Find the volume generated by revolving the region shown below about $x=3$ using Washer method.
We have outer radius as a function of $y$ as: $$R(y)=3$$ Inner radius as a function of $y$ as: $$r(y)=3-\sqrt{y}$$ So the required volume is: $$\int_{0}^{9}\pi (9-(3-\sqrt{y})^2)dy=\frac{135\pi}{2}$$ But the answer is $\frac{572\pi}{2}$. Where i went wrong?