0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Mattos Aug 30 '16 at 3:38
  • $\begingroup$ no, in the solid, only $\phi$. $\endgroup$ – user119144 Aug 30 '16 at 3:41
1
$\begingroup$

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

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 $$ \int_0^{2\pi} \int_{\pi/6}^{\pi/3} \int_0^{2 \cos \phi} \rho^2 \sin \phi \; d \rho \, d \phi \, d \theta = \frac{2\pi}{3} $$

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

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.