Volume of a revolution for an ellipse?

Question

How do you even integrate an ellipse? The question is...

Rotating the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ about the $x$-axis generates an ellipsoid. Compute its volume. The way i know how to rotate a function around the $x$ axis is by multiplying the integrand of $f(x)^2$ by $\pi$

$$\text{volume} = \pi \int f(x)^2$$

Answer 1

Volume of revolution of a curve about the x-axis is calculated by

$$V= \int A(x)\,dx $$

where $A(x)$ is the cross sectional area

$$A(x)= \pi y^2$$

$$dV=\pi y^2\,dx$$ substituting $y^2=b^2(1-{x^2\over a^2})$ we get,

$$V=\int_{-a}^{a} \pi b^2(1-{x^2\over a^2})dx$$

On integrating we get $$V=\frac{4}{3}\pi ab^2$$

Answer 2

The equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{b^2}=1.$

My method is to do it in a spherical polar coordinate.
We define $x=ar\sin\theta\cos\varphi, y=br\sin\theta\sin\varphi, z=br\cos\theta$, then $\text{d}V=\text{d}x\text{d}y\text{d}z=ab^2 r^2\sin\theta\text{d}r\text{d}\theta\text{d}\varphi$. The volume \begin{align} V = \iiint\text{d}V = \int^1_0r^2\text{d}r\int^{2\pi}_0\text{d}\varphi\int^\pi_0ab^2\sin\theta \text{d}\theta = \frac{4ab^2\pi}{3}. \end{align}