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?

Question

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

Answer 1

The Jacobian is not correct. It should be $J=u$. Then, we have

$$\begin{align} \int_0^4\int_0^y \sqrt{x^2+y^2}\,dx\,dy&=\int_0^1\int_0^4 u^2\sqrt{v^2+1}\,du\,dv\\\\ &=\int_0^4 u^2 \,du\,\int_0^1\sqrt{v^2+1}\,dv\\\\ &=\frac{64}{3}\,\left(\frac12\left(\sqrt 2+\log(1+\sqrt 2)\right)\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{32\left(\sqrt 2+\log(1+\sqrt 2)\right)}{3}} \end{align}$$

Answer 2

By switching to polar coordinates, the wanted integral equals

$$ \int_{0}^{\pi/4} \int_{0}^{\frac{4}{\cos\theta}}\rho^2\,d\rho\,d\theta = \int_{0}^{\pi/4}\frac{1}{3}\left(\frac{4}{\cos\theta}\right)^3\,d\theta $$ and by using Weierstrass substitution $\theta=2\arctan\frac{t}{2}$ that boils down to $$ \frac{32}{3}\left(\sqrt{2}+2\,\text{arctanh}\tan\frac{\pi}{8}\right)=\color{red}{\frac{32}{3}\left[\sqrt{2}+\log(1+\sqrt{2})\right]}$$ as wanted.