I found the volume of the solid by doing a double integration, however, it was deemed wrong and I am unsure as to why.
The problem statement is to find the volume that lies under the double cone $ z^2=4x^2 + 4y^2 $ inside the cylinder $ x^2 + y^2 = x $ and above the plane $ z=0 $.
To setup my region, $R$, I achieved $$R = \Big\{ (r,\theta) \mid 0\le r \le cos(\theta), 0 \le \theta \le \frac{\pi}{2} \Big\},$$
To set up my integral, I found $z=2r$ by changing $x = rcos(\theta)$ and $y = rsin(\theta)$ and simplifying which achieved: $$ \int_{0}^{\frac{\pi}{2}} \int_{0}^{cos(\theta)} (2r)rdrd\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{cos(\theta)} 2r^2drd\theta $$
From here, I simplified the first integral which resulted in $$ \int_{0}^{\frac{\pi}{2}} \frac{2cos^3(\theta)}{3}d\theta = \frac{2}{3} \int_{0}^{\frac{\pi}{2}} (1-sin^2(\theta))cos\theta d\theta$$
I then used u-sub with $u=sin\theta$, and I achieved my final answer to be $\frac{4}{9}$.
The only mistake I feel like I may have made is the bounds of the radius, $\theta$ is definitely in range $[0,\frac{\pi}{2}]$ because it is above the $z$-plane so it is only the top half of the cone.