Volume of ellipsoid

Question

How do I calculate the volume of the intersection of ellipsoid $x^2/36+y^2/49+z^2/49\leq 1$ and subspace $x/6+y/7+z/7\leq1$ ?

Answer 1

If the 36 were replaced with 49, and the 6 were replaced with 7, you have a plane cutting a sphere, and the volume of the resulting pieces of the sphere can be found in standard ways that I could elaborate on if you'd like.

To get the volume you are looking for, all the components of this sphere will have been contracted by $6/7$ in one dimension to obtain that ellipse. Such a contraction has the effect of scaling volume by $6/7$.

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Added: Consider a sphere with radius 7, centered at the origin. Also there is a horizontal plane at $z=7-a$, which slices the sphere $a$ units below the sphere's north pole. Let's find the volume of the "lens" that lies above. Using the method of discs,

$$V_{\text{lens}}=\int_{z=7-a}^{z=7}\pi(\sqrt{49-z^2})^2\,dz$$

This is an easy integral. From here, the volume of the rest of the sphere is $\frac{4}{3}\pi(7^3)-V_{\text{lens}}$.

For our purposes, $a$ should be the distance between $\left(\frac{7}{\sqrt{3}},\frac{7}{\sqrt{3}},\frac{7}{\sqrt{3}}\right)$ and $\left(\frac{7}{{3}},\frac{7}{{3}},\frac{7}{{3}}\right)$. (Imagine a lens cap that is sliced off from the sphere using the plane $x/7+y/7+z/7=1$; this distance would be its "altitude".) This distance should be easy to compute.

So now you have $\frac{4}{3}\pi(7^3)-V_{\text{lens}}$, the volume of the complement of a lens cut off from a sphere. Your actual problem asked for a version of this object that has been compressed by a factor of $6/7$ in the $x$-direction. So the volume will be compressed in the same way, and the result is $\frac{6}{7}\cdot\left(\frac{4}{3}\pi(7^3)-V_{\text{lens}}\right)$.