• RECALL: $dA=dx\,dy=r\,dr\,d\phi$

Where did the factor of $r$ come from in the above expression? Consider a “coordinate rectangle” bounded by curves of the form $r=\hbox{constant}$, $\phi=\hbox{constant}$. If this “rectangle” is small enough, its sides have length $dr$ and $r\,d\phi$, so that its area is the product $(dr)(r\,d\phi)$. 1)

For orthogonal coordinates in the plane, that is, one whose level curves (like $r=\hbox{constant}$) are perpendicular, this method can be used to quickly determine the correct form of $dA$. This method also works fine for the volume element $dV$ in three dimensions, yielding \begin{equation} dV = (dx)(dy)(dz) = (dr)(r\,d\phi)(dz) = (dr)(r\,d\theta)(r\sin\theta\,d\phi) \end{equation} in Cartesian, cylindrical, and spherical coordinates respectively.

But what do we do if the coordinates are not orthogonal? We start by reviewing the 1-dimensional case, which is just the method of substitution.

• RECALL: $x=x(u) \qquad\Longrightarrow\qquad \displaystyle\Int f\,dx = \Int f \> {dx\over du} \, du$

When making such a substitution, of course, one must replace $x$ everywhere by $u$, so that both $f$ and the integration limits must be rewritten in terms of $u$.

Consider now a change of variables in two dimensions \begin{equation} x=x(u,v) \qquad y=y(u,v) \end{equation} We can therefore write the position vector as \begin{equation} \rr = \rr(u,v) = x(u,v)\,\xhat + y(u,v)\,\yhat \end{equation} But this is just a special case of a parametric surface!

We already know how to find the surface element of such surfaces. First compute \begin{equation} d\SS = d\rr_1 \times d\rr_2 = \left( \Partial{\rr}{u} \times \Partial{\rr}{v} \right) \, du \, dv \end{equation} and then take the magnitude of $d\SS$ to obtain $\dS$. We have \begin{equation} \Partial{\rr}{u} \times \Partial{\rr}{v} = \left| \matrix{ \xhat& \yhat&  \zhat \cr \noalign{\smallskip} \Partial{x}{u}& \Partial{y}{u}& 0\cr \noalign{\smallskip} \Partial{x}{v}& \Partial{y}{v}& 0\cr } \right| = \zhat \left| \matrix{ \Partial{x}{u}& \Partial{y}{u}\cr \noalign{\smallskip} \Partial{x}{v}& \Partial{y}{v}\cr } \right| \end{equation}

This determinant is important enough to have its own name; we define the Jacobian of the transformation from $u,v$ to $x,y$ to be 2) \begin{equation} \Jacobian{x}{y}{u}{v} = \left| \matrix{\Partial{x}{u}& \Partial{y}{u}\cr \noalign{\smallskip} \Partial{x}{v}& \Partial{y}{v}\cr} \right| = \left| \matrix{\Partial{x}{u}& \Partial{x}{v}\cr \noalign{\smallskip} \Partial{y}{u}& \Partial{y}{v}\cr} \right| \end{equation}

Putting this all together, we obtain \begin{equation} dA = \left| \Jacobian{x}{y}{u}{v} \right| \, du \, dv \end{equation} Note that there are two sets of vertical bars here — this expression contains the absolute value of the Jacobian, since areas are positive.

Not surprisingly, when this procedure is applied to the polar coordinate transformation \begin{equation} x=r\cos\phi \qquad y=r\sin\phi \end{equation} we get (check this yourself!) \begin{equation} \Jacobian{x}{y}{r}{\phi} = r \end{equation} so that, as we already knew, $dA=r\,dr\,d\phi$.

