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 = -1.

Question

I don't understand the washer thickness topic which is why im having trouble with this question

Answer 1

A circle of radius R has area πR^2

so, a disc of thickness dx and radius R has a volume

v = πR^2 dx

Now, if it has a hole in it of radius r, then the hole has volume

v = πr^2 dx

So, the washer (a disc with hole) has volume

v = π(R^2-r^2) dx

Sketch the region. It is a curvy triangular area, which is to be revolved around the line y = -1, so

v = ∫[1,e] π(R^2-r^2) dx
where R=2 and r=y+1=ln(x)+1
v = ∫[1,e] π(4-(lnx)^2) dx

You can also consider the volume to be a set of nested cylinders (shells). If you take a sheet of material of length c and width h, and thickness dy, its volume is ch dy. Now, roll it up so that its length is the circumference (c=2πr), and you have a cylinder of volume

v = 2πrh dy

Now you can see that the volume of the rotated region is

v = ∫[0,1] 2πrh dy
where r=1+y and h=x-1 = e^y - 1

v = ∫[0,1] 2π(y+1)(e^y-1) dy

Answer 2

Oops the volume is

v = ∫[1,e] π(4-(lnx+1)^2) dx