<s>[INST] Calculate the integral using Jacobian The problem asks you to  calculate the following integral using Jacobian
$$\int{4x^2+y^2}dA$$ and it tells you to substitute $y+2x=u,y-2x=v,0<u<2,-2<v<0$, and the Jacobian is $\cfrac{1}{4}$.
Could anyone here help me out? Thank you.
 [/INST] First, we determine $u$ and $v$ in term of $x$ and $y$. Using elimination/ substitution, it's easy to find that $x=\frac{1}{4}(u-v)$ and $y=\frac{1}{2}(u+v)$. Just adding and subtract $u$ and $v$. Then
\begin{align}
\int(4x^2+y^2)\,dA&=\int\int(4x^2+y^2)\,dx\,dy\\
&=\int_{-2}^0\int_0^2\left(4\left(\frac{1}{4}(u-v)\right)^2+\left(\frac{1}{2}(u+v)\right)^2\right)|J|\,du\,dv\\
&=\int_{-2}^0\int_0^2\frac{1}{4}\left((u-v)^2+(u+v)^2\right)\frac{1}{4}\,du\,dv\\
&=\frac{1}{16}\int_{-2}^0\int_0^2\left(2u^2+2v^2\right)\,du\,dv\\
&=\frac{1}{16}\int_{-2}^0\int_0^22\left(u^2+v^2\right)\,du\,dv\\
&=\frac{1}{8}\int_{-2}^0\left[\frac{1}{3}u^3+v^2u\right]_0^2\,dv\\
&=\frac{1}{8}\int_{-2}^0\left(\frac{8}{3}+2v^2\right)\,dv\\
&=\frac{1}{8}\left[\frac{8}{3}v+\frac{2}{3}v^3\right]_{-2}^0\\
&=\frac{1}{8}\left(\frac{16}{3}+\frac{16}{3}\right)\\
&=\frac{4}{3}
\end{align}
</s>