Suppose you had the surface integral $\iint \limits_{A} = x^{3}(1-x^{4}-y^{4})dx \ dy$ where $A$ is the region defined by $x \geq 0, \; y \geq 0, \; x^{4}+y^{4} \leq 1$.
How would you solve this using the substitution $x=\sqrt{R \ cos \theta}, \; y=\sqrt{R \ sin \theta}$?
Obviously this substitution gives us $x^{3}(1-x^{4}-y^{4})=(R \ cos \theta)^{\frac{3}{2}}(1-R^{2})$, but what would this transformation make $dx \ dy$ become?
Also, is there any difference between a double integral and a surface integral, or are they the same thing?
Not much of an answer (too long for comment), but to expand on my previous comment: $$\mathrm{d}x\,\mathrm{d}y=|J|\,\mathrm{d}r\,\mathrm{d}\theta=\begin{vmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \theta}\\[1ex]\dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial \theta}\end{vmatrix}\,\mathrm{d}r\,\mathrm{d}\theta=\frac{1}{2\sqrt{2\sin2t}}\,\mathrm{d}r\,\mathrm{d}\theta$$