Volume of $n$ dimensional ellipsoid Let $c_1,c_2,...,c_n$ be positive constants. Consider the $n$ dimensional ellipsoid given by $\{(x_1,...,x_n)|\sum_{k=1}^n\frac{x_k^2}{c_k^2}<1\}$. Prove that it's $n$ dimensional volume is $\frac{\pi^{n/2}}{\Gamma(n/2+1)}\prod_{k=1}^nc_k$

I'm know from http://en.wikipedia.org/wiki/Volume_of_an_n-ball that the volume of the $n$ ball is exactly the above formula with $c_1=c_2=...=c_n=r$, which at least confirms the formula in this special case. It seems we can argue by scaling in each coordinates, but how to make this rigorous?
 A: This comes out of substituting $\frac {x_i}{c_i}=y_i$ and using the volume of an $n$ dimensional ball in $n$ space plus the Jacobian transformation of volumes.
A: You know the formula for a ball. Note that the ellipsoid is just the ball scaled by a linear transformation. This linear transformation is diagonal with elements $c_1,c_2,\dots, c_n$. It is a standard fact (proved in Folland's Real Analysis, for example), that if $V$ is a Lebesgue measurable set and $L$ is a linear transformation, then the measure of $L(V)$ is $m(V)\cdot \det(L)$. It is easy to see that the determinant of the linear transformation of interest is $\prod_1^n c_i$, and this gives your desired result. 
A: 1) confirm the result in 2-D case
2) Assume the formula holds for n-D, then calculate the volumne in (n+1)-D using the techniques in the proof part of your wiki link, except for one small trick.
Suppose your ellipsoid in (n+1)-D has the axis $c_1,c_2,...c_{n+1}$.
Then the n-D ellipsoid with $x_{n+1}=a$ is 
$$
\left(x_1,x_2,...x_n|\sum_{i=1}^n \frac{x_i^2}{c_i^2}<1-\frac{a^2}{c_{n+1}^2} \right)
$$
 the same as 
$$
\left(x_1,x_2,...x_n|\sum_{i=1}^n \frac{x_i^2}{d_i^2}<1 \right)
$$
where $d_i=c_i\sqrt{1-a^2/c_{n+1}^2}$
3) So it holds for arbitrary dimension
