Today, we're going to look at integrals over regions in the plane that have circular symmetry. This leads us naturally to the topic of polar coordinates.

Here's a very natural question: What's the volume under the graph of $f(x,y)=1-(x^2+y^2)$ and over the $xy$-plane?

Here's the domain of integration, namely $x^2+y^2\leq1$, with arrows indicating how we might integrate with respect to $y$ first.

Using some symmetry, this leads to the integral

$$ \begin{aligned} &4\int_0^1 \int_0^{\sqrt{1-x^2}} (1-x^2-y^2) \, dy = 4\int_0^1 \left(y-x^2y-\frac{1}{3}y^3\right)\Big|_0^{\sqrt{1-x^2}} \, dx \\ &= 4\int_0^1 \left(\sqrt{1-x^2} - x^2 \sqrt{1-x^2} -\frac{1}{3}(1-x^2)^{3/2}\right) \, dx = \color{red}{??} \end{aligned} $$An alternative way to describe the position of points in the plane

A few polar points with various descriptions

The relationship with $x$ and $y$

A rectangle is easy to describe in Cartesian coordinates because it's bound by two lines of constant $x$ and two lines of constant $y$: $$\{(x,y):a\leq x \leq b \text{ and } c\leq y \leq d\}.$$

The regions that are easy to describe in polar coordinates are bound by lines of constant $\theta$ and arcs of constant $r$: $$\{(r,\theta):a\leq r \leq b \text{ and } \alpha \leq \theta \leq \beta\}.$$

- The unit disk $$\{(r,\theta):r \leq 1 \text{ and } 0 \leq \theta \leq 2\pi\}.$$
- Top half of a disk of radius $R$ centered at the origin $$\{(r,\theta):r \leq R \text{ and } 0 \leq \theta \leq \pi\}.$$
- An annulus with inner radius $a$ and outer radius $b$: $$\{(r,\theta):a \leq r \leq b \text{ and } 0 \leq \theta \leq 2\pi\}.$$

There are three essential parts to translate from Cartesian to polar:

$$\color{green}{\iint_D} \color{blue}{f(x,y)} \, \color{red}{dA} \: \to \: \color{green}{\int_{\alpha}^{\beta} \int_a^b} \color{blue}{F(r,\theta)} \: \color{red}{r \, dr \, d\theta}.$$- The bounds $\displaystyle \iint_D \to \int_{\alpha}^{\beta} \int_a^b$
- The function $f(x,y) \to F(r,\theta)$
- The area element $dA \to r\,dr\,d\theta$

The bounds of integration typically look like $\displaystyle \int_{\alpha}^{\beta} \int_a^b$.

In this case of the whole unit disk, this boils down to $\displaystyle \int_{0}^{2\pi} \int_0^1$.

To translate $f(x,y)$ to $F(r,\theta)$, we can replace

- $x=r\cos(\theta)$
- $y=r\sin(\theta)$

For example, $f(x,y)=x-y^3$ would become $F(r,\theta)=r\cos(\theta)-r^3\sin^3(\theta)$.

Often though, we'll see an $x^2+y^2$ and we can replace that directly with $r^2$.

For example, $f(x,y)=(x^2+y^2)\sin(x^2+y^2)$, then

$$F(r,\theta) = r^2\sin(r^2).$$The result is independent of $\theta$.

The final step is the easiest to do, though slightly tricky to see why you should do it. The $dA$ is always replaced by $r\,dr\,d\theta$. That is

$$dA \to r\,dr\,d\theta.$$To understand where the extra $r$ comes from, you need to understand that the volume represented by a polar integral is approximated by a sum:

$$\iint_d f(x,y) \, dA \approx \sum_{i=1}^m \sum_{j=1}^n f(r_i \cos(\theta_j),r_i \sin(\theta_j)) \Delta A_{i,j}.$$In that sum that approximates the volume, the $\Delta A_{i,j}$ term is not constant but depends on how far we are away from the origin, as illustrated in the figure.

