0
$\begingroup$

This is the integral to evaluate the volume of the cone $z=\sqrt{x^2+y^2}$ inside the sphere $x^2+y^2+z^2=z$ How can I rewrite this integral in cylindrical coordinates. Thank you.

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

$\endgroup$
2
  • $\begingroup$ You don't actually need the integral in spherical coordinates. Just the description of the volume is sufficient. $\endgroup$ – Shubham Johri Nov 6 '20 at 9:18
  • $\begingroup$ I know that, but there is a question on a practice final exam that asks to specifically set up this in cylindrical coordinates. $\endgroup$ – Math Whiz Nov 6 '20 at 9:54
1
$\begingroup$

The sphere and cone intersect at $(0,0,0)$ and the curve $x^2+y^2=1/4,z=1/2$. The volume looks like a cone with a hemispherical top. In cylindrical coordinates, the lower boundary is $z=r$ and the upper boundary is the sphere $z^2-z+r^2=0$, which gives $z=\frac12+\sqrt{\frac14-r^2}$. The intersection curve is $r=z=\frac12$. The required integral is$$\int_0^{2\pi}\int_0^{1/2}\int_r^{\frac12+\sqrt{\frac14-r^2}}r~dz~dr~d\theta$$

$\endgroup$

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.