Polar Coordinates

Section 8.2 Polar Coordinates

Subsection 8.2.1 Definition of Polar Coordinates

I have already introduce the polar coordinate transformation in Example 8.1.4. In this section, I want to further investigate this transformation and do a number of examples with it.

First, though, I’ll start with some motivation. Polar coordinates (and the other two important coordinate systems in Section 8.3) are useful because they describe circles and, therefore, anything with circular construction or symmetry of any type. Circles are common and natural shapes and many applications of mathematics deal with circls, circular motion, object with circular symmetry, etc. When a situation has some kind of symmetry to it, it’s often best to choose a coordinate system that suits the symmetry.

This can be demontrated by talking about motion that is confined to circles. Say there is an object in \(\RR^2\) that is restricted to a circe of a particular radius \(R\text{.}\) It lives on the locus which is that circle. Applications will often want a description of this locus, to solve problems of movement, intersection, force, etc. To describe the circle in normal cartesian coordinates, I use the equation \(x^2 + y^2 = R^2\text{.}\) This isn’t a terrible description, but it’s also not the easier description; we may need square roots to solve for individual variables when we use this locus, for example. However, in polar coordinates, the locus of the circle is just \(r=R\text{.}\) This is as simple as it gets. The coordinate system suits the task.

Let me recap the details of the polar coordinate system, though they should be familiar from previous courses. In polar coordinates, a point in \(\RR^2\) is described by a radius \(r\) (which is the distant from the point to the origin) and an angle \(\theta\) (which is the angle between the \(x\)-axis and the line from the origin to the point, measure counterclockwise). This is show in Figure 8.2.1.

Figure 8.2.1. Polar Coordinates

Let me also recall the change of coordinates. Here is the change where \(r\) and \(\theta\) are the independent variables (which, in multiple integration, is used for changing from cartesian to polar coordinates).

\begin{align*} x \amp = r \cos \theta\\ y \amp = r \sin \theta\\ \det J \amp = r \\ dx dy \amp = r dr d\theta \end{align*}

The reverse change of variables is also useful, though less often for integration, so I won’t calculate the Jacobian determinant here.

\begin{gather*} r = \sqrt{x^2 + y^2} \\ \theta = \arctan \left( \frac{y}{x} \right) \end{gather*}

Subsection 8.2.2 Properties of Polar Coordinates

I want to start with one notable weakness of polar coordinates. Unlike cartesian coordinates, which are defined and work well everywhere in \(\RR^2\text{,}\) polar coordaintes are not defined at the origin. The radius. \(r=0\text{,}\) is fine, but how would angle be defined at the origin? There is no arc to measure here, no line from the origin to a point to determine an angle. Angle simply can’t be determined. This fact breaks the coordinate system at the origin. This isn’t too much of a problem, but it is something that I always need to keep in mind when using polar coordinates.

One way to understand a coordinate system is to understand the paths in \(\RR^2\) where one of the coordinates is constant. In cartesian coordinates, this gives the horizontal lines \(y=c\) and the vertical lines \(x=c\) for the conventional square grid pattern of coordinate lines.

For polar coordinates, the loci \(r = c\) are circle, as I remarked earlier. The loci \(\theta = c\) are rays out from the original. This gives a spider-web pattern of coordinate lines, as shown in Figure 8.2.2. When I think of polar coordinates visually, this spider-web pattern is what I see.

Figure 8.2.2. The Polar Coordinate Grid

I want to make on final comments about the \(r\) Jacobian for polar coordinates. In cartesian coordinates, both \(x\) and \(y\) have unit of length. Therefore, the differential \(dx dy\) has units of (infinitesimal) area, as it should for integration. (Recall we wrote this as \(dA\) when we were working in general, without specific coordinates). However, in polar coordinates, \(r\) has units of distance, but \(\theta\) has no units. (Radian angles can naturally be define as ratio of distances; the units cancel in the fraction and an angle has no natural unit.). Therefore, the differential \(dr d\theta\) has units only of distance. This doesn’t make sense for integration, since this should be an infinitesimal area. The term with the Jacobian determinant is \(r dr d\theta\text{;}\) the additional \(r\) gives this units of area.

