Find the volume of the area in the first quadrant enclosed by x=ln(3), y=1, and y=e^(2/3) when it is rotated around the line y=-1.
I assume you meant y = e^(2/3 x), otherwise this is just a cylinder.
If so, then the curve intersects x = ln3 at (ln3, 3^(2/3)
Using discs (washers) of thickness dx, we have
v = ∫[0,ln3] π(R^2-r^2) dx
where R = 3^(2/3) + 1 and r = 1+2=2
v = ∫[0,ln3] π((e^(2/3 x)+1)^2-2^2) dx
Using cylinders of thickness dy, you get
∫[1,3^(2/3)] 2πrh dy
where r=y+1 and h=ln3 - 3/2 lny
∫[1,3^(2/3)] 2π(y+1)(ln3 - 3/2 lny) dy