Polar coordinates

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We integrate over regions in polar coordinates.

We are currently interested in computing integrals of functions over various regions in $\R ^2$ and $\R ^3$ via $\underbrace {\iint _R F(x,y) \d A}_{\text {double integral}} \quad \text {and}\quad \underbrace {\iiint _R F(x,y,z) \d V}_{\text {triple integral}}$ Some regions like rectangles and boxes are easy to describe using $(x,y)$ -coordinates (a.k.a. rectangular coordinates). However, other regions like circles and other things with rotational symmetry are easier to work with in polar coordinates. Recall that in polar coordinates, $\begin{align*} x(\theta) &= r(\theta) \cdot \cos(\theta)\\ y(\theta) &= r(\theta) \cdot \sin(\theta) \end{align*}$

where $r(\theta )$ is a function of $\theta$ . When working with parametric equations of this form, it is common to notate $(r \cdot \cos (\theta ), r\cdot \sin (\theta )) \quad \text {as}\quad (r,\theta )$ and state that we are working in polar coordinates.

An ordered pair consisting of a radius and an angle $(r,\theta )$ can be graphed as $\begin{align*} x &= r\cdot \cos(\theta)\\ y &= r\cdot \sin(\theta) \end{align*}$

meaning:

Coordinates of this type are called polar coordinates.

Consider the point $(5, 2\pi /3)$ in polar coordinates. What is this point when expressed in $(x,y)$ -coordinates? $(x,y) = \left (\answer {5\cos (2\pi /3)}, \answer {5 \sin (2\pi /3)}\right )$

Consider the point $(-1, -5)$ in $(x,y)$ -coordinates. What is this point when expressed in polar coordinates with $0\le \theta <2\pi$ ? $(r,\theta ) = \left (\answer {\sqrt {26}}, \answer {\arctan (5)+\pi }\right )$

Double integrals in polar coordinates

The basic form of the double integral is:

which can be interpreted as

Over some region, sum up products of heights and areas.

Of course if you want to evaluate the integral (and honestly, who doesn’t?) you have to change $R$ to a region defined in $(x,y)$ -coordinates, and change $\d A$ to $\d x\d y$ or $\d y\d x$ leaving iterated integrals $\int _a^b\int _c^d F(x,y) \d y \d x\quad \text {and}\quad \int _c^d \int _a^b F(x,y) \d x\d y$

Now consider representing a region $R$ with polar coordinates.

Let $R$ be the region in the first quadrant bounded by the curve. We can approximate this region using the natural shape of polar coordinates: Portions of sectors of circles. In the figure, one such region is shaded, shown below:

From the picture above, we see that: $\begin{align*} \d A &= \d r \cdot (r \d \theta) \\ &= r \d r \d \theta \end{align*}$

Recalling that the determinant of a $2\times 2$ matrix gives the area of a parallelogram, we could also deduce the correct formula for $\d A$ by setting $\begin{align*} x(r,\theta) = r \cos(\theta)\\ y(r,\theta) = r \sin(\theta) \end{align*}$

and computing:

$\begin{align*} \d A &= \left| \det \begin{bmatrix} \pp[x]{r} \d r & \pp[y]{r} \d r\\ \pp[x]{\theta} \d \theta & \pp[y]{\theta} \d \theta \end{bmatrix}\right|\\ &= \left| \det \begin{bmatrix} \answer[given]{\cos(\theta)} \d r & \answer[given]{\sin(\theta)} \d r\\ \answer[given]{-r\sin(\theta)} \d \theta & \answer[given]{r\cos(\theta)} \d \theta \end{bmatrix}\right|\\ &= r \d r \d \theta \end{align*}$

So to evaluate $\iint _R F\d A,$ replace $\d A$ with $r\d r\d \theta$ and convert the function $z=F(x,y)$ to a function of polar coordinates: $F(r\cos (\theta ),r\sin (\theta ))$ Finally, find bounds $g_1(\theta )\leq r\leq g_2(\theta )$ and $\alpha \leq \theta \leq \beta$ that describe $R$ . Let’s state this as a theorem:

