Question

Use the Shell Method to compute the volume V of the solid obtained by rotating the region enclosed by the graphs of the functions
y = x^2, y = 8 − x^2,and x = 1/2
about the y-axis.

Here is how I set up the integral:

2 pi integral sign [-2,2] (1/2 - x)[(8-x^2)- (x^2)]dx


Since x=1/2 is greater than a (a= -2), I subtracted x from 1/2. Is this correct?

Also, since the graph of y= 8- x^2 is above the graph of y=x^2, I subtracted x^2 from 8-x^2. Is this correct?

I got (64 pi)/3, but the answer is wrong.

Answer 1

Why do you integrate over [-2,2]?
The curves intersect at (-2,4) and (2,4) but what does the line x = 1/2 have to do with it? There's a region left of the line, and a region right of the line. Since we are rotating about the y-axis, I think we want the region from x=1/2 to x=2, otherwise the axis of rotation is within the solid being generated.

So, with that in mind, if what I have said is correct, then

v = ∫[1/2,2] 2πrh dx
where r = x and h=(8-x^2)-x^2
v = 2π∫[1/2,2] x(8-2x^2) dx = 225π/16

Now, the way you set up your integral makes me think you are rotating around the line x = 1/2, and not the y-axis. If so, you probably need to integrate over [1/2,2], not [-2,2] and use the factor (x-1/2) not (1/2-x).