Volume of Ellipsoid Around X Axis

Question

Given the Ellipsoid $x^2+4y^2+4z^2\le 4$ How Can I calculate it's volume using the volume of revolution around $x$ axis?

Answer 1

Cartesian equation of ellipsoid is $$\frac{x^2}{4}+y^2+z^2=1$$ In $xy$ plane where $z=0$ its equation is $$\frac{x^2}{4}+y^2=1\to y^2=1-\frac{x^2}{4}$$ The volume of the solid of rotation around $x$-axis is $$V=\pi\int_{-2}^2 \left(1-\frac{x^2}{4}\right)\,dx=\frac{8}{3}\pi$$

The volume of an ellipsoid having semiaxis $a,b,c$ is

$V=\frac{4}{3}\pi abc=\frac{4}{3}\pi \cdot 2\cdot 1\cdot 1=\frac{8}{3}\pi$

Hope this helps

Answer 2

At each x, $y^2+ z^2= 4- x^2$. That is circle with center at (x, 0, 0) and radius $\sqrt{4- x^2}$ so area $\pi(4- x^2)$. Taking an "infinitesimal" thickness, dx, the volume of such disk is $\pi(4- x^2)dx$. The volume of the ellipsoid is $\pi \int_{-2}^2 (4- x^2)dx$.