# shell method #2

• Oct 24th 2009, 10:39 PM
abilitiesz
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?
• Oct 25th 2009, 06:03 AM
Scott H
First, we note that $y=x^2$ defines a parabola facing up, while $x=y^2$ defines a parabola facing right. These two curves intersect at $(0,0)$ and $(1,1)$, defining a small region between these two points.

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

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

Or, alternatively, we may use the Shell Method:

$\int_0^1 2\pi(x+1)(\sqrt{x}-x^2)\,dx.\;\;\;\;\;\;\;\;\;\;(\mbox{formula used: }2\pi rh\,dx)$
• Oct 25th 2009, 06:05 AM
tom@ballooncalculus
(Scrubbed)

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