Change of variables

A disc described in $x$ and $y$ has curved limits of integration, $y=\pm \sqrt{a^2-x^2}$, and the iterated integrals they produce are usually far harder than the question deserves.

The previous lesson built the double integral $\underset{R}{\iint }fdA$ as a limit of sums over small rectangles and evaluated it by Fubini's theorem as two single integrals, first over rectangles and then over regions whose limits are functions. It ended on a mismatch: a disc is a rectangle in no Cartesian sense, and its square roots leak into every inner integral. Polar coordinates describe a disc as a rectangle, and this lesson finds the price of switching to them, uses them on an integral that has no elementary antiderivative, and then generalises to any substitution in two variables. It assumes the double integral and iterated integration, and the cross product of two vectors.

Regions that are rectangles in $r$ and $\theta$

Polar coordinates, met in the lesson on the chain rule as a change of variables for derivatives, locate a point by its distance $r$ from the origin and the angle $\theta$ its radius makes with the positive $x$ axis:

Every point other than the origin gets exactly one pair with $r>0$ and $0\le \theta <2\pi$; the origin, $r=0$ at any angle, is a single point with no area, so an integral does not notice it.

The payoff is in how regions look. The disc $x^2+y^2\le a^2$ is $0\le r\le a$, $0\le \theta \le 2\pi$: constant limits on both variables, a rectangle in the $r\theta$ plane. The annulus between the circles of radius $1$ and $3$, the shape of a washer, is $1\le r\le 3$ with $\theta$ running all the way round. A sector, the slice of a pizza of radius $15$ between the directions $\theta =0$ and $\theta =\frac{\pi }{4}$, is $0\le r\le 15$, $0\le \theta \le \frac{\pi }{4}$, and the upper half disc is the same with $0\le \theta \le \pi$. In Cartesian coordinates each of these needs square roots, and the annulus needs splitting into four pieces as well.

Polar regions with variable limits work exactly like the general regions of the previous lesson. The circle $x^2+y^2=2x$, of radius $1$ centred at $(1,0)$, becomes $r^2=2r\cos \theta$, that is $r=2\cos \theta$, so the disc it bounds is $0\le r\le 2\cos \theta$ for $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$. The inner limit depends on the outer variable, and the order of integration is forced: $r$ inside, $\theta$ outside.

Rewriting the integrand is the easy half. The hard half is $dA$: it is tempting to write $drd\theta$, and it is wrong.

The area of a polar rectangle

A grid of lines of constant $r$ and constant $\theta$ cuts the plane into polar rectangles: the region $r_0\le r\le r_0+\Delta r$, ${\theta }_0\le \theta \le {\theta }_0+\Delta \theta$, a curved box bounded by two arcs and two radii. Unlike the squares of a Cartesian grid, they grow with distance from the origin, because a fixed angle sweeps a longer arc at a larger radius.

The exact area comes from the sector. A sector of radius $r$ and angle $\Delta \theta$ is the fraction $\frac{\Delta \theta }{2\pi }$ of a disc of area $\pi r^2$, so its area is $\frac{1}{2}{r}^2\Delta \theta$. The polar rectangle is the sector of radius $r_0+\Delta r$ with the sector of radius $r_0$ removed, so its area is

where $\bar{r}$ is the radius midway across the box. No approximation has been made. The box from $r=2$ to $r=2.1$ with $\Delta \theta =0.1$ has area $\frac{1}{2}(4.41-4)(0.1)=0.0205$, and indeed $2.05\times 0.1\times 0.1=0.0205$. The box is almost a rectangle of width $\Delta r$ and length $\bar{r}\Delta \theta$, the arc length, and the factor $\bar{r}$ is the stretch that turns an angle into a length.

Now repeat the construction of the double integral with this grid. Cut the region into polar rectangles, sample $f$ at the midpoint of each, and add. The sum is

which is exactly a Riemann sum, over a rectangle in the $r\theta$ plane, for the function $f(r\cos \theta ,r\sin \theta )r$. As the grid refines it converges both to $\underset{R}{\iint }fdA$ and to the iterated integral of that function, so

The area element is $dA=rdrd\theta$. The first check is the disc itself: $\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{a}{\int }}rdrd\theta =2\pi \cdot \frac{a^2}{2}=\pi a^2$. Forgetting the $r$ gives $2\pi a$, the circumference, which is not even measured in the right units.

