Find the volume of the solid of revolution obtained by rotating the region

That is bounded by y=x^3 and y=x^2 about the x-axis?

I got pi/7, but that doesn't seem correct. I can't figure out what I'm doing wrong

Might as well use the washer method. The axis of rotation is horizontal, so slice vertically, implying dx.

V = ∫ A(x) dx from 0 to 1

A(x) = πR^2 - πr^2
A(x) = π(R^2 - r^2)

First find R, the outer radius of the washer. The outer radius is defined by x^2, so R = x^2. The inner radius, r, is defined by x^3, so r = x^3.

A(x) = π((x^2)^2 - (x^3)^2)
A(x) = π(x^4 - x^6)

V = π ∫ (x^4 - x^6) dx from 0 to 1
V = π[(1/5)x^5 - (1/7)x^7] from 0 to 1
V = π[(1/5)(1)^5 - (1/7)(1)^7] <--- Note that the lower limit will be 0.
V = 2π/35

Done!