# Thread: Shell Method Setup

1. ## Shell Method Setup

I am having trouble setting up the 3rd equation in this question. It asks me to use the shell method to find the volume of the region bounded by y=x^2 and y=x+2 revolved around

x=2
x=-1
x axis How?
y=4

I know i have the limits of integration correct but i cant figure out what the correct radius or shell height are. Any guidance is appreciated.

2. ## Re: Shell Method Setup

Originally Posted by petenice
I am having trouble setting up the 3rd equation in this question. It asks me to use the shell method to find the volume of the region bounded by y=x^2 and y=x+2 revolved around
I presume you have drawn a graph? y= x^2 and y= x+ 2 intersect at (-1, 1) and (2, 4).

x=2
Using the shell method, you would rotate a vertical line, at a given "x" around x= 2 so its radius would be 2- x. The shell height will be $x+2- x^2$.

x=-1
This is on the left side so the radius is x- (-1)= x+ 1.

x axis How?
Now you are rotating around a horizontal axis (y= 0) so your variable will be y, not x. The radius will be y. As for the length of the "shell", you will need to do it in two parts. from y= 0 up to y= 1, it will be $\sqrt{x}- (-\sqrt{x})= 2\sqrt{x}$ and from y= 1 up to y= 4, it will be [tex]\sqrt{y}- (y- 2)= \sqrt{y}- y+ 2[tex].

y=4
Again, this is a horizontal axis so your variable will be y. The radius will by 4- y.

I know i have the limits of integration correct but i cant figure out what the correct radius or shell height are. Any guidance is appreciated.

3. ## Re: Shell Method Setup

Originally Posted by HallsofIvy
Now you are rotating around a horizontal axis (y= 0) so your variable will be y, not x. The radius will be y. As for the length of the "shell", you will need to do it in two parts. from y= 0 up to y= 1, it will be $\sqrt{x}- (-\sqrt{x})= 2\sqrt{x}$ and from y= 1 up to y= 4, it will be $\sqrt{y}- (y- 2)= \sqrt{y}- y+ 2$.

Im assuming thats supposed to be $\sqrt{y}- (-\sqrt{y})= 2\sqrt{y}$ correct? I think i got it. Took me a while to understand where you got that $2\sqrt{y}$ but i understand now! Thanks so much.