Finding Volume of Rugby Ball

Question

I am asked to find the volume of rugby ball whose surface is given by the ellipsoid:

$$\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{9} = 1$$

I am having trouble figuring out which coordinate system I should use. Is it possible to solve the triple integral of the volume by just using cartesian co-ordinates, without making conversions to the spherical or cylindrical coordinate system?

Answer 1

The shape is a unit sphere that has been scaled by factors of $2,2$, and $3$ in the $x, y$, and $z$ directions. The volume is scaled by the same factors. So: $$ V = 2\cdot2\cdot3\cdot\frac43\pi1^3 = 16\pi $$

Answer 2

You can use spherical coordinates by making the transformation $u=\frac{x}{2}$, $v=\frac{y}{2}$ and $w=\frac{z}{3}$. So you will be integrating the Jacobian over the unit ball!

Answer 3

Notice,

In general, the volume of ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ is $$=\frac{4\pi}{3}(abc)$$

Hence, the volume of ellipsoid $\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}=1\iff \frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{1^2}{3^2}=1$ is $$=\frac{4\pi}{3}(2\cdot2\cdot 3)$$ $$=\frac{48\pi}{3}$$ $$=\color{blue}{16\pi}$$

Answer 4

You can also find this volume by rotating the region inside the ellipse $\frac{y^2}{4}+\frac{z^2}{9}=1$ about the z-axis:

$\displaystyle V=2\int_0^3\pi(R(z))^2dz=2\pi\int_0^3 4\big(1-\frac{z^2}{9}\big)dz=8\pi\bigg[z-\frac{z^3}{27}\bigg]_0^3=16\pi.$