Let's examine the single variable case again, from a slightly different perspective than we have previously used. Suppose we start with the problem $$\int_0^1 x^2\sqrt{1-x^2}\,dx;$$ this computes the area in the left graph of figure 17.7.1. We use the substitution $x=\sin u$ to transform the function from $x^2\sqrt{1-x^2}$ to $\sin^2u\sqrt{1-\sin^2u}$, and we also convert $dx$ to $\cos u\,du$. Finally, we convert the limits 0 and 1 to 0 and $\pi/2$. This transforms the integral: $$\int_0^1 x^2\sqrt{1-x^2}\,dx = \int_0^{\pi/2}\sin^2u\sqrt{1-\sin^2u} \cos u\,du.$$ We want to notice that there are three different conversions: the main function, the differential $dx$, and the interval of integration. The function is converted to $\sin^2u\sqrt{1-\sin^2u}$, shown in the right-hand graph of figure 17.7.1. It is evident that the two curves pictured there have the same $y$-values in the same order, but the horizontal scale has been changed. Even though the heights are the same, the two integrals $$\int_0^1 x^2\sqrt{1-x^2}\,dx\qquad\hbox{and}\qquad \int_0^{\pi/2}\sin^2u\sqrt{1-\sin^2u}\,du $$ are not the same; clearly the right hand area is larger. One way to understand the problem is to note that if both areas are approximated using, say, ten subintervals, that the approximating rectangles on the right are wider than their counterparts on the left, as indicated. In the picture, the width of the rectangle on the left is $\Delta x=0.1$, between $0.7$ and $0.8$. The rectangle on the right is situated between the corresponding values $\arcsin(0.7)$ and $\arcsin(0.8)$ so that $\Delta u=\arcsin(0.8)-\arcsin(0.7)$. To make the widths match, and the areas therefore the same, we can multiply $\Delta u$ by a correction factor; in this case the correction factor is approximately $\cos u=\cos(\arcsin(0.7))$, which we compute when we convert $dx$ to $\cos u\,du$.

Now let's move to functions of two variables. Suppose we want to convert an integral $$\int_{x_0}^{x_1}\int_{y_0}^{y_1} f(x,y)\,dy\,dx$$ to use new variables $u$ and $v$. In the single variable case, there's typically just one reason to want to change the variable: to make the function "nicer'' so that we can find an antiderivative. In the two variable case, there is a second potential reason: the two-dimensional region over which we need to integrate is somehow unpleasant, and we want the region in terms of $u$ and $v$ to be nicer—to be a rectangle, for example. Ideally, of course, the new function and the new region will be no worse than the originals, and at least one of them will be better; this doesn't always pan out.

As before, there are three parts to the conversion: the function itself must be rewritten in terms of $u$ and $v$, $dy\,dx$ must be converted to $du\,dv$, and the old region must be converted to the new region. We will develop the necessary techniques by considering a particular example, and we will use an example we already know how to do by other means.

Consider $$\int_{-1}^1\int_0^{\sqrt{1-x^2}} \sqrt{x^2+y^2}\,dy\,dx.$$ The limits correspond to integrating over the top half of a circular disk, and we recognize that the function will simplify in polar coordinates, so we would normally convert to polar coordinates: $$\int_{0}^\pi\int_0^1 \sqrt{r^2}\;r\,dr\,d\theta={\pi\over3}.$$ But let's instead approach this as a substitution problem, starting with $x=r\cos\theta$, $y=r\sin\theta$. This pair of equations describes a function from "$r$-$\theta$ space'' to "$x$-$y$ space'', and because it involves familiar concepts, it is not too hard to understand what it does. In figure 17.7.2 we have indicated geometrically a bit about how this function behaves. The four dots labeled a–d in the $r$-$\theta$ plane correspond to the three dots in the $x$-$y$ plane; dots a and b both go to the origin because $r=0$. The horizontal arrow in the $r$-$\theta$ plane has $r=1$ everywhere and $\theta$ ranges from 0 to $\pi$, so the corresponding points $x=r\cos\theta$, $y=r\sin\theta$ start at $(1,0)$ and follow the unit circle counter-clockwise. Finally, the vertical arrow has $\theta=\pi/4$ and $r$ ranges from 0 to 1, so it maps to the straight arrow in the $x$-$y$ plane. Extrapolating from these few examples, it's not hard to see that every vertical line in the $r$-$\theta$ plane is transformed to a line through the origin in the $x$-$y$ plane, and every horizontal line in the $r$-$\theta$ plane is transformed to a circle with center at the origin in the $x$-$y$ plane. Since we are interested in integrating over the half-disk in the $x$-$y$ plane, we will integrate over the rectangle $[0,\pi]\times[0,1]$ in the $r$-$\theta$ plane, because we now see that the points in this rectangle are sent precisely to the upper half disk by $x=r\cos\theta$ and $y=r\sin\theta$.