
Suppose we have a function $f(x, y)$ and we rewrite it as
$f(u(x, y), v(x, y))$, i.e. perform a change of variables,
then the caculation of area using double integral can be
carried out with these new variables $u, v$. But we have to
figure out the relation between $dudv$ and $dxdy$.
Here is the thing, with the variables $x, y$, $dxdy$ is the area
of the parallelogram
between the vectors $(dx, 0)$ and $(0, dy)$, under the new variables,
these vectors are transformed into two new vectors $(u_x dx, v_x dx)$,
$(u_y dy, v_y dy)$, and $dudv$ is the area of the parallelogram
between these two new vectors.
\[
\begin{array}{cc}
(dx, 0) & (u_x dx, v_x dx) \\
(0, dy) & (u_y dy, v_y dy)
\end{array}
\]
To get the area of this parallelogram, we need the
determinant of this matrix:
\begin{align*}
M &=
\begin{pmatrix}
u_x dx & u_y dy \\
v_x dx & v_y dy
\end{pmatrix} \\
&=
\begin{pmatrix}
u_x & u_y \\
v_x & v_y
\end{pmatrix}
\begin{pmatrix}
dx & 0 \\
0 & dy
\end{pmatrix} \\
&=
J
\begin{pmatrix}
dx & 0 \\
0 & dy
\end{pmatrix},
\end{align*}
where $J$ stands for the Jacobian matrix of $f$. So the
determinant is
\begin{align*}
\det(M) &=
\det(J) dxdy
\end{align*}
1 comments:
Remember in $R^2$, abs of determinant is equal to area of the parallelogram between these two vectors.
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