# Thread: Shell method, can someone explain this problem?

1. ## Shell method, can someone explain this problem?

I know how to do very basic and simple shell method problems, but this one is difficult for me. Every problem that I have had to do so far, the graphs always stay on one side of the x and y axis. Meaning they never cross it. This one does, however, and I don't understand how to work with that. So the information given is:
x= y^4/4 -y^2/2

x=y^2/2

There is a graph of this shown, and the graphs intersect at (0,0) and (2,2).
But the graph of y^4/4 - y^2/2 goes over to the left side of the y axis for a bit, and then goes back to the right. I need to find the volume of the region that would be formed by rotating the graph about the x axis. So how do I find the radius and the height for this? Maybe I'm making it more complicated than it is...

2. Hi

I just want to be sure to have correctly understood
You want to evaluate the volume of the solid bounded by the two graphs and rotating around x-axis right ?

3. Yeah, that's what I meant. Sorry I'm a little slow on the reply. So yeah, it's being rotated about the x axis. But in order to use the shell method, I need to know the radius and the height, right? The fact that the graph crosses over to the left side of the y axis is making me unsure of what the radius and height would be.

4. First you need to define the integration boundaries in x. You can show that it goes from x=-1/4 to x=2

Then you need to separate the calculation into 2 parts :
- from x=-1/4 to x=0 the curve is bounded only by x= y^4/4 -y^2/2. This means that you need to inverse this expression in order to get y function of x.
You will get 2 functions : one (f(x)) for the lower part and one (g(x)) for the upper part
Due to the fact that the rotation is around x axis, the radius is y and the height is x
The volume is
$\displaystyle V_1 = \int_{-\frac{1}{4}}^{0} \:\int_{0}^{2\pi}\:\int_{f(x)}^{g(x)} y\:dy\:d{\theta}\:dx$

- from x=0 to x=2 the curve is bounded in the lower part by x=y^2/2 and in the upper part by x= y^4/4 -y^2/2. You also need to inverse these expressions to calculate the volume V2

5. $\displaystyle x_1 = \frac{y^2}{2}$

$\displaystyle x_2 = \frac{y^4}{4} - \frac{y^2}{2}$

$\displaystyle V = 2\pi \int_0^2 y(x_1 - x_2) \, dy$