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!

  • $\begingroup$ 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} $\endgroup$ – MBYagbasan Nov 23 '14 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 $\frac{1}{\sin\phi}=\frac{\rho}{r}$.

  • $\begingroup$ 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 $r=4\sin\phi$? $\endgroup$ – Marcus Fermat Nov 23 '14 at 5:04
  • $\begingroup$ 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? $\endgroup$ – Marcus Fermat Nov 23 '14 at 5:13
  • $\begingroup$ @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. $\endgroup$ – user84413 Nov 23 '14 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.