0:31

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

• Thanks! I understand the fact that $r\ge2$ but how do you conclude that $r\le 4$. Given that $r=\rho \sin\phi$, shouldn't we have that $r=4\sin\phi$? Commented Nov 23, 2014 at 5:04
• And also, why is the integrand that way? don't we just substitute ${\rho}^2$ for $r^2+z^2$ and then multiply by r because of the Jacobian of the Transformation? Commented Nov 23, 2014 at 5:13
• @MarcusFermat I am using that $r=\rho\le4$ in the xy-plane, and that $dV=\rho^2 \sin\phi d\rho\ d\phi d\theta$ in spherical coordinates. The solid is the region outside the cylinder $r=2$ and inside the top half of the sphere $\rho=4$ which lies to the right of the xz-plane. Commented Nov 23, 2014 at 20:20