[Math] Proving the Volume of an Ellipsoid

calculusintegrationvolume

This is the question:

The solid generated by rotating the region inside the ellipse with equation
$$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$
around the $x$-axis is called an ellipsoid.

(a) Show that the ellipsoid has volume $\displaystyle \frac{4}{3} \pi a b^2.$

(b) What is the volume if the ellipse is rotated around the $y$-axis.

(I want to accomplish this using integrals and basic plane geometry. To give you an idea of how much I know about integrals I'm $4$ weeks into my calculus II course, which is my first exposure to integrals)

I started by drawing an ellipse on a Cartesian plane. The ellipse went from $a$ to $b$, I then rotated this ellipse around the $x$-axis to get an ellipsoid. Now to get the volume I have to find a cross-sectional area, so I noticed that the ellipsoid is really made up of a bunch of circles stacked along the $x$-axis. Where I'm stuck at right now is how I can find the radius of these circles, which would give me my integrand.

Best Answer

I'll use disk integration, where the volume is given by; $$\pi \int_{x_1}^{x_2} f(x)^2 \mathrm{d}x$$

First we solve for $y$ as a function of $x$, which is trivial so I'll just give the result; $$y = b\sqrt{1-\left (\frac{x}{a}\right )^2}$$

Now the interesting part. The ellipse doesn't go from $a$ to $b$ but from $-a$ to $a$. So we construct;

$$\pi \int_{-a}^{a} b^2 \left (1-\left(\frac{x}{a}\right )^2 \right) \mathrm{d}x=\pi \left( -\frac{b^2}{a^2} \int_{-a}^{a} x^2 \mathrm{d}x + b^2 \int_{-a}^{a} \mathrm{d}x \right )=\pi\left (-\frac{2ab^2}{3}+2ab^2\right )=\frac43 \pi a b^2$$

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