Answer
The region bounded by
$x= 0 ,\ \ x=\frac{1}{y^2} ,\ \ y=1,\ \ y=4 $
rotates about $y=-2$
Work Step by Step
Given
$$
2 \pi \int_1^4 \frac{y+2}{y^2} d y
$$
Rewrite the integral as the following
$$
2 \pi \int_1^4 (y+2)\left(\frac{1}{y^2}\right) d y
$$
Compare with
$$V= \int_a^b 2\pi r(y) h(y)dy$$
Here $r(y)= y+2,\ \ h(y)=\frac{1}{y^2}$. This represents the volume of the generated solid when the region bounded by
$$x= 0 ,\ \ x=\frac{1}{y^2} ,\ \ y=1,\ \ y=4 $$
rotate about $y=-2$
