I am trying to find the correct answer for the volume of solid of revolution generated by revolving the region bounded by the following curves around the y-axis: $$ y = \sqrt{25-x^{2}}; x = 3; y = 0$$
My solution: when $x = 3$, $y =4$. I made the first equation as a function with respect to $y$ so that it is produced as $ x = \sqrt{25-y^{2}}$. I then used the equation: $$\pi \int_{0}^{4}\left ( \sqrt{25-x^{2}} \right )^{2}dy$$ After working carefully and checking my arithmetic, I got an answer of $\frac{236\pi}{3}$. When I check the memo, it says the correct answer is $\frac{128\pi}{3}$.
Can someone please help me find where I erred, I have a feeling I made a mistake before the integral step.
Since every point of the region satisfies $x\ge 3$, you need to have$$\pi\int_{0}^{4}\left(\left(\sqrt{25-y^2}\right)^2\color{red}{-3^2}\right)dy=\frac{128}{3}\pi.$$