1 Changing variables in multiple integrals

When the method of substitution is used to solve an integral of the form    ${\int \; <!--nolimits\; -->}_a^{b}f\left(x\right)dx$   three parts of the integral are changed, the limits, the function and the infinitesimal $dx$ . So if the substitution is of the form $x=x\left(u\right)$ the $u$ limits, $c$ and $d$ , are found by solving $a=x\left(c\right)$ and $b=x\left(d\right)$ and the function is expressed in terms of $u$ as $f\left(x\left(u\right)\right)$ .

Figure 28

Figure 28 shows why the $dx$ needs to be changed. While the $\delta u$ is the same length for all $u$ , the $\delta x$ change as $u$ changes. The rate at which they change is precisely $\frac{d}{du}x\left(u\right)$ . This gives the relation

$\delta x=\frac{dx}{du}\delta u$

Hence the transformed integral can be written as

${\int \; <!--nolimits\; -->}_a^{b}f\left(x\right)dx={\int \; <!--nolimits\; -->}_c^{d}f\left(x\left(u\right)\right)\frac{dx}{du}du$

Here the $\frac{dx}{du}$ is playing the part of the Jacobian that we will define.

Another change of coordinates that you have seen is the transformations from cartesian coordinates $\left(x,y\right)$ to polar coordinates $\left(r,\theta \right)$ .

Recall that a double integral in polar coordinates is expressed as

$\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->f\left(x,y\right)dxdy=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->g\left(r,\theta \right)rdrd\theta$

Figure 29

We can see from Figure 29 that the area elements change in size as $r$ increases. The circumference of a circle of radius $r$ is $2\pi r$ , so the length of an arc spanned by an angle $\theta$ is $2\pi r\frac{\theta }{2\pi }=r\theta$ . Hence the area elements in polar coordinates are approximated by rectangles of width $\delta r$ and length $r\delta \theta$ . Thus under the transformation from cartesian to polar coordinates we have the relation

$\delta x\delta y\to r\delta r\delta \theta$

that is, $r\delta r\delta \theta$ plays the same role as $\delta x\delta y$ . This is why the $r$ term appears in the integrand. Here $r$ is playing the part of the Jacobian.