I want to find the area enclosed in the ellipse given by $(2x+5y-3)^2 + (3x - 7y + 8)^2 = 1$.
I elected to change variables; that is, compute the area within $u^2 + v^2 = 1$ with $u = 2x+5y-3$ and $v=3x - 7y + 8$. Hence we seek
$$\iint_{u^2 + v^2 < 1} \mathrm dx \, \mathrm dy$$
We can say $\mathrm du \, \mathrm dv = |J| \mathrm dx \, \mathrm dy$, where $J = \det \big(\frac{\partial(u,v)}{\partial(x,y)}\big)$. As $\frac{\partial(u,v)}{\partial(x,y)} = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$, $|J|=29$, so $$\mathrm du \, \mathrm dv = 29 \, \mathrm dx \, \mathrm dy \rightarrow \mathrm dx \, \mathrm dy = \frac{1}{29} \, \mathrm du \, \mathrm dv$$
Which means $\iint_{u^2 + v^2 < 1} \mathrm dx \, \mathrm dy = \iint_{u^2+v^2<1} \frac{1}{29} \, \mathrm du \, \mathrm dv$.
We change variables again, into polar coordinates, giving $$\int_0^{2\pi}\int_0^1\frac{1}{29} r \, \mathrm dr \, \mathrm d\theta = \boxed{\frac{\pi}{29}}$$
Is this reasoning correct? I'm mainly concerned about the validity of going from $(u,v)$ to $(r,\theta)$ the same way I'd do it if it were a $(x,y)\rightarrow(r,\theta)$ transformation.