As an example, let's return to our basic natural question: What's the volume under the graph of $f(x,y)=1-(x^2+y^2)$ and over the $xy$-plane?

Since the domain where $f(x,y)>0$ is exactly the unit disk $D$, I guess the answer should be

$$\iint_D (1-(x^2+y^2)) \, dA.$$We've just got to express that in polar coordinates.

Let's translate. Our new function (expressed in polar coordinates) is

$$F(r,\theta) = 1-r^2.$$Our bounds of integration (since we're integrating over the unit disk) are $0<r<1$ and $0<\theta<2\pi$.

Thus, our new integral is

$$\iint_D (1-(x^2+y^2)) \, dA = \int_0^{2\pi} \int_0^1 (1-r^2)\,r\,dr\,d\theta.$$Once we've expressed our integral in polar coordinates, it turns out to be very easy to integrate.

$$\begin{align} \int_0^{2\pi} \int_0^1 (1-r^2)\,r\,dr\,d\theta &= 2\pi \int_0^1 (r-r^3) dr \\ &=2\pi\left(\frac{1}{2}r^2 - \frac{1}{4}r^4\right)\Big|_0^1 = \frac{\pi}{2}. \end{align}$$This is very much like example 13.3.1 from our textbook. Our mission is find the signed volume between the graph of $f(x,y)=1-(2x+y)$ and over the top half of the disk of radius 2. In the image shown below, note that the graph lies partly above and partly below the $xy$-plane. By "signed" volume, we mean the volume of the portion above minus the volume of the portion below.

The region we are integrating over can be described as

$$\{(r,\theta): 0\leq r \leq 2, 0\leq \theta \leq \pi\}.$$These inequalities yield our bounds of integration.

Our integral now becomes

$$\begin{align} \iint_D (1-(2x+y))\,dA &= \int_0^{\pi} \int_0^2 (1-2r\cos(\theta)-r\sin(\theta)) \,r \,dr \, d\theta \\ &= \int_0^{\pi}\left(\frac{1}{2}r^2-\frac{2}{3}r^3\cos(\theta)-\frac{1}{3}r^3\sin(\theta)\right)\Big|_0^2 \, d\theta \\ &= \int_0^{\pi}\left(2-\frac{16}{3}\cos(\theta)-\frac{8}{3}\sin(\theta)\right) d\theta \\ &= 2\pi - \frac{16}{3}. \end{align}$$For our final example, we're going to prove the fabulous fact that

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}.$$We'll accomplish this by evaluating an associated double integral over the plane using polar coordinates.

As a first step, we'll express the square of our integral as an integral over the whole plane where the integrand has rotational symmetry. I guess we want to show that this is equal to $\pi$.

$$\begin{align} \left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)^2 &= \left(\int_{-\infty}^{\infty} e^{-x^2} dx\right) \left(\int_{-\infty}^{\infty} e^{-\color{red}x^2} d\color{red}x\right) \\ &= \left(\int_{-\infty}^{\infty} e^{-x^2} dx\right) \left(\int_{-\infty}^{\infty} e^{-\color{red}y^2} d\color{red}y\right) \\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx\,dy \stackrel{\color{gray}?}{=} \pi \end{align}$$The integral we are interested in is well approximated by the integral over a disk of radius $R$, where $R$ is very large. We get the exact value by letting $R\to\infty$. Also, the extra $r$ from the $r\,dr\,d\theta$ helps us because it allows us to use the substitution $u=-r^2$.

$$\begin{align} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx\,dy &\approx \int_0^{2\pi} \int_0^R e^{-r^2}r\,dr\,d\theta \\ &=2\pi \int_0^{-R^2} e^u \left(-\frac{1}{2}\right) \, du \\ &= -\pi\left(e^{-R^2}-e^0\right) \to \pi \\ & \text{as }R\to\infty. \end{align}$$