# Change the integral from spherical coordinates to cylindrical coordinates

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.

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

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

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$$\int_0^{2\pi}\int_0^{1/2}\int_r^{\frac12+\sqrt{\frac14-r^2}}r~dz~dr~d\theta$$