Find $$\iint_D(x+y)\,dx\,dy$$
While $D$ is the domain bounded by the lines:
$$x+y=1,\, x+y=3,\, y=5x,\, y=10x$$
My Work:
First off, I thought of changing the variables, where $u=x+y$, and $v=\frac{y}{x}$ ($x$ is outside my domain so I didn't care about $x=0$, would love to hear feedback about this step).
So now my bounds are $u=1, u=3, v=5, v=10$, and I get a rectangular domain.
To calculate the Jacobian that I need to multiply my integral with, I'll calculate first the inverse,
$$J^{-1}
=
\left[ {\begin{array}{cc}
u_x & v_x \\
u_y & v_y \\
\end{array} } \right] = \left[ {\begin{array}{cc}
1 & \frac{-y}{x^2} \\
1 & \frac{1}{x} \\
\end{array} } \right] = \frac{1}{x} +\frac{y}{x^2}=\frac{x+y}{x^2} \Longrightarrow J = \frac{x^2}{x+y}=\frac{x^2}{u}$$
So I got: $$\iint_{D^{*}}u\frac{x^2}{u}=\iint_{D^*}(\frac{u}{v+1})^2=\int^3_1du\int^{10}_5(\frac{u}{1+v})^2dv=\int^3_1du\int^{10}_5u^2*(1+v)^{-2}dv = \int^3_1-\frac{u^2}{1+v} |^{10}_5du= \int^3_1\frac{5u^2}{66}du=\frac{5*3^3}{198} - \frac{5}{198}=\frac{65}{99}$$
EDITS:
Fixed the integral after the help comment from MathLover that $x=\frac{u}{1+v}$
Updated the solution, had mistake in calculations