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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!

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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

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

enter image description here

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  • $\begingroup$ Is it possible for me to know why is $r$ between 0 and 1 instead of being between 0 and $\sqrt{a^2 + b^2 + c^2}$? Edit: Sorry I thought you were converting it into standard spherical coordinates, I got it now, thank you! $\endgroup$ – Derp Mar 28 '18 at 12:15
  • $\begingroup$ @Derp You are welcome! $\endgroup$ – user Mar 28 '18 at 12:20
  • $\begingroup$ Why is r = 1 as upper bound $\endgroup$ – Gathdi Jun 8 at 7:02
  • $\begingroup$ @Gathdi Simply because tha radius is equal to 1. $\endgroup$ – user Sep 22 at 13:21
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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.

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