Fubini Let $F:\R ^2\to \R$ be continuous on the region $R=\{(r,\theta ):\text {$\alpha \leq \theta \leq \beta $ and $g_1(\theta )\leq r\leq g_2(\theta )$}\}$ Then: $\iint _R F(x,y)\d A = \int _\alpha ^\beta \int _{g_1(\theta )}^{g_2(\theta )} F\big (r\cos (\theta ),r\sin (\theta )\big ) r\d r\d \theta .$

Write down a double integral in polar coordinates that will compute the area of a circle of radius $a$ . $\iint _R \d A = \int _{\answer {0}}^{\answer {2\pi }} \int _{\answer {0}}^{\answer {a}}r \d r \d \theta$

Compute the area of the lemniscate given by: $r(\theta ) = \sqrt {\cos (2\theta )}$

When first exploring problems like this, we should try to plot them if we can. $\graph [polar,xmin=-2, xmax=2, ymin=-1.5, ymax=1.5]{r=\sqrt {\cos (2\theta )}}$ Now either by reasoning about cosine, or by experimenting with the plot, we should convince ourselves that if $0\le \theta \le \pi /4$ , the area is given by: $\begin{align*} 4\int_{0}^{\pi/4} \int_{0}^{\sqrt{\cos(2\theta)}}\answer[given]{r} \d r \d \theta &= 4\int_{0}^{\pi/4} \eval{\answer[given]{r^2/2}}_{\answer[given]{0}}^{\answer[given]{\sqrt{\cos(2\theta)}}} \d \theta\\ &= 4\int_{0}^{\pi/4} \answer[given]{\cos(2\theta)/2} \d \theta\\ &= \int_{0}^{\pi/4} \answer[given]{2\cos(2\theta)} \d \theta\\ &= \eval{\answer[given]{\sin(2\theta)}}_{\answer[given]{0}}^{\answer[given]{\pi/4}} \\ &= \answer[given]{1} \end{align*}$

Finally, let’s derive the volume of a sphere using a double integral in polar coordinates.

Find the volume of a sphere with radius $a$ .

The sphere of radius $a$ , centered at the origin, has equation $\answer [given]{x^2+y^2+z^2}=a^2$ Solving for $z$ , we have $z=\answer [given]{\sqrt {a^2-x^2-y^2}}$ . This gives the upper half of a sphere. We wish to find the volume under this top half, then double it to find the total volume.

The region we need to integrate over is the circle of radius $a$ , centered at the origin. $R =\{(r,\theta ):\text {$\answer [given]{0}\leq r\leq \answer [given]{a}$ and $\answer [given]{0}\leq \theta \leq \answer [given]{2\pi }$}\}$ Thus, the volume of a sphere with radius $a$ is: $2\iint _R\sqrt {a^2-x^2-y^2}\d A$

$\begin{align*} &= 2\int_{\answer[given]{0}}^{\answer[given]{2\pi}}\int_{\answer[given]{0}}^{\answer[given]{a}}\sqrt{a^2-(r\cos\theta)^2-(r\sin\theta)^2}r\d r\d\theta\\ &=2\int_{\answer[given]{0}}^{\answer[given]{2\pi}}\int_{\answer[given]{0}}^{\answer[given]{a}}r\sqrt{a^2-r^2}\d r\d\theta\\ &=2\int_{\answer[given]{0}}^{\answer[given]{2\pi}}\eval{\answer[given]{\frac{-2}{6}(a^2-r^2)^{3/2}}}_{0}^a \d\theta\\ &= \int_{\answer[given]{0}}^{\answer[given]{2\pi}}\left(\answer[given]{\frac{2}{3}a^3}\right)\d \theta\\ &=\eval{\answer[given]{\frac{2}{3}a^3\theta}}_{\answer[given]{0}}^{\answer[given]{2\pi}}\\ &=\answer[given]{\frac{4}{3}\pi a^3}. \end{align*}$

The formula for the volume of a sphere with radius $r$ is given as $4/3\pi r^3$ . We have justified this formula with our calculation!

One may wonder how polar coordinates could be extended to triple integrals…read on!