Spiral of Archimedes area and sketch in polar coordinates 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.
 A: Sketching this in polar coordinates is pretty straightforward.  Draw your axes and you know that the radial value is equal to the angle, so your curve would start at the origin when $\theta$ is zero, it would be $\frac{\pi}{2}$ at 90 degrees, etc.
As for calculating the area, you are correct that you integrate it, but I'm not sure what it means physically because the curve does not close on itself since $f(0) \neq f(2\pi)$ (perhaps this is the insight they are trying to get from you).
A: 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*}
A: commenting on Daves's answer "but I'm not sure what it means physically because the curve does not close on itself since f(0)≠f(2π)f(0)≠f(2π)"
The area calculated is the area bounded by the curve and the terminal vector. ie the curve is subtended by the sweeping vector, which in this case is at 2pi- the positive x axis. . 
