# spherical to cylindrical coordinates, volume

A solid is it limited by the surfaces whose equations in spherical coordinates are:

$r=2\cos(\phi) \qquad , \phi =\frac{\pi}{6}, \qquad \phi=\frac{\pi}{3}$.

then write an expression to calculate the volume of solid in cylindrical coordinates.

The figure is a sphere with center in (0,0,1) and radius 1, between two cones. So, I have $0<\theta <2\pi$, but i don't know how write the height z and the radius $r$.

Any help is appreciated.

• Is one of those arguments supposed to be $\theta$? And what do you mean by 'the figure is a sphere with center..'? What figure are you talking about? – Mattos Aug 30 '16 at 3:38
• no, in the solid, only $\phi$. – user119144 Aug 30 '16 at 3:41

$$\rho = 2 \cos \phi \iff \rho^2 = 2 \rho \cos \phi \iff x^2 + y^2 + z^2 = 2 z \iff x^2 + y^2 + (z-1)^2 = 1$$

So like you said, it's a sphere at $(0,0,1)$ with $r=1$ and bounded by the cones $\phi = \pi/6$ and $\phi = \pi/3$.

For cylindrical coordinates, $0 \le \theta \le 2\pi$.

$\phi = \pi/6$ intersects the top half of the sphere $z = 1 + \sqrt{1-r^2}$.

To be precise, $\phi = \pi/6$ corresponds to $z = \sqrt{3} r$, which equals $1 + \sqrt{1-r^2}$ when $r = \sqrt{3}/2$

Similarly $\phi = \pi/3$ corresponds to $z = r / \sqrt{3}$ and intersects the lower half of the sphere $1-\sqrt{1-r^2}$ when $r = \sqrt{3}/2$

Thus the integral is $$\int_0^{2\pi} \int_0^{\sqrt{3}/2} \int_{r/\sqrt{3}}^{\sqrt{3}r} r \; dz \, dr \, d \theta + \int_0^{2\pi} \int_{\sqrt{3}/2}^1 \int_{1 - \sqrt{1-r^2}}^{1 + \sqrt{1-r^2}} r \; dz \, dr \, d \theta = \frac{\pi}{2} + \frac{\pi}{6}$$

and this agrees with the (better) spherical coordinates $$\int_0^{2\pi} \int_{\pi/6}^{\pi/3} \int_0^{2 \cos \phi} \rho^2 \sin \phi \; d \rho \, d \phi \, d \theta = \frac{2\pi}{3}$$

• thx a lot, very usefull – user119144 Aug 30 '16 at 14:44
• Sure, definitely use spherical coordinates here if you can . . . cylindrical was tricky – user288742 Aug 30 '16 at 14:46