Given the general equation of the ellipsoid $\frac + \frac + \frac =1$, I am supposed to use a 3D Jacobian to prove that the volume of the ellipsoid is $\frac\pi abc$ 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 $E = \<(x, y, z): 0\le x \le a, 0\le y \le b\sqrt<1-\frac>, 0\le z \le c\sqrt<1-\frac - \frac> \>$ 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: $$8 \int_0^a \int_0^
HINT
and with the limits
Remember also that in this case
$$dx\,dy\,dz=r^2abc\sin \phi \,d\phi \,d\theta \,dr$$
answered Mar 28, 2018 at 12:10 157k 12 12 gold badges 82 82 silver badges 149 149 bronze badges$\begingroup$ Is it possible for me to know why is $r$ between 0 and 1 instead of being between 0 and $\sqrt$? Edit: Sorry I thought you were converting it into standard spherical coordinates, I got it now, thank you! $\endgroup$
Commented Mar 28, 2018 at 12:15 $\begingroup$ @Derp You are welcome! $\endgroup$ Commented Mar 28, 2018 at 12:20 $\begingroup$ Why is r = 1 as upper bound $\endgroup$ Commented Jun 8, 2019 at 7:02 $\begingroup$ @Gathdi Simply because tha radius is equal to 1. $\endgroup$ Commented Sep 22, 2019 at 13:21$\begingroup$ The addition of r into the definition of x, y, and z made me uneasy as well, so hopefully this explanation helps: The definition of x, y, and z (as given here) essentially take a sphere of radius r and scale it by a, b, and c. Once you understand that we actually started with the definition of a sphere (which includes r), and then scaled to a, b, and c, it becomes more clear why r must scale from 0 to 1. Hope that helps others. $\endgroup$
Commented Jan 13, 2020 at 15:50 $\begingroup$Perhaps this was the intended solution: Consider the linear map $L:\mathbb R^3\to\mathbb R^3$ given by $L(x,y,z)=(ax,by,cz)$. This $L$ maps the ball $B=\<(x,y,z);x^2+y^2+z^2\leq1\>$ to the ellipse $E=\$. In fact, we can think of $L$ as a diffeomorphism $B\to E$. We can now compute the volume of $E$ as the integral $$ \int_E1 = \int_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.
answered Mar 28, 2018 at 12:12 Joonas Ilmavirta Joonas Ilmavirta 26k 10 10 gold badges 61 61 silver badges 111 111 bronze badgesTo subscribe to this RSS feed, copy and paste this URL into your RSS reader.
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