Lets say we have the following coordinates which describe an ellipse in 3 dimensions:
$x = r\sin(a)\cos(b)$
$y = \frac{1}{\sqrt2}r\sin(a)\sin(b)$
$z = r\cos(a)$
Where:
$4\sqrt2\ge r \ge 0$
$\pi \ge a \ge 0$
$2\pi \ge b \ge 0$
These coordinates will be used to calculate the volume of the ellipse, but that is not my question. My question is how to calculate the functional determinant of this in an easy way? The long way would be to calculate it by taking the determinant of a 3x3 matrix which would take extremely long calculations.
I'll give you the start of the calculation
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl}J = \displaystyle \left|\frac{\partial \left(x , y , z\right)}{\partial \left(r , a , b\right)}\right|&=&\displaystyle \left|\begin{array}{ccc}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial a}&\frac{\partial x}{\partial b}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial a}&\frac{\partial y}{\partial b}\\ \frac{\partial z}{\partial r}&\frac{\partial z}{\partial a}&\frac{\partial z}{\partial b} \end{array}\right|\\ &=&\displaystyle \left|\begin{array}{ccc}\sin \left(a\right) \cos \left(b\right)&r \cos \left(a\right) \cos \left(b\right)&{-r} \sin \left(a\right) \sin \left(b\right)\\ \frac{1}{\sqrt{2}} \sin \left(a\right) \sin \left(b\right)&\frac{1}{\sqrt{2}} r \cos \left(a\right) \sin \left(b\right)&\frac{1}{\sqrt{2}} r \sin \left(a\right) \cos \left(b\right)\\ \cos \left(a\right)&{-r} \sin \left(a\right)&0 \end{array}\right| \end{array}$$
In the second row of this determinant, you can factor $\frac{1}{\sqrt{2}}$. In the second column, you can factor $r$, and in the last column, you can factor $r \sin \left(a\right)$. It gives
$$J = \frac{1}{\sqrt{2}} {r}^{2} \sin \left(a\right) \left|\begin{array}{ccc}\sin \left(a\right) \cos \left(b\right)&\cos \left(a\right) \cos \left(b\right)&{-\sin } \left(b\right)\\ \sin \left(a\right) \sin \left(b\right)&\cos \left(a\right) \sin \left(b\right)&\cos \left(b\right)\\ \cos \left(a\right)&{-\sin } \left(a\right)&0 \end{array}\right|$$
It should now be easy to finish this calculations.
Edit: the rules I used are the following
$$\left|\begin{array}{ccc}X a&X b&X c\\ d&e&f\\ g&h&i \end{array}\right| = X \left|\begin{array}{ccc}a&b&c\\ d&e&f\\ g&h&i \end{array}\right|$$
and
$$\left|\begin{array}{ccc}X a&b&c\\ X d&e&f\\ X g&h&i \end{array}\right| = X \left|\begin{array}{ccc}a&b&c\\ d&e&f\\ g&h&i \end{array}\right|$$
The same rule works with any row or any column instead of the first row or the first column. It also works for smaller or larger determinants.