Question

Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.

Answer 1

The volume is just a stack of discs, each of radius y and thickness x, so

v = ∫[a,b] πr^2 dx

y(1) = 0
y(e) = 1

so,

v = ∫[1,e] πr^2 dx

so, what's r? If we were revolving around y=0, r would just be y. But the axis is 3 units farther away than that, and the area between y=0 and y=-3 is empty, so we really have washers with holes in them, so

v = ∫[1,e] π(R^2-r^2) dx

where R=y+3 and r=3.

v = ∫[1,e] π((ln(x)+3)^2-3^2) dx
= π∫[1,e] ln(x)^2 + 6 ln(x) dx