Answer
The generated solid when the region bounded by
$x= y ,\ \ x=\sqrt{y-1} ,\ \ y=1,\ \ y=5 $
rotates about $x-$axis
Work Step by Step
Given
$$
\int_1^5 2 \pi y \sqrt{y-1} d y
$$
Compare with
$$V= \int_a^b 2\pi r(y) h(y)dy$$
Here $r(y)= y,\ \ h(y)= \sqrt{y-1}$. This represents the volume of the generated solid when the region bounded by
$$x= y ,\ \ x=\sqrt{y-1} ,\ \ y=1,\ \ y=5 $$
rotates about $x-$axis.
