# Math Help - Integration with cylindrical shells

1. ## Integration with cylindrical shells

Consider the given curves to do the following.
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = -4.

I'm getting 302pi/60
I set it up as 2pi*(the integral from 0 to 1) of (y-4)((y^2)-(y^.5))dy

2. Originally Posted by Hellacious D
Consider the given curves to do the following.
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = -4.

I'm getting 302pi/60
I set it up as 2pi*(the integral from 0 to 1) of (y-4)((y^2)-(y^.5))dy
$V = 2\pi \int_0^1 (y+4)(\sqrt{y} - y^2) \, dy$

confirmed w/ washers ...

$V = \pi \int_0^1 (\sqrt{x}+4)^2 - (x^2+4)^2 \, dx
$

3. Thank you, but I need to be able to do this with shells.

4. Originally Posted by Hellacious D
Thank you, but I need to be able to do this with shells.
I did it two ways ... the first integral is done with shells.

5. $V = 2\pi \int_0^1 (y+4)(\sqrt{y} - y^2) \, dy$
Thanks, but can you explain how you knew to add 4 and subtract y^2? For some reason, I can't wrap my head around this method when I'm rotating about axes other than x or y.

6. Originally Posted by Hellacious D
$V = 2\pi \int_0^1 (y+4)(\sqrt{y} - y^2) \, dy$
Thanks, but can you explain how you knew to add 4 and subtract y^2? For some reason, I can't wrap my head around this method when I'm rotating about axes other than x or y.
sketch a representative horizontal rectangle within the given region.

the right end of the rectangle is $x = \sqrt{y}$

the left end of the rectangle is $x = y^2$

the height of the rectangle is (right - left) = $\sqrt{y} - y^2$ and the thickness of the rectangle is $dy$

the rectangle is a distance of $y$ above the x-axis

the vertical distance from the rectangle to the line $y = -4$ is $y - (-4) = y+4$ , so the radius of revolution is $y+4$

the volume of a single shell is ...

dV = $2\pi$(radius of revolution)(height of rectangle)(rectangle thickness)

$dV = 2\pi (y+4)(\sqrt{y} - y^2) \, dy$