What am I doing wrong in this solid of revolution problem?

Question

I have to calculate the volume of the solid of revolution that comes out after rotating the curves $y=\sqrt{x-1}$, $x=5$, $y=0$ around the axis $x=5$. What I did was: $$V=\int_{0}^{2}\pi(5-(y^2+1))^2dy=\frac{496\pi}{15}$$ The answer should be $\dfrac{256\pi}{15}$

Answer 1

After translating picture such that $x=5$ becomes $y$-axis, you get area bounded by $y=\sqrt{x+4}$, $x=0$, $y=0$. Now we express $x=y^2-4$, and the result is $\int_0^2\pi(y^2-4)^2dy$.