Why is my washer method volume negative for y=x^2 and y=√x?

[AriaNM]

I’m working through a calculus problem about finding the volume of a solid of revolution using the washer method, and I’ve hit a wall. My integral is set up for the region bounded by y = x² and y = √x rotated around the x-axis, but when I evaluate it, I get a negative result for the volume, which obviously can’t be right. I think my error is in determining the correct limits of integration or perhaps misidentifying the outer and inner radii for the washers, but I’m not seeing it.

[GeorgeJ]

You're on the right track. The region bounded by y=x^2 and y=√x intersects at (0,0) and (1,1), so the x-limits are 0 to 1. When you revolve about the x axis the outer radius is the top curve y=√x and the inner radius is the bottom curve y=x^2. So R = √x and r = x^2. The volume is pi times the integral from 0 to 1 of (R^2 − r^2) dx = pi ∫0^1 ((√x)^2 − (x^2)^2) dx = pi ∫0^1 (x − x^4) dx. That evaluates to pi(1/2 − 1/5) = 3π/10. So that should be the positive value.

[Edward.L]

I did this once and I think I swapped the radii by mistake, which produced a negative. Make sure over [0,1] the top function really is √x. It sits above x^2 there, so the outer radius is √x and not the other way around.

[JasonDA]

Could the issue be the limits? The two curves meet only at x=0 and x=1, so the correct x-range is 0 to 1. If you chose a different interval you’d be integrating over a region that isn’t the one you described, which can give a wrong sign.

[KennethVA]

I tried a check with the y-version using shells. For horizontal slices, rotate around the x-axis, radius is y and height is the difference between the x-values from the curves, which are x_right = sqrt(y) and x_left = y^2, so height = sqrt(y) − y^2. Integrating from y=0 to 1 gives volume 2π ∫0^1 y (sqrt(y) − y^2) dy = 3π/10. It matches, and it helped me see the limits and which curve is on the outside.