Polar Coordinates¶
We begin by recalling the definition of polar coordinates in the plane. In particular, we consider the area of a small "polar rectangle" in the plane. Understanding how the area of a polar rectangle is related to the area of a Cartesian rectangle will be a key first step toward integrating in polar coordinates.
Video Lecture
The Polar Transformation¶
The transformation of the plane that maps a point's polar coordinates $(r,\theta)$ to its Cartesian coordinates $(x,y)$ is given by
$$
T(r,\theta) = \begin{cases} x = r\cos\theta, \\ y = r\sin\theta. \end{cases}
$$
This transformation is well-defined and single-valued for any representation of the point in polar coordinates; that is, for any choice of $r$ and $\theta$ that represents the point.
The same is not true of the inverse transformation, so we need to set some conventions.
First, we assume that $r > 0$ for all points other than the pole. The pole is, of course, the only point for which $r=0$.
Next, the usual convention is to assume that $0 \leq \theta < 2\pi$. This is in fact the convention of our book when presenting polar coordinates. However, this convention doesn't jive with the book's definition of the inverse transformation for polar coordinates: the book defines $\theta = \tan^{-1}(\tfrac{y}{x})$. There are a couple of problems with this. The first problem is that the range of $\arctan$ is $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$. The second problem is that $x$ could be $0$.
There are many ways to fix these problems. We want the book's definition, $\theta = \arctan(\tfrac{y}{x})$, to be true at least some of the time. So for now let's choose to restrict $\theta$ to the interval $-\frac{\pi}{2} \leq \theta < \tfrac{3\pi}{2}$.
The inverse transformation is then given by
$$
T^{-1}(x,y) = \begin{cases} r = \sqrt{x^2 + y^2}, \\[1ex] \theta = \overline{\arctan} \left(\frac{y}{x}\right), \end{cases}
$$
where
$$
\overline{\arctan} \left(\frac{y}{x}\right) = \begin{cases}
\arctan\left(\frac{y}{x}\right) & \text{if}\ x > 0, \\[0.5 ex]
\arctan\left(\frac{y}{x}\right) + \pi & \text{if}\ x < 0, \\[0.5 ex]
\frac{\pi}{2} & \text{if}\ x = 0\ \text{and}\ y > 0, \\[0.5 ex]
-\frac{\pi}{2} & \text{if}\ x = 0\ \text{and}\ y < 0.
\end{cases}
$$
Notice that $\theta$ is not defined if $x$ and $y$ are both $0$, but this is okay since $r=0$ corresponds to only one point. That is, the pole does not have an angle.
Note: As we said above, this is not the only way to define the inverse transformation. Many people want $0 \leq \theta < 2\pi$. This can easily be arranged by "rotating" the definition of $\overline{\arctan}$, or by using $\text{arccot}$ instead of $\arctan$. Recall, $\text{arccot}(u) = \arctan\left(\tfrac{1}{u}\right)$. If you are a person that wants $\theta$ to be in a more conventional interval, then modifying the definition of $T^{-1}$ is an exercise.
In the video below, we discuss how to compute the Jacobian of the transformation $T$, and what it means.
Recommended Exercise. Compute the Jacobian of $T^{-1}$ assuming that $x > 0$. Before you look at the answer below, ask yourself if your answer makes sense.
Video Lecture
Jacobian of $T^{-1}$
$$
J(T^{-1}) = \left\vert \frac{\partial(r,\theta)}{\partial(x,y)}\right\vert = \frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{r}.
$$
Examples and Exercises¶
Please try the following examples and exercises on your own before watching / reading the solutions. If you need a hint to get started, you should watch the beginning of the video solution (for those that have videos), then pause it and try to finish the example on your own.
1. Example¶
Compute $\displaystyle \int\!\!\!\int_R 3x + 4y^2\, dA$ where $R$ is the region in the $xy$-plane bounded between $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$, with $y \geq 0$.
Check your answer
$$\frac{15\pi}{2}$$
Partial Video Solution
2. Example¶
Find the volume of the solid that is bounded between the plane $z=0$ and the surface $z = 1 - x^2 - y^2$.
Video Solution
3. Example¶
The graph of the polar function $r = \cos(2\theta)$ is sometimes referred to as the "four-leaved rose."
(a) Sketch the graph in the polar plane, paying very close attention to how the graph is traced out as $\theta$ increases.
(b) Write down a double-integral to represent the area of one leaf of the rose.
(c) Finally, compute this integral.
Video Solution to parts (a) and (b)
Answer to part (c)
The area of one leaf of the rose is $\frac{\pi}{8}$.
4. Exercise¶
Generalize Example 3 to find a formula for the area enclosed by any polar curve $r= \rho(\theta)$ for $\alpha \leq \theta \leq \beta$. Start by finding a double integral, then notice that you can actually compute the inner integral.
Answer
The area is given by
$$
\begin{eqnarray*}
\text{Area}(D) & = & \int\!\!\!\int_D 1\, dA \\ & = & \int_\alpha^\beta \int_0^{\rho(\theta)} r\,dr\,d\theta \\[1ex]
& = & \int_\alpha^\beta \Big[ \frac{1}{2}r^2\Big]_{r = 0}^{\rho(\theta)}\, d\theta \\[1ex]
& = & \int_\alpha^\beta \frac{1}{2} \Big[\rho(\theta)\Big]^2\, d\theta.
\end{eqnarray*}
$$
Notice that this is the same formula that you learned in Calculus II.
Discussion¶
Questions? You can ask them here.