Change of Variables

For more general transformations $x = x(u,v)$ and $y = y(u,v)$ we can use the jacobian function in the linearalgebra package, which is loaded by default. Let $f(x,y)=x+y$, and we will calculate


\begin{displaymath}
\int\int_R  \left( x+y\right) \mathrm{d}A.
\end{displaymath}


\begin{maximasession}
f(x,y) := x + y;
\maximaoutput*
\i1. f(x,y) := x + y; \\
\o1. f\left(x , y\right)\mathbin{:=}x+y \\
\end{maximasession}

Let's make the transformation $x=u^3-v^4$ and $y = 5uv$.


\begin{maximasession}[x,y]: [u^3 - v^4, 5 * u * v];
\maximaoutput*
\i2. [x,y]: [...
...4, 5 * u * v]; \\
\o2. \left[ u^3-v^4 , 5 u v \right] \\
\end{maximasession}

We need the Jacobian:


\begin{maximasession}
J: jacobian([x,y], [u,v]);
J: determinant(J);
\maximaoutpu...
...r } \\
\i4. J: determinant(J); \\
\o4. 20 v^4+15 u^3 \\
\end{maximasession}

We were lucky in this example because the Jacobian is positive as long as $u$ is positive (or just not terribly negative). If this had not happened then we would need to be careful about the sign of $J$, which in principle could be a problem because Maxima does not integrate absolute value. Nevertheless, we finish up with the integration. It takes the form (we will make up random limits of integration, but in a given problem we would need to determine these)


\begin{displaymath}
\int_3^4\int_1^2 f\left( x(u,v),y(u,v) \right)   \left\ver...
...al(x,y)}{\partial(u,v)} \right\vert \mathrm{d}u \mathrm{d}v.
\end{displaymath}


\begin{maximasession}
integrate(integrate(f(x,y) * J, u,1,2), v,3,4);
\maximaout...
...) * J, u,1,2), v,3,4); \\
\o5. -{{113349305}\over{252}} \\
\end{maximasession}



G. Jay Kerns 2009-12-01