I was teaching today about the Washer Method vs the Shell Method and I got stuck on this volumen question:
Find the volume of the solid obtained by rotating the region bounded by $y=x-x^2$ and $y=0 $ about the line y=2.
So I managed to set up the washer method as: $\pi \displaystyle \int_{0}^{1}(2^2)-(2-(x-x^2))^2dx$ and the shell method as follows: $2\pi \displaystyle \int_{0}^{1/4}(2-y)(1/2-\sqrt{y+1/4})dy$ but both give me different answers. I got stumped and couldn't figure it out, what exactly went wrong?
Using the shell method, solving $y=x-x^2$ for $x$ gives $x = \frac{1}{2}(1\pm\sqrt{1-4y}\,)$, so that for a given value of $y$, the height of the shell is $$\frac{1}{2}(1+\sqrt{1-4y}\,)-\frac{1}{2}(1-\sqrt{1-4y}\,) = \sqrt{1-4y}.$$ Thus the integral should be $$2\pi\int_0^{1/4} (2-y)\sqrt{1-4y}\,dy,$$ which gives the same answer as you got using the washer method.