I want to calculate the integral
$$
\iint_D(x-y)dxdy
$$
where D is the triangle made up of the vertices (0,0), (-2,1) and (-1,3). (Graph)
My idea was to do this substitution
$$
\begin{equation}
\begin{cases}
u = y + \frac x2 \qquad dxdy=\frac6{25}dudv \\
v = y – 2x
\end{cases}
\end{equation}
$$
and then end up with a triangle in the uv-plane with vertices in (0,0), (0,5) and ($\frac52$,5) (Graph). However, when trying to calculate this I ended up horribly wrong.
When I check the example solution they use a different substitution.
$$
\begin{equation}
\begin{cases}
u = x + 2y \qquad dxdy = \frac15 dudv \\
v = 3x + y
\end{cases}
\end{equation}
$$
My questions are
- Is my substitution correct? If not, what did I do wrong?
- I get the feeling the provided example solution is easier to calculate. Is there some hint that I should be getting to use this substitution instead?
I realize there are several ways to solve these kind of problems (right?), so just wanted to know if "my way" will work (if actually calculate the integral correctly…).
Best Answer
Adjusting the substitution: $$x=\frac{2}{5} (u-v)$$ $$y=\frac{4u+v}{5}$$
$$\frac{\partial x}{\partial u}=\frac{2}{5} ,\frac{\partial x}{\partial v}=-\frac{2}{5}, \frac{\partial y}{\partial u}=\frac{4}{5},\frac{\partial y}{\partial v}=\frac{1}{5}$$
So the Jacobian is $$\frac{2}{5}\times\frac{1}{5} - (\frac{-2}{5})\times\frac{4}{5}=\frac{10}{25}$$
and not $6/25$.