4
$\begingroup$

Find the volume of the ellipsoid $$11x^2+9y^2+15z^2-4xy+10yz-20xz=80$$

Tried to make the expression simpler as $$\frac{X^2}{A^2}+\frac{Y^2}{B^2}+\frac{Z^2}{C^2}=1$$ where $X, Y, Z$ are liner expressions of $x, y, z$. Then I would use the volume of a ball $$\frac43\pi R^3$$ and write the volume of ellipsoid as $$\frac43\pi ABC$$ But I'm stuck at the first step… any help?

$\endgroup$
  • 1
    $\begingroup$ Do you now anything about eigenvectors, or how to diagonalize a matrix? Without these concepts, you'll just have to use brute force algebra, which will be a tedious process. $\endgroup$ – bubba Oct 16 '16 at 3:59
  • $\begingroup$ @bubba Yes, I have knowledge about linear algebra, could you tell me more or give a detailed solution about it? $\endgroup$ – Nan Oct 16 '16 at 4:05
  • $\begingroup$ What have you tried? Where are you stuck on? Don't expect us to do everything from start to finish. $\endgroup$ – GohP.iHan Oct 16 '16 at 4:54
2
$\begingroup$

Because the ellipsoid has an equation involving only quadratic terms, it is centred at the origin and we can write the equation in matrix form as $$\begin{bmatrix}x\\y\\z\end{bmatrix}^T\begin{bmatrix}11&-2&-10\\-2&9&5\\-10&5&15\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=80$$ The right-hand side is 80, so the eigenvalues $k_i$ of the coefficient matrix $A$ in the middle will be scaled by the same factor; the true eigenvalues $\lambda_i$ will be recovered when the $k_i$ are divided by 80. Now to find the eigenvalues of $A$: $$\det(A-kI)=0$$ $$\begin{vmatrix}11-k&-2&-10\\-2&9-k&5\\-10&5&15-k\end{vmatrix}=0$$ $$(11-k)(9-k)(15-k)+100+100-100(9-k)-25(11-k)-4(15-k)=0$$ $$1485-399k+35k^2-k^3+200-900+100k-275+25k-60+4k=0$$ $$-k^3+35k^2-270k+450=0$$ We do not need to know the roots $k_1,k_2,k_3$ of this polynomial, only the product of the roots. By Viète's formulas we have $$k_1k_2k_3=-\frac{450}{-1}=450$$ Dividing by $80^3$ we recover the product of the true eigenvalues: $$\lambda_1\lambda_2\lambda_3=\frac{450}{80^3}=\frac9{10240}$$ Each eigenvalue is the square of the reciprocal of the length of one of the ellipsoid's semi-axes (see for example here, page 18). Therefore the product of the ellipsoid's semi-axis lengths is $$abc=\lambda_1^{-\frac12}\lambda_2^{-\frac12}\lambda_3^{-\frac12}=(\lambda_1\lambda_2\lambda_3)^{-\frac12}=\sqrt{\frac{10240}9}=\frac{32\sqrt{10}}3$$ Finally, the ellipsoid's volume works out as $$V=\frac43\pi abc=\frac43\pi\left(\frac{32\sqrt{10}}3\right)=\frac{128\sqrt{10}\pi}9=141.2919\dots$$

$\endgroup$

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.