multiple integralmultivariable-calculus
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 are very close. The integral in cylindrical coordinates has the right value, but the lower limit for $z$ should be $r$, not $0$. Notice that in spherical coordinates the angle $\phi$ should be integrated from $0$ to $\pi/4$. This restrict us to the volume inside the cone.
Addendum: The volume is the region between the cone and the sphere in octant I ($x,y,z\ge 0$). See the figure below.
In cylindrical coordinates we have $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta.$$
In spherical coordinates we have $$\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi.$$
I think your method is correct (of converting first to cylindrical, and then to spherical), but you did make one mistake. Here I will convert directly to spherical from Cartesian using the transformation:
$$ \begin{align*} x &= \rho \sin \phi \cos \theta \\ y &= \rho \sin \phi \sin \theta \\ z &= \rho \cos \phi \end {align*} $$
So the equation $y = (x^2+y^2)^2 + z^4$ becomes:
$$ \rho \sin \phi \sin \theta = \rho^4 \left( \sin^4 \phi + \cos^4 \phi \right) $$
Which you can solve for $\rho$:
$$ \rho = \sqrt[3]{\frac{\sin \phi \sin \theta}{\sin^4\phi + \cos^4\phi}} $$
Note it is $\sin\phi\sin\theta$ in the numerator, and not $\sin^2\theta$.
Now to compute the volume, you should be able to integrate the volume differential $(\rho^2 \sin\phi)$ over the region where $\rho$ is between 0 and the expression above (and the appropriate bounds for $\theta$ and $\phi$).
Best Answer
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$$\int_0^{2\pi}\int_0^{1/2}\int_r^{\frac12+\sqrt{\frac14-r^2}}r~dz~dr~d\theta$$