I want to show that the radius of a circle $r_c$, that has an area $A_c$ that is twice the area $A_p$ of a polar diagram, is equal to the quadratic mean of all values $r_p$ of the polar diagram.
I interpreted this as wanting to show: $$ r_c = \sqrt{\frac{1}{2\pi} \int_{0}^{2\pi} r_p^2(\theta) d\theta} $$
I know that I can define the following using polar coordinates: $$ A_p = \frac{1}{2} \int_{0}^{2\pi} r_p^2(\theta) d\theta $$
However, after solving the following, I get a different answer: \begin{align} A_c &= 2A_p \\ \pi r_c^2 &= 2 \frac{1}{2} \int_{0}^{2\pi} r_p^2(\theta) d\theta \\ r_c &= \sqrt{\frac{1}{\pi} \int_{0}^{2\pi} r_p^2(\theta) d\theta} \end{align}
I feel like I must have made a really silly mistake somewhere! This line of reasoning suggests that this would only be possible if the original areas were equal.