Here is an integral:
\begin{align*}
Q &= \int_0^1 \int_0^1 x^2 y \; dx dy \\
&= 1/6
\end{align*}
Easy enough, but what if we change the variables like this:
\begin{align*}
u &= x \\
v &= xy
\end{align*}
then
\begin{align*}
J &=
\begin{pmatrix}
1 & 0 \\
y & x
\end{pmatrix} \\
dudv &= |\det(J)| \; dxdy \\
&= x\; dxdy \\
dxdy &= (1/x) dudv
\end{align*}
Substitute everything into $Q$:
\begin{align*}
Q &= \iint_? uv (1/x) \; dudv \\
&= \iint_? v \; dudv
\end{align*}
The only thing we need now is the boundary. We acquire this by
mapping everying point $(x, y)$ to $(u, v)$ and transform the
1x1 square area into a triangle (see the figure).
And from there we have
\begin{align*}
Q &= \int_0^1 \int_0^u v \; dvdu \\
&= 1/6
\end{align*}

## Saturday, April 5, 2014

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