What is the area bounded by the curve $r^2 + \theta^2 = 1$?

Question

What is the area bounded by the curve $r^2 + \theta^2 = 1$? (given in polar coordinates)

My approach was to calculate the definite integral: $$\frac12 \int_0^1 (1-\theta^2) \, d\theta$$

Integration limits are $0$ to $1$ because this is the domain of $r$. The final answer I got is $\frac13$ while the answer in the book is $\frac23$.

Shouldn't I multiply the integral by $\frac12$ as written above? Is there something else wrong in my way?

Answer 1

The reason is that the range of $\theta$ is not $[0,1]$ but $[-1,1].$

Answer 2

There is a symmetry you are failing to exploit. If $(r,\theta)$ is on the graph so is $(-r,-\theta)$ so that your formula only counts half the area.