area of part of Archimedes's spiral

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 $$\int_{\pi/2}^{3\pi/2}\int_0^\theta r\ dr\ d\theta=\frac{13\pi^3}{24}$$

The integral for finding the area in polar coordinate is different from what you have.

Please use the correct formula and you will get the correct answer.

$$ A = (1/2) \int r^2 d\theta $$ where in your case $ r=\theta $