Subsection 8.2.3 Common Regions of Integration in Polar Coordinates

I said ealier that polar coordinates are good for situations with some kind of circular symmetry or description. I want to go over some of these regions. First, I can integrate over a circle, obviously. To integrate over a circle of radius \(R\text{,}\) the bounds are \(r \in [0,R]\) and \(\theta \in [0,2\pi]\text{.}\)

\begin{equation*} \int_0^R \int_0^{2\pi} f(r,\theta) r dr d \theta. \end{equation*}

I can integrate over various subsets of circles. A wedge is a portion of a circle like a pizza slice. The radius is not limit, but the angle is. The bounds are \(r \in [0,R]\) and \(\theta \in [\theta_1, \theta]\text{,}\) for some range of angles.

\begin{equation*} \int_0^R \int_{\theta_1}^{\theta_2} f(r,\theta) r dr d \theta. \end{equation*}

A region formed of a larger circle with a smaller circle (with the same centre) removed is called an annulus. In an annulus, the angle is unchanges, but the radius has limit. The region doesn’t start at \(r=0\text{,}\) but rather starts as the inner radius \(r = R_1\) and ends at a larger radius \(r = R_2\text{,}\) with \(R_1 \lt R_2\text{.}\) The bounds are \(r \in [R_1,R_2]\) and \(\theta \in [0,2\pi]\text{.}\)

\begin{equation*} \int_{R_1}^{R_2} \int_0^{2\pi} f(r,\theta) r dr d \theta. \end{equation*}

Finally, I could have a wedge-section of an annulus if I restrict both the radius and the angle. I might call this an arc. The bounds for this arc \(r \in [R_1,R_2]\) and \(\theta \in [\theta_1,\theta_2]\text{.}\)

\begin{equation*} \int_{R_1}^{R_2} \int_{\theta_1}^{\theta_2} f(r,\theta) r dr d \theta. \end{equation*}

Figure 8.2.3 shows these four types of region for integration with polar coordinates.

Figure 8.2.3. Some Regions of Integration for Polar Coordinates

Subsection 8.2.4

In many of these examples, particular when the bound become constant in polar constants, are seperable integrals. (Recall the definition from Definition 7.1.3.) For these integral, I’ll often split up integral into two separate single-variable integrals. I won’t always make not of this, but when the \(dr d\theta\) suddenly splits up, look to see if this is what I’ve done.

Example 8.2.4.

Consider the function \(f(x,y) = x + y\) on the arc in the first quadrant between radii \(2\) and \(4\text{.}\) I set up the integral over the arc with \(r \in [2,4]\) and \(\theta \in \left[ 0, \frac{\pi}{2} \right]\text{.}\) Then I proceed to setup an calculate the iterated integral. Note that I must remember to include the Jacobian \(r\) in \(r dr d \theta\) in the setup of the integral.

\begin{align*} \int_D x + y dA \amp = \int_0^{\pi/2} \int_2^4 r (\cos \theta + \sin \theta) r dr d\theta\\ \amp = \int_0^{\pi/2} \left. \frac{r^3}{r} (\cos \theta + \sin \theta) \right|_2^4 d\theta\\ \amp = \frac{56}{3} \int_0^{\pi/2} (\cos \theta + \sin \theta) d \theta\\ \amp = \left. \frac{56}{3} (\sin \theta - \cos \theta) \right|_0^{\pi/2}\\ \amp = \frac{56}{3} (1 - 0 - 0 + 1) = \frac{112}{3} \end{align*}

Example 8.2.5.

Now consider the function \(e^{x^2 + y^2}\) on the circle of radius \(R\) centered at the origin. Notice here that the integrand also has circular symmetry — changing to polar coordinates will help the integrand as well as the domain.

