# Volume of ellipsoid outside sphere

I have the ellipsoid $\frac{x^2}{49} + y^2 + z^2 = 1$ and I want to calculate the sum of the volume of the parts of my ellipsoid that is outside of the sphere $x^2+y^2+z^2=1$

How to do this? I know the volume of my sphere, $\frac{4\pi}{3}$, and that I probably should set up some double- or triple integral and transform the coordinates to spherical coordinates and evaluate but I have to admit I'm stuck on how to set this up.

Hints.

1) If $(x,y,z)$ satisfies $x^2+y^2+z^2\leq 1$ then $$\frac{x^2}{49} + y^2 + z^2 \leq x^2+y^2+z^2\leq 1.$$ What does this inequality mean?

2) $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$ is the equation of an ellipsoid centered at the origin of semi-principal axes of length $a$, $b$, $c$, and its volume is $\frac{4\pi (a\cdot b \cdot c)}{3}$.

The ellipsoid is simply the sphere, stretched in the $x$-direction by a linear factor of $\sqrt{49}=7$.

Therefore the volume of the ellipsoid is $7$ times the volume of the sphere, so the volume of the ellipsioid minus the sphere (which is completely contained in it) is $6$ times the volume of the sphere.

If you do not have a good perception of objects in 3D and want a pure analytical solution:

Let $A(z)$ be the area of a slice of the ellipsoid outside the sphere, at height $z$. At height $z$, the ellipsoid is the ellipse $$\frac{x^2}{49(1-z^2)}+\frac{y^2}{1-z^2}=1,$$ which has area $7(1-z^2)\pi$, and the sphere is the disc $$\frac{x^2}{(1-z^2)}+\frac{y^2}{1-z^2}=1,$$ which has area $\pi(1-z^2)$. And since $A(z)$ equals the area of this ellipse minus the area of the disc: $$A(z)=6(1-z^2)\pi$$

To compute the total volume, just integrate $A(z)$ between heights $z=-1$ and $z=1$: $$V=\int_{-1}^1 A(z)\; dz = \int_{-1}^1 6(1-z^2)\pi\; dz = 8\pi$$

Indeed, this equals $6$ times the volume of the sphere.