A disc of varying density

The previous lesson computed the mass of a flat plate as the integral of its surface density, $M=\underset{R}{\iint }\rho dA$. A round plate whose density depends only on the distance from the centre is what polar coordinates were made for.

Example. A thin disc of radius $10$ cm is machined thinner towards its rim, so that its surface density is $\rho =3-0.1r$ grams per square centimetre, with $r$ in centimetres. Find its mass.

The disc is $0\le r\le 10$, $0\le \theta \le 2\pi$, and the density does not involve $\theta$, so the $\theta$ integral contributes a factor $2\pi$:

which is about $733.04$ grams. The same integral in Cartesian coordinates is $\underset{-10}{\overset{10}{\int }}\underset{-\sqrt{100-x^2}}{\overset{\sqrt{100-x^2}}{\int }}(3-0.1\sqrt{x^2+y^2})dydx$, whose inner integral alone needs a logarithm; evaluated numerically, it gives $733.03$. A uniform disc of density $3$ would weigh $300\pi$, about $942$ grams, so the thinning removes more than a fifth of the metal, most of it near the rim where the rings are longest.

Now you. A washer occupies the annulus $1\le r\le 3$ and has surface density $\rho =x^2+y^2$. Find its mass, to two decimal places.

The Gaussian integral

The function $e^{-x^2}$ has no antiderivative that can be written with the functions of Calculus I; Joseph Liouville proved in the 1830s that no combination of powers, exponentials, logarithms and trigonometric functions will do. Yet its integral over the whole line has an exact value, and polar coordinates find it. The idea, usually credited to Siméon Denis Poisson, is to make the problem two dimensional on purpose.

Example. Evaluate $I=\underset{-\infty }{\overset{\infty }{\int }}{e}^{-x^2}dx$.

The integral converges, since $e^{-x^2}\le e^{-|x|}$ once $|x|\ge 1$, and it is positive. Write it twice, once with $x$ and once with $y$ as the dummy variable, and multiply. A product of a function of $x$ alone and a function of $y$ alone integrates to the product of the two integrals, which is Fubini's theorem read backwards, so

The integrand depends only on $x^2+y^2=r^2$, and the whole plane is $0\le r<\infty$, $0\le \theta \le 2\pi$. Switching coordinates,

so $I=\sqrt{\pi }$, about $1.7724539$. Simpson's rule on $-10\le x\le 10$ agrees to ten decimal places. The step that made it work is the factor $r$: $e^{-r^2}$ has no antiderivative either, but $r{e}^{-r^2}$ does, by the substitution $u=r^2$ of Calculus I.

The switch to the whole plane deserves one more line, because $I^2$ was defined through squares and the polar integral through discs. Over the square $-R\le x,y\le R$ the double integral is ${\left(\underset{-R}{\overset{R}{\int }}{e}^{-x^2}dx\right)}^2$. That square contains the disc of radius $R$ and sits inside the disc of radius $R\sqrt{2}$, and the integrand is positive, so the square's integral lies between the two disc integrals, $\pi (1-e^{-R^2})$ and $\pi (1-e^{-2R^2})$. At $R=2$ the three are $3.084$, $3.112$ and $3.141$. As $R$ grows both bounds tend to $\pi$, and the squeeze settles it.

Now you. Use the same squaring to evaluate $\underset{-\infty }{\overset{\infty }{\int }}{e}^{-x^2/2}dx$.

That value is why the normal distribution of Probability and Statistics carries its odd constant. The standard normal density is $\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}$, and a probability density must have total area $1$; dividing by $\sqrt{2\pi }$ is exactly what makes it so. Pierre-Simon Laplace had found $\sqrt{\pi }$ by another route in 1774, in work on probability.

The Jacobian

Polar coordinates are one substitution among many. In general, let $x=x(u,v)$ and $y=y(u,v)$ carry a region $S$ of the $uv$ plane onto the region $R$ of the $xy$ plane, one to one except perhaps along boundaries, with continuous partial derivatives. The question is the same as before: how large is the image of a small rectangle $\Delta u$ by $\Delta v$?

