1. ## shell method #2

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

y=x^2
x=y^2

I'm having trouble deciphering on what to do. When I tried to solve it I got 3pi/10. But, I checked it and the answer was incorrect. Can someone show me the steps on how to setup the problem?

2. First, we note that $\displaystyle y=x^2$ defines a parabola facing up, while $\displaystyle x=y^2$ defines a parabola facing right. These two curves intersect at $\displaystyle (0,0)$ and $\displaystyle (1,1)$, defining a small region between these two points.

To find the volume of revolution about the line $\displaystyle x=-1$, we may use the Disc Method:

$\displaystyle \int_0^1(\pi(\sqrt{y}+1)^2-\pi(y^2+1)^2)\,dy.\;\;\;\;\;\;\;\;\;\;(\mbox{formu la used: }(\pi {r_2}^2-\pi {r_1}^2)\,dy)$

Or, alternatively, we may use the Shell Method:

$\displaystyle \int_0^1 2\pi(x+1)(\sqrt{x}-x^2)\,dx.\;\;\;\;\;\;\;\;\;\;(\mbox{formula used: }2\pi rh\,dx)$

3. (Scrubbed)

Sorry, yes, rotation unnecessary... I read x = -1 as y = -x!