The integral is annoying in conventional Cartesian coordinates.

\begin{equation*} \int_D f(x,y) dA = \int_{-R}^R \int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}} e^{x^2 + y^2} dy dx \end{equation*}

This is essentially impossible. However, in polar coordinates it improves greatly.

\begin{align*} \int_D f(x,y) dA \amp = \int_0^{2\pi} \int_0^R e^{r^2} r dr d\theta\\ \amp = \int_0^{2\pi} d\theta \int_0^R e^{r^2} r dr\\ \amp = 2\pi \left. \frac{e^{r^2}}{2} \right|_0^R = \frac{2\pi}{2} (e^{R^2} - 1) = \pi (e^{R^2} - 1) \end{align*}

Example 8.2.6.

Now recall the integral I did to calcluate the volume of a sphere of radius \(R\) in Example 7.3.3.

\begin{equation*} 8 \int_0^{R} \int_0^{\sqrt{R^2 - x^2}} \sqrt{R^2 - x^2 - y^2} dy dx \end{equation*}

This was a trickly integral. However, if I change to polar coordinates and integrate over the whole circle instead of just a quarter, the integral improves a great deal. The \(r\) from the Jacobian, which sits outside the \(\sqrt{R^2 - r^2}\text{,}\) is extremely convenient, allowing a substitution \(u = R^2 - r^2\) to work for the \(r\) integral. (In the previous approach, trig substitutions were the only approach. In a sense, changing to polar coordinates is sort-of doing the trig substitution in a more holistic way.)

\begin{align*} \amp = 2 \int_0^{2\pi} \int_0^R \sqrt{R^2 - r^2} r dr d\theta\\ \amp = 2 \int_0^{2\pi} d\theta \int_0^R \sqrt{R^2 - r^2} r dr d\theta\\ \amp = 4\pi \left. (R^2 - r^2)^{\frac{3}{2}} \frac{2}{3} \frac{-1}{2} \right|_0^R\\ \amp = \frac{4\pi}{3} (R^2)^{\frac{3}{2}} = \frac{4\pi R^3}{3} \end{align*}

Example 8.2.7.

Recall the integral for the volume of a cone of height \(h\) and radius \(R\) from Example 7.3.4. I didn’t even do the integral, since I didn’t want to deal with the complicated trig substitution which would have been required.

\begin{equation*} \int_D f(x,y) dA = 4 \int_0^R \int_0^{\sqrt{R^2 - x^2}} \left( h - \frac{h}{r} \sqrt{x^2 + y^2} \right) dy dx \end{equation*}

Now I can use polar coordinates to make the integral much more accessible. I’m integrating over a circle the base of the cone.

\begin{align*} \amp = \int_0^{2\pi} \int_0^R \left( h - \frac{hr}{R} \right) r dr d\theta\\ \amp = 2\pi \left. \left( \frac{hr^2}{2} - \frac{hr^3}{3R} \right) \right|_0^R\\ \amp = 2\pi \left( \frac{hR^2}{2} - \frac{hR^2}{3} \right) = \frac{\pi R^2 h}{3} \end{align*}

Example 8.2.8.

To show how some of these techniques work together, consider integrating the function \(f(x,y) = x^2\) over the ellipse \(D = \frac{x^2}{4} + \frac{y^2}{9} = 1\text{.}\)

\begin{equation*} \int_D x^2 dA \end{equation*}

I’m going to do two changes of variables. First, I make the change of variables give by \(x=2u\) and \(y=3v\text{.}\) This change of variables has Jacobian \(|J| = 6\) so that \(dx dy = 6 du dv\text{.}\) The ellipse \(D\) becomes the unit circle \(C\text{.}\)

\begin{equation*} \int_C (2u)^2 6 du dv = 24 \int_C u^2 du dv \end{equation*}

Now, since I am integrating over the unit circle, I can change to polar coordinates.

\begin{align*} 24 \int_C u^2 du dv \amp = 24 \int_0^{2\pi} \int_0^1 r^2 \cos^2 \theta r dr d\theta\\ \amp = 24 \int_0^{2\pi} \cos^2 \theta d \theta \int_0^1 r^3 dr\\ \amp = 24 \left. \left( \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right) \right|_0^{2\pi} \left. \left( \frac{r^4}{4} \right) \right|_0^1\\ \amp = 24 \pi \frac{1}{4} = 6 \pi \end{align*}

Example 8.2.9.

