Volume of a cone inside a sphere

Question

Using spherical coordinates I have to find the volume of a cone $z=\sqrt{x^2+y^2}$ inscribed in a sphere $(x-1)^2+y^2+z^2=4.$

I can`t find $\rho$ because the center of sphere is displaced from the origin.

---

I tried solving it using Mathematica, but i did something wrong somewhere enter image description here

Answer 1

Not an answer (so please don't downvote), but a figure that should help you visualize the problem:

Clearly you want to integrate over $\theta$ from $0 \to 2 \pi$, and $\phi$ from $0 \to \phi_0$, where the limit is based on the half angle of the cone.

The radius $\rho$ is a function of both these variables, and can be derived from the Pythagorean theorem.

Answer 2

HINT

Eliminating $z$ between the two given equations and simplifying one obtains

$$ x^2-x+\frac12+y^2-2=0 $$

the cylinder radius $ R=\frac{\sqrt 7}{2}$

$$ (x-\dfrac12)^2+y^2=R^2$$

with parametrization

$$ x= \dfrac12 +R \cos \theta, y= R \sin \theta $$

Due to symmetry volume of one cone nappe need be considered.

The sketch can be used to define easier parametrized limits of integration. Cone common apex is origin of spherical coordinates.

Answer 3

$x=\rho\cos\theta\sin\phi\\ y=\rho\sin\theta\sin\phi\\ z=\rho\cos\phi$

Plug these substitutions into the given equations.

$x^2 - 2x + 1 + y^2 + z^2 = 4$

$\rho^2 - 2\rho\cos\theta\sin\phi - 3 = 0$

We have $\rho$ as a quadratic, so use the quadratic formula.

$\rho = \cos\theta\sin\phi +\sqrt {\cos^2\theta\sin^2\phi +3}$

Answer 4

1. Another variant is to use little "shifted" spherical coordinates $$\begin{array}{} x-1=\rho\cos\theta\sin\phi\\ y=\rho\sin\theta\sin\phi\\ z=\rho\cos\phi \end{array}$$ Then your first equation goes to $\rho^2 = 4$, Jacobian will be same.

2. Another way is to stay on Cartesian coordinates and find projection of intersection of sphere and cone which is $\left( x-\frac{1}{2} \right)^2+y^2=\frac{7}{4}$. Now volume inside cone bounded by sphere is $$\int\limits_{\frac{\sqrt{7}-1}{2}}^{\frac{\sqrt{7}+1}{2}}\int\limits_{-\sqrt{\frac{7}{4}-\left( x-\frac{1}{2} \right)^2}}^{\sqrt{\frac{7}{4}-\left( x-\frac{1}{2} \right)^2}}\int\limits_{\sqrt{x^2+y^2}}^{\sqrt{4-(x-1)^2-y^2}}dxdydz$$