Volume of Ellipsoid using Triple Integrals

Question

Given the general equation of the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1$, I am supposed to use a 3D Jacobian to prove that the volume of the ellipsoid is $\frac{4}{3}\pi abc$

I decided to consider the first octant where $0\le x\le a, 0\le y \le b, 0 \le z \le c$

I then obtained $8\iiint _E dV$ where $E = \{(x, y, z): 0\le x \le a, 0\le y \le b\sqrt{1-\frac{x^2}{a^2}}, 0\le z \le c\sqrt{1-\frac{x^2}{a^2} - \frac{y^2}{b^2}} \}$

I understood that a 3D Jacobian requires 3 variables, $x$, $y$ and $z$, but in this case I noticed that I can simple reduce the triple integral into a double integral:

$$8 \int_0^a \int_0^{b\sqrt{1-\frac{x^2}{a^2}}} c\sqrt{1-\frac{x^2}{a^2} - \frac{y^2}{b^2}} dydx$$ which I am not sure what substitution I should do in order to solve this, any advise on this matter is much appreciated!

Answer 1

HINT

Let use spherical coordinates with

• $x=ra\sin\phi\cos\theta$
• $y=rb\sin\phi\sin\theta$
• $z=rc\cos\phi$

and with the limits

• $0\le \theta \le \frac{\pi}2$

• $0\le r \le 1$

• $0\le \phi \le \frac{\pi}2$

Remember also that in this case

$$dx\,dy\,dz=r^2abc\sin \phi \,d\phi \,d\theta \,dr$$

Answer 2

Perhaps this was the intended solution: Consider the linear map $L:\mathbb R^3\to\mathbb R^3$ given by $L(x,y,z)=(ax,by,cz)$. This $L$ maps the ball $B=\{(x,y,z);x^2+y^2+z^2\leq1\}$ to the ellipse $E=\{(x,y,z);x^2/a^2+y^2/b^2+z^2/c^2\leq1\}$. In fact, we can think of $L$ as a diffeomorphism $B\to E$. We can now compute the volume of $E$ as the integral $$ \int_E1 = \int_{L(B)}1 = \int_B1\cdot\det(L) = \det(L)\int_B1, $$ because the determinant is constant. The integral over the ball is the volume of the ball, $\frac43\pi$, and the determinant of $L$ is…

This argument shouldn't be hard to finish. (Let me know if you have issues with it.) This way you can use the Jacobian (of a linear function) to turn the integral into an integral over something that is already known.

Answer 3

$$ V = 8 \int_0^a dx \int_0^{b\sqrt{1-\frac{x^2}{a^2}}} c\sqrt{1-\frac{x^2}{a^2} - \frac{y^2}{b^2}} dy \\ = 8 \int_0^a dx \int_0^{b\sqrt{1-\frac{x^2}{a^2}}}c\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2(1-\frac{x^2}{a^2})}}dy $$ let $y = u \sin\theta$, $u = b\sqrt{1-\frac{x^2}{a^2}}$, then $dy = u\cos\theta d\theta$, and we have $$ \int_0^{b\sqrt{1-\frac{x^2}{a^2}}}c\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2(1-\frac{x^2}{a^2})}}dy \\ = \int_0^{\frac{\pi}{2}}{bc(1-\frac{x^2}{a^2})\cos^2\theta}d\theta = bc(1-\frac{x^2}{a^2})\int_0^{\frac{\pi}{2}}(\frac{1+\cos(2\theta)}{2})d\theta = \frac{\pi}{4}bc(1-\frac{x^2}{a^2}) $$ The above is actually $\frac{1}{4}$ area of ellipse for any $x\in [-a,a]$, as discussed here, while $(1-\frac{x^2}{a^2})$ is a scale factor from equation: $$ \frac{y^2}{\displaystyle\left(b \sqrt{1-\frac{x^2}{a^2}} \right)^2} + \frac{z^2}{\displaystyle\left(c \sqrt{1-\frac{x^2}{a^2}} \right)^2} = 1 $$ So that $$ V = 2\pi bc\int_0^a(1-\frac{x^2}{a^2})dx = 2\pi bc(a-\frac{a^3}{3a^2}) = \frac{4\pi}{3}abc $$

BTW, the answer from Jacobi determinant is so intuitive.