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.

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

  • $\begingroup$ 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

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


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