Linear approximation answers it. Moving from $(u_0,v_0)$ to $(u_0+\Delta u,v_0)$ moves the image point by approximately $⟨x_u,y_u⟩\Delta u$, and moving in $v$ moves it by approximately $⟨x_v,y_v⟩\Delta v$, with the partials taken at $(u_0,v_0)$. So the small rectangle lands on an approximate parallelogram with those two edges. The cross product measures parallelograms: its length is the base times the height. Treating the edges as vectors in space with third component $0$,

so the image has area approximately $|J|\Delta u\Delta v$, where

This is the Jacobian determinant of the substitution, often written $\frac{\partial (x,y)}{\partial (u,v)}$, after Carl Gustav Jacob Jacobi, who studied such determinants in an 1841 paper. It is the local area scale factor: near each point, the substitution multiplies small areas by $|J|$. The Riemann sum argument of the polar case then gives the change of variables formula,

For polar rectangles the area was exact; here the parallelogram is only approximate, and proving that its error vanishes faster than $\Delta u\Delta v$ belongs to analysis.

Check it against polar coordinates, with $u=r$ and $v=\theta$. The partials are $x_r=\cos \theta$, $y_{\theta }=r\cos \theta$, $x_{\theta }=-r\sin \theta$ and $y_r=\sin \theta$, so $J=r{\cos }^2\theta +r{\sin }^2\theta =r$, the factor the sectors gave. In one variable, substitution $x=g(u)$ gives $dx=g^{\prime }(u)du$, and $g^{\prime }(u)$ is the factor by which the substitution stretches length. The Jacobian is its two dimensional version, with one difference: area is never negative, so the formula takes $|J|$, where one variable absorbs a negative $g^{\prime }$ by swapping the limits.

Ellipses and parallelograms

The substitution $x=au$, $y=bv$, with $a,b>0$, carries the unit disc $u^2+v^2\le 1$ onto the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$. Its Jacobian is $a\cdot b-0\cdot 0=ab$, a constant, since the disc is stretched by $a$ across and $b$ up. So the area of the ellipse is $ab$ times the area of the unit disc, $\pi ab$, which for $a=b$ is the disc again.

A linear substitution does the same for a parallelogram, whose sides are two pairs of parallel lines. Each pair consists of two level lines $u=c_1$ and $u=c_2$ of some linear function $u$, so two such functions turn the parallelogram into a rectangle.

Example. Let $R$ be the parallelogram bounded by $x+y=0$, $x+y=3$, $x-2y=0$ and $x-2y=3$. Evaluate $\underset{R}{\iint }(x+y)(x-2y)dA$.

Set $u=x+y$ and $v=x-2y$, so that $R$ becomes the square $0\le u\le 3$, $0\le v\le 3$. Subtracting the equations gives $3y=u-v$, so $y=\frac{u-v}{3}$ and $x=\frac{2u+v}{3}$. Then

and $|J|=\frac{1}{3}$. The integral is $\frac{1}{3}\underset{0}{\overset{3}{\int }}\underset{0}{\overset{3}{\int }}uvdudv=\frac{1}{3}\cdot \frac{9}{2}\cdot \frac{9}{2}=6.75$. Two checks: the area of $R$ is $\frac{1}{3}\times 9=3$, and the corners $(0,0)$, $(2,1)$, $(1,-1)$ and $(3,0)$ give edges $⟨2,1⟩$ and $⟨1,-1⟩$, whose cross product has length $|-2-1|=3$. In $x$ and $y$ the region needs splitting into three pieces. There is a shortcut, too: the Jacobian of $u,v$ with respect to $x,y$ is $1\cdot (-2)-1\cdot 1=-3$, and the Jacobian of the inverse substitution is its reciprocal.

Now you. Let $R$ be bounded by $y-x=0$, $y-x=2$, $y+2x=0$ and $y+2x=3$. Find the area of $R$ and evaluate $\underset{R}{\iint }(y-x)(y+2x)dA$.

Out of the plane

Changing variables fits the coordinates to the region: polar coordinates for discs, annuli and sectors, a stretch for an ellipse, a linear map for a parallelogram. The price is the factor $|J|$ by which the new coordinates distort area, and for polar coordinates that price, the $r$ in $rdrd\theta$, is what evaluated the Gaussian integral.

Everything so far happens in the plane. Solids have volume and mass too. The next lesson repeats the construction in space: triple integrals over solids, cylindrical and spherical coordinates, and the volume elements that play the part of $rdrd\theta$.