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Can someone tell me how to find the volume of an ellipsoid of dimension $\mathbb{R}^3$ by using linear algebra? I know the formula is $\frac{4}{3}\pi abc$. I am given the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$

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  • $\begingroup$ I think that this has very little to do with linear algebra, anyway. $\endgroup$ – Giuseppe Negro Mar 13 '14 at 22:08
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    $\begingroup$ @GiuseppeNegro Linear transformation, Jacobian determinant... not unreasonable. $\endgroup$ – user127096 Mar 13 '14 at 22:09
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The ellipsoid satisfying

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$

can be seen as the image of the unit sphere under the linear transformation

$$(x,y,z) \mapsto \left(ax, by, cz\right)$$

In turn, we can view this transformation as the action of the matrix

$$A = \left[\begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$$

Note that $\det(A) =abc$; then it follows from the properties of the determinant that

$$\operatorname{Volume}_{\text{ ellipsoid}} = \operatorname{Volume}_{\text{ sphere}} \cdot \det(A) = \left(\frac 4 3 \pi \cdot 1^3\right)(abc)$$

as desired.

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