Evaluate $\int \int \sqrt{(x^2 + y^2)}$ $dx dy$ over a triangle with corners (0, 0), (4, 4), and (4, 0) using $x = u$, $y = uv$. Help? 
So far I have:
$$a = x = 4\\ b = y = 0\\ c = y = x$$
$$y=x \rightarrow uv = u\rightarrow v=u/u \rightarrow v=1$$
$$x = 4 \rightarrow u = 4$$
$$y = 0\rightarrow uv = 0 \rightarrow u = 0 \rightarrow v = 0$$
$$\sqrt {(x^2 + y^2)} = (u^2 +uv^2)^{(1/2)}$$
\begin{align}  \|J\| & =det\left[ {\begin{array}{cc}
   \frac {∂x}{∂u}   & \frac{∂x}{∂v} \\
   \frac{∂y}{∂u}   & \frac{∂y}{∂v}\\
  \end{array} } \right] = \left[ {\begin{array}{cc}
   1   & v \\
   0   & u\\
  \end{array} } \right] = u = 4
\end{align}
\begin{align*}
\int_{0}^{4}\int_{0}^{1} \sqrt{(u^2 + uv^2)} + 4 ~dvdu = \int_{0}^{4}\int_{0}^{1} (u^2 + uv^2)^{1/2} + 4 ~dvdu
\end{align*}
I can already see this is incorrect due to the given answer.
This gives me J = 0, however J is supposed to = 1? and the answer to the problem is supposed to be:
$$\frac{32(\sqrt2+ln(1+\sqrt2))}3$$
according to the website: https://www.whitman.edu/mathematics/calculus_online/section15.07.html

The Jacobian is not correct.  It should be $J=u$.  Then, we have
$$\begin{align}
\int_0^4\int_0^y \sqrt{x^2+y^2}\,dx\,dy&=\int_0^1\int_0^4 u^2\sqrt{v^2+1}\,du\,dv\\\\
&=\int_0^4 u^2 \,du\,\int_0^1\sqrt{v^2+1}\,dv\\\\
&=\frac{64}{3}\,\left(\frac12\left(\sqrt 2+\log(1+\sqrt 2)\right)\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{32\left(\sqrt 2+\log(1+\sqrt 2)\right)}{3}}
\end{align}$$
