PROBLEM:
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5. (Use disk method)
$$ xy = 3, y = 1, y = 4, x = 5 $$
So first I graphed this out, and came up with a region, like so (represented by the white region):
The formula using for disk method is of the form:
$$ \pi \int (r(x/y))^2*(dx/y) $$
In disk method, when rotating around a vertical axis, the differential of dy is used.
Setting this integral up, our limits are y = 1, to y = 4, as given in the problem statement.
My main question is how do we represent the r in the disk method equation, when the axis is meant to be around x = 5, and not the y-axis.
The best I could think of would be:
$$ \pi \int_{1}^{4} (\frac {y} {3} + 5)^{2} dy $$
However, something tells me this isn't the right approach, because I'm not seeing the logic behind whether or not I should add, subtract, or subtract from the x = 5.
Thank you.
Useful picture! The region being rotated is the "white" region, and it is being rotated about its right boundary line.
Imagine taking an up and down slice of the solid of revolution, perpendicular to the $y$-axis, at height $y$ above the $x$-axis. The cross-section is a disk whose radius is the horizontal distance from the curve $xy=3$ to the line $x=5$.
The radius of the slice at height $y$ is therefore $5-x$, that is, $5-\frac{3}{y}$. The area of cross-section is therefore $\pi\left(5-\frac{3}{y}\right)^2$. Thus the volume of the solid is $$\int_1^4 \pi\left(5-\frac{3}{y}\right)^2\,dy.$$ To finish, expand the square and integrate term by term.