Changes of variables in double integrals
Compute the integral of the function $f(x,y)=x^2+y^2$ on the region between the $x$ axes and the $45$ degree line, with radius between $1$ and $2$.
We take the polar coordinates as new variables: $$\begin{array}{l} u(x,y)=r(x,y)=\sqrt{x^2+y^2} \\ v(x,y)=\theta(x,y)=\arctan\Big(\dfrac{y}{x}\Big)\end{array}$$ or similarly: $$\begin{array}{l} x=r\cdot\cos\theta \\ y=r\cdot\sin\theta \end{array}$$ Then $J=\begin{bmatrix} \cos\theta & -r\cdot\sin\theta \\ \sin\theta & r\cdot\cos\theta \end{bmatrix}$ and $|J|=r$.
By inspection of the drawing, we can see that the new integration region is $\begin{array}{c} r\in[1,2] \\ \theta\in[0,\pi/4] \end{array}$ and $\widehat{f}(r,\theta)=r^2$.
Then: $$\int_R (x^2+y^2) \ dxdy = \int_0^{\pi/4}\int_1^2 r^2\cdot r\cdot drd\theta=\int_0^{\pi/4}\int_1^2 r^3drd\theta$$ $$\int_0^{\pi/4} \Big[\dfrac{r^4}{4}\Big]_1^2 \ d\theta=\int_0^{\pi/4} \dfrac{15}{4}d\theta=\dfrac{15}{16}\pi$$
$$\displaystyle \int_R (x^2+y^2) \ dxdy = \dfrac{15}{16}\pi$$