Volume of Ellipsoid using Triple Integrals 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!
 A: 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.
A: 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
$$dx\,dy\,dz=r^2abc\sin \phi \,d\phi \,d\theta \,dr$$

