How do i set bounds for the washer method with y=x^2 and y=2x around the x-axis?

[Robert_J]

I’m working through a problem where I need to find the volume of a solid formed by rotating the region between y = x² and y = 2x around the x-axis. I set up my washer method integral with outer radius 2x and inner radius x², but when I evaluate it from 0 to 2, my answer doesn’t match the solution in the book. I think my understanding of the bounds might be off, or maybe I’m misidentifying the radii.

[Nora_M]

Your bounds and setup look correct. The curves meet at x=0 and x=2, so on [0,2] the outer radius is 2x and the inner radius is x^2. The volume is pi integral from 0 to 2 of [(2x)^2 - (x^2)^2] dx = pi ∫0^2 (4x^2 - x^4) dx = pi[(4/3)x^3 - (1/5)x^5] from 0 to 2 = pi(32/3 - 32/5) = (64/15) pi ≈ 13.38. If your book has a different number, the most common slip is arithmetic in evaluating the antiderivative.

[Mia32]

I ran into a similar pitfall once. I forgot that the inner radius is x^2, so its square is x^4. If I accidentally used (x^2)^1 or forgot the squares altogether, the integrand looked like 4x^2 - x^2 which is off. Small algebra mistakes can throw the result off.

[Scarlett.L]

Another thing I check is the sign of the radii. It’s all positive here, so we’re fine. If you somehow treated the inner radius as 2x and outer as x^2, you’d flip things and end up with a totally different value. It helps to plug in a test x like 1 and see which radius is bigger then compare radii squares.

[Justin54]

One quick check: did you confirm the intersection points are 0 and 2? If you used a different interval, the answer will be off. Also, if the book uses a different method, the numeric might look different but should match the exact value (64 pi / 15) if done correctly.