This not only tells us how to find area by doing an integral with respect to $u$ and $v$ (simply integrate the above expression), it tells us how to transform any integral in $x$ and $y$ to one involving $u$ and $v$ — hence the name “change of variables”. We therefore have \begin{equation} \DInt{R} f \, dx \, dy = \DInt{D} f \left| \Jacobian{x}{y}{u}{v} \right| \, du \, dv \end{equation} where $R$ is some region in the $xy$-plane, and $D$ is that same region expressed in terms of $u$ and $v$. We must also reinterpret $f$ on the right as a function of $u$ and $v$; this can be written formally as $f\Big(x(u,v),y(u,v)\Big)$, but this simply means to use what you know to rewrite all occurrences of $x$ and $y$ in terms of $u$ and $v$.

To change variables in a double integral,

1. Rewrite $dA$ in terms of the new variables using the above formula.
2. Rewrite the function being integrated in terms of the new variables.
3. Change the limits from the old variables ($R$) to the new ones ($D$).

A similar formula holds for triple integrals. Given $x$, $y$, $z$ as functions of $u$, $v$, $w$, the corresponding infinitesimal box is a parallelepiped with sides \begin{equation} d\rr_1 = \Partial{\rr}{u}\,du \qquad d\rr_2 = \Partial{\rr}{v}\,dv \qquad d\rr_3 = \Partial{\rr}{w}\,dw \end{equation} The volume of such a parallelepiped is given by the triple product \begin{equation} dV = \left| \left( d\rr_1 \times d\rr_2 \right) \cdot d\rr_3 \right| \, du \, dv \, dw \end{equation} Defining the Jacobian for transformations of three variables to be \begin{equation} \JACOBIAN{x}{y}{z}{u}{v}{w} = \left| \matrix{ \Partial{x}{u}& \Partial{y}{u}& \Partial{z}{u}\cr \noalign{\smallskip} \Partial{x}{v}& \Partial{y}{v}& \Partial{z}{v}\cr \noalign{\smallskip} \Partial{x}{w}& \Partial{y}{w}& \Partial{z}{w}\cr } \right| = \left| \matrix{ \Partial{x}{u}& \Partial{x}{v}& \Partial{x}{w}\cr \noalign{\smallskip} \Partial{y}{u}& \Partial{y}{v}& \Partial{y}{w}\cr \noalign{\smallskip} \Partial{z}{u}& \Partial{z}{v}& \Partial{z}{w}\cr } \right| \end{equation} we obtain \begin{equation} dV = \left| \JACOBIAN{x}{y}{z}{u}{v}{w} \right| \,du\,dv\,dw \end{equation}

A useful example, found in most textbooks, is to use the Jacobian to derive the volume element $dV$ in spherical coordinates, which you should already know. (Many texts use a left-handed convention for spherical coordinates, which results in the determinant being negative. The absolute value signs occur in the change of variables formula precisely to prevent negative area and volume elements when transforming between right-handed and left-handed coordinate systems.)

Knowing $dV$ we can convert any integral using \begin{equation} \TInt{V} f \, dx \, dy \, dz = \TInt{W} f \> \left| \JACOBIAN{x}{y}{z}{u}{v}{w} \right| \, du \, dv \, dw \end{equation} for appropriate regions $V$ and $W$ (and with $f$ rewritten along the lines above). The above method for double integrals adapts easily to triple integrals.

In practice, one is often given $u$ and $v$ in terms of $x$ and $y$, not the other way around. Rather than solving for $x$ and $y$, which is often difficult, there is a simpler way. Inverting the roles of the two sets of variables in the derivation above, we must have not only \begin{equation} dx\,dy = \left| \Jacobian{x}{y}{u}{v} \right| du\,dv \end{equation} but also \begin{equation} du\,dv = \left| \Jacobian{u}{v}{x}{y} \right| dx\,dy \end{equation} from which we can conclude that \begin{equation} \Jacobian{x}{y}{u}{v} = {1\Bigg/\Jacobian{u}{v}{x}{y}} \end{equation} But the expression on the right can be computed without solving for $x$ and $y$! This gives another way to compute the (left-hand) Jacobian in this case. A similar formula holds in three dimensions. If you use this method, don't forget to take the reciprocal!