# Three different methods for the volume of an ellipsoid, only one works [closed]

The ellipsoid is represented by rotating $(x/3)^2+y^2=1$ around the $x$-axis. I've tried 3 different method and only the third one seems to work out.

What're wrong with the first and second way?

• It is in your best interest that you type your questions (using MathJax) instead of posting pictures. Jul 23 '18 at 16:34
• Thank you. I'll give it a try. Jul 25 '18 at 5:47

We can calculate the integral in several ways.

1) Spherical coordinates

We have that for $x=r\sin \theta$ and $y=r \cos \theta$ by the ellipse'e equation we have

$$\left(\frac{r\sin \theta}{3}\right)^2+r^2 \cos^2 \theta=1\implies r=\frac{9}{\sqrt{9\cos^2 \theta+\sin^2 \theta}}$$

and therefore

$$V=\int_0^{2\pi} d\phi \int_0^{\pi} d\theta \int_0^{\frac{9}{\sqrt{9\cos^2 \theta+\sin^2 \theta}}} \rho^2 \sin \theta \, d\rho=12 \pi$$

2) Disk method

$$V=\int_{-1}^{1} \pi (9-9y^2)dy=12\pi$$

3) Shell method

$$V=\int_{0}^{3} 2\pi x \cdot2\sqrt{1-\frac{x^2}{9}}dx=12\pi$$

4) Pappus theorem

Since the centroid of half ellipse is far $d=\frac{4a}{3\pi}$ from a principal axis and the area of half ellipse is equal to $A=\frac{\pi ab}{2}$ we have

$$V=2\pi d \cdot A=2 \pi \frac{4\cdot 3}{3\pi}\frac{\pi \cdot 3\cdot 1}{2}=12 \pi$$

• I think you confused r and rho for method 1, and also the axis of rotation for methold 2&3. Jul 25 '18 at 6:16
• @NKYu For method 1, $r$ is the radius on the ellipse and $\rho$ is the spherical coordinate varing from $0$ to $r$. For method 2 and 3 it depends how we call the coordinates but it is of course a minor issue.
– user
Jul 25 '18 at 6:31