Let $D = \mathbf{\Phi}\left(\mathcal R\right)$ where $\mathbf{\Phi}\left(u,v\right)=\left(u^2,u+v\right)$ and $\mathcal R=[4,8]\times [0,4]$.
Calculate $$\iint\limits_D y \ dA$$
Note: It is not necessary to describe $D$.
I think you set the $x$ and $y$ bounds of the integral to $[0,4]$ and $[4,8]$ respectively and then set $x=u^2$ and $y=u+v$, but I'm not exactly sure. A walkthrough of the problem will be helpful.
Your approach is incorrect, because the region in coordinates $(x,y)$ is $\mathcal{D}$, not $\mathcal{R}$. In other words, $\mathcal{D}$ is the region you obtain by transforming $\mathcal{R}$ via $\Phi$, which you are implicitly told that it refers to coordinates $(u,v)$. The idea is there, though, and I'll try my best to explain it without solving the whole thing.
I realized that my previous answer was totally wrong, as it basically assumed that the change in $(x,y)$ is linear as in $(u,v)$ which is in general NOT true, and not true in this case. I am sorry to have posted a low quality answer and now answering properly.
Revisiting the question, the underlying trick is that you can transform the domain in coordinates $(x,y)$ to coordinates $(u,v)$ by knowing the effect of the transformation $\Phi$. Here I show how to do that.
Basically, the transformation $\mathcal{D}=\Phi(\mathcal{R})$ implies that differential areas in $\mathcal{D}$ relate to differential areas of $\mathcal{R}$ through the rate of change in $\Phi$, i.e. its Jacobian. The Jacobian of $\Phi$ corresponds to $$\begin{align} J&=\left|\det\left(\begin{bmatrix}\frac{dx}{du}&\frac{dx}{dv}\\\frac{dy}{du}&\frac{dy}{dv}\end{bmatrix}\right)\right|\\ &=\left|\det\left(\begin{bmatrix}2u&0\\1&1\end{bmatrix}\right)\right|\\ &=|2u|=2u \end{align}$$ where the last equality comes from the fact that $u$ is already positive. Then, $$\int_{\mathcal{D}}y\,\mathrm{d}A = \int_{\mathcal{R}}J\cdot y(u,v)\,\mathrm{d}A_{\mathcal{R}}=\int_{\mathcal{R}}2u(u+v)\,\mathrm{d}u\mathrm{d}v$$ where you know the integration limits.