Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 \le \theta \le 2\pi$) and to the left of the $y$-axis.
The area the question wants is between $\theta = \pi/2$ and $\theta = 3\pi/2$ for the graph $r=\theta$. Therefore, I computed the integral $\int_{\pi/2}^{3\pi/2} \theta \,d\theta = \pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?
Best Answer
Recall that the area element in polar coordinates is given by $r\ dr\ d\theta$. For any value of $\theta,r$ ranges from $0\to\theta$. Further, $\theta$ ranges from $\pi/2\to3\pi/2$.
The answer is $$\int_{\pi/2}^{3\pi/2}\int_0^\theta r\ dr\ d\theta=\frac{13\pi^3}{24}$$