The problem: Find the volume of a solid formed by y=$\sqrt{x}$ and $y=x$ around $x=6$.
I am fairly sure this is right but wanted to receive verification just for peace of mind.
The sketch of region:
Setup (using shell method): $ V= 2\pi[\int_{0}^1 (6-x)(\sqrt(x)-x)dx + \int_{1}^6 (6-x)(x-\sqrt(x)dx] $ From there, I factored out a negative from the second integrand to match the first one and then my answer came out to be approximately $$2\pi(14.35) \approx 90.14 $$ Is this correct?
On another point, I kept on trying to do the washer method but came out with a negative value for a volume which makes no sense in this context. I know it splits up into two integrals but I cannot figure out where I went wrong. Your assistance is appreciated!
I think that you should consider, as the rotating section, the bounded planar region given by $$\{(x,y): x\leq y\leq \sqrt{x}\}.$$ Hence the volume should be (shell method and washer method): $$V=2\pi \int_{x=0}^1(6-x)(\sqrt{x}-x)dx=\pi\int_{y=0}^1 ((6-y^2)^2-(6-y)^2)dy=\frac{28\pi}{15}.$$