The equation of ellipsoid is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ The hint for substitution is given as $x=au$, $y=bv$ and $z=cv$. I know that the projection of the ellipsoid on $xy-$plane is an ellipse with the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$. But I don't know what will be the limits of $z$ in integration. After transformation the Jacobian of transformation will be $J=abc$. I don't know how to proceed this one and compute the volume of ellipsoid. Please explain in detail about the limits before and after substitution with the help of the mathematica command that helps in plotting the ellipsoid along with its shadow. THanks in advance!

  • 1
    $\begingroup$ If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$. $\endgroup$ – A.Γ. Jan 8 at 17:08
  • $\begingroup$ After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$ $\endgroup$ – saulspatz Jan 8 at 17:08
  • $\begingroup$ @A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir! $\endgroup$ – Noor Aslam Jan 8 at 17:11
  • $\begingroup$ The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult? $\endgroup$ – David K Jan 8 at 17:38

I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.

You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{$\color{blue}{*}$}$$ So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.

If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!

You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(\color{blue}{*})$ above to get: $$\int_{-a}^{a} \int_{-b\sqrt{1-\tfrac{x^2}{a^2}}}^{b\sqrt{1-\tfrac{x^2}{a^2}}} \ldots \,\mbox{d}y \,\mbox{d}x$$ Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives: $$\int_{-a}^{a} \int_{-b\sqrt{1-\tfrac{x^2}{a^2}}}^{b\sqrt{1-\tfrac{x^2}{a^2}}} \int_{-c\sqrt{1-\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}}}^{c\sqrt{1-\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}}} 1 \,\mbox{d}z \,\mbox{d}y \,\mbox{d}x$$ The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to: $$8\int_{0}^{a} \int_{0}^{b\sqrt{1-\tfrac{x^2}{a^2}}} c\sqrt{1-\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}} \,\mbox{d}z \,\mbox{d}y \,\mbox{d}x$$ And you could take it from there.

Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:

The hint for substitution is given as $x=au$, $y=bv$ and $z=c\color{red}{w}$.

the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$: $$\iiint_E 1 \,\mbox{d}z \,\mbox{d}y \,\mbox{d}x = abc \iiint_S 1 \,\mbox{d}w \,\mbox{d}v \,\mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.

  • $\begingroup$ I know the volume of the unit sphere is $\frac{4\pi}{3}$, can I compute this $\frac{4\pi}{3}$, from the integral putting limits on it $\int\int\int_{S}dwdvdu$. I shall be very thankfull! Sir $\endgroup$ – Noor Aslam Jan 8 at 17:57
  • $\begingroup$ You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations. $\endgroup$ – StackTD Jan 8 at 18:03
  • $\begingroup$ Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8\int_{0}^{1}\int_{0}^{\sqrt{1-u^2}}\int_{0}^{\sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8\int_{0}^{1}\int_{0}^{\sqrt{1-u^2}}\int_{0}^{\sqrt{1-u^2-v^2}}dwdvdu=\frac{32\pi}{3}$ $\endgroup$ – Noor Aslam Jan 8 at 18:11
  • $\begingroup$ It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches. $\endgroup$ – StackTD Jan 8 at 18:15
  • $\begingroup$ sir I have multiplied 8 with $\frac{4\pi}{3}$ $\endgroup$ – Noor Aslam Jan 8 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.