Figure 8.2.10. An Offset Circle in Polar Coordinates

Say I wanted to integrate the function \(x^2 + y^2\) over the circle of radius 1 centered at \((1,0)\text{.}\) The integrand obviously lends itself to polar coordinates. There is circular symmetry in the domain, but the offset is a little confusing.

The equation of this circle is \((x-1)^2 + y^2 = 1\text{.}\) I can make this a polar locus by replacing \(x\) and \(y\) with their polar equivalent. Doing so produces \(r^2 \cos^2 \theta - 2 r \cos \theta + 1 + r^2 \sin^2 \theta =1\text{,}\) which simplifies into \(r^2 - 2r \cos \theta = 0\) or \(r = 2 \cos \theta\) for \(\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]\text{.}\) I can actually describe this in terms of polar coordinates; I just have to allow one of the bounds to be variable (like I did with variable bounds for cartesian coordinates). I will let the bound for \(\theta\) be constant, so that \(\theta\) will be the outside variable. Then I can use the expression \(r = 2 \cos \theta\) as an higher bound for \(r\) in terms of \(\theta\text{.}\) The lower bound for \(r\) will stil be \(0\text{.}\) Figure 8.2.10 shows, in the straight lines, the range of the radius from \(0\) to \(2 \cos \theta\) as \(\theta\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\text{.}\)

\begin{align*} \int_D f(x,y) dA \amp = \int_{-\pi/2}^{\pi/2} \int_0^{2\cos \theta} r^2 r dr d\theta\\ \amp = \int_{-\pi/2}^{\pi/2} \left. \frac{r^4}{4} \right|_0^{2 \cos \theta} d \theta\\ \amp = 4 \int_{-\pi/2}^{\pi/2} \cos^4 \theta d \theta\\ \amp = 4 \left. \frac{1}{32} \left( 12 \theta + 8 \sin (2\theta) + \sin (4\theta) \right) \right|_{-\pi/2}^{\pi/2}\\ \amp = \frac{1}{8} \left(12 \left( \frac{\pi}{2} - \frac{-\pi}{2} \right) \right) + 0 + 0 = \frac{3\pi}{2} \end{align*}

Example 8.2.11.

Figure 8.2.12. Circle of Radius 2 with Offset Circle of Radus 1 Removed

What if \(D\) is the circle of radius \(2\) excluding the circle of radius \(1\) centered at \((1,0)\) and we want to integrate \(\sqrt{x^2 + y^2}\) over this region? (This region is show in Figure 8.2.12 Here, we can think of the entire circle of radius 2 as \(D_1\) and the removed circle as \(D_2\text{;}\) then the integral will be the integral over \(D_1\) subtracting the integral over \(D_2\text{.}\) To describe \(D_2\text{,}\) I’ll use the smae bounds as the previous example.

The first part is an integral over \(D_1\text{.}\)

\begin{align*} \int_{D_1} \sqrt{x^2 + y^2} dA \amp = \int_0^{2\pi} \int_0^2 r r dr d\theta\\ \amp = \left. 2\pi \frac{r^3}{3} \right|_0^2 = \frac{16\pi}{3} \end{align*}

The second part is an integral over \(D_2\text{.}\)

\begin{align*} \int_{D_2} \sqrt{x^2 + y^2} dA \amp = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_0^{2 \cos \theta} r^2 dr d\theta\\ \amp = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \left. \frac{r^3}{3} \right|_0^{2 \cos \theta} d\theta\\ \amp = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \left( \frac{8 \cos^3 \theta}{3} \right) = \frac{8}{3} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos^3 \theta d \theta\\ \amp = \frac{8}{3} \left. \left( \frac{1}{3} (2 + \cos^2 + \theta) \sin \theta \right) \right|_{\frac{-\pi}{2}}^{\frac{\pi}{2}} = \frac{32}{9} \end{align*}

The result of the original integral is the difference of the two parts.

\begin{equation*} \int_{D_1 \backslash D_2} \sqrt{x^2 + y^2} dA = \frac{16\pi}{3} - \frac{32}{9} = \frac{48\pi - 32}{9} \end{equation*}

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