This is an exercise from Apostol's Calculus, Volume 1. It asks us to sketch the graph in polar coordinates and find the area of the radial set for the function:
$$f(\theta) = \theta$$
On the interva $0 \leq \theta \leq 2 \pi$. I think to find the area we should just integrate $\theta \ d\theta$ from 0 to $2\pi$ like any other function? Is that right? Also I'm not sure how to think about sketching a function in polar coordinates.
The problem is the book gives the answer as $4\pi^3/3$ which is not what I get if I just integrate the function.
Best Answer
First, to sketch such a graph, you want to consider the distance from the origin as the angle from the $x$-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of $x$ and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a few values of the angle and find the radius, i.e., the distance from the origin at the angle. So, doing that we obtain the following graph:
Then, to calculate the area of the radial set, you must integrate $\frac{1}{2} r^2$, where the radius is the value of the function. So we have, \begin{align*} \text{Area} &= \frac{1}{2}\int_0^{2 \pi} \theta^2 \, d\theta \\ &= \left. \frac{1}{2} \cdot \frac{\theta^3}{3} \right|_0^{2 \pi} \\ &= \frac{(2 \pi)^3}{6}\\ &= \frac{4 \pi^3}{3}. \end{align*}