Triple integral between $z^2 = {(x - 1)}^2 + y^2$ and the unit sphere.

Question

I have to compute $\int_D f$, where $D$ is the region in ${(0 , \infty)}^3$ between the cone $z^2 = {(x - 1)}^2 + y^2$ and the sphere $x^2 + y^2 + z^2 = 1$, and $f : D \to \mathbb{R}$ is given by $f(x , y , z) = z \sqrt{x^2 + y^2}$.

My attempt: If I use spherical coordinates $(\rho , \varphi , \theta) \in (0 , \infty) \times (0 , 2 \pi) \times (0 , \pi)$ given by $$ \left\{ \begin{array} xx = \rho \sin \varphi \cos \theta \\ y = \rho \sin \varphi \sin \theta \\ z = \rho \cos \varphi \end{array} \right. $$ then we can start to deduce $\rho \leq 1$ from the equation of the sphere, and $\varphi < \frac{\pi}2$ is deduced from $z > 0$. We can use this to see furthermore that $\theta < \frac{\pi}2$ because $x , y > 0$. The equation of the cone says $$ {\rho}^2 {\cos}^2 \varphi = {\rho}^2 {\sin}^2 \varphi - 2 \rho \sin \varphi \cos \theta + 1. $$ This helps appearently nothing. I am wondering if spherical coordinates are helpful to fix this integral.

Answer 1

Cylindrical coordinates is easier.

Equation of the surface of the cone, $z^2 = (x - 1)^2 + y^2$

Equation of the surface of the sphere, $x^2 + y^2 + z^2 = 1$

At intersection,

$z^2 = (x - 1)^2 + y^2 = 1 - x^2 - y^2 \implies x^2 - x + y^2 = 0$

This is the projection in xy plane, of the intersection of cone with the sphere. It is circle of radius $\frac{1}{2}$ at $(\frac{1}{2}, 0)$ in xy-plane.

Now in polar coordinates, $x = r \cos\theta, y = r\sin\theta$ so the equation of the circle becomes,

$r^2 - r \cos\theta = 0 \implies r = \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$ as we are only supposed to consider region in first octant.

Bounds of $z$ simply come from the equations of cone and the sphere. It is bound below by the cone and above by the sphere. So,

$ \sqrt {1 + r^2 - 2 r\cos\theta} \leq z \leq \sqrt{1-r^2}$

So integral becomes,

$\displaystyle \int_0^{\pi/2} \int_0^{\cos\theta} \int_{\sqrt {1 + r^2 - 2 r\cos\theta}}^{\sqrt{1-r^2}} r^2 z \ dz \ dr \ d\theta$

Given the inner most integral will return $z^2$, the integral becomes simpler to evaluate.