Calc II problem - finding volume of a solid revolving about the line x= -1?

Question

Hi, I'm trying to find the volume of the solid generated by revolving the area bound between the curves y= x^2, x=0, and y = x+2 (in the first quadrant only) about the line
x= -1.

I have difficulty determining the radius when the axis of rotation is something other than the x- or y-axis - it always flips me out.

Q: What would be the radius for this problem & how did you determine it?

I decided to use the Shell method
(V = 2(pi)rhw) for this problem and I THINK the height is (x+2 - x^2) with bounds from 0 to -1.

Please let me know how radius was determined & if I'm basically on the right track! Thanks!!!! :>)

Thanks, everyone! I think I got it now. Special thanks to Northsta. - this helps me a lot :>)

Answer 1

Let's use the shell method.

First let's find out where the two curves meet.

x² = x + 2
x² - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, 2

But since we are limiting it to the part of the curve in the first quadrant our limits of integration are [0, 2].

Since we are revolving around x = -1, the radius is:
r = x - (-1) = x + 1

The height is:
h = (x + 2) - x² = 2 + x - x²

V = ∫(2πrh) dx = 2π∫[(x + 1)(2 + x - x²)] dx

= 2π∫(2 + 3x - x³) dx

= 2π[2x + (3/2)x² - x^4/4] | [Evaluated from 0 to 2]

= 2π[4 + 6 - 4] = 2π(6) = 12π

Answer 2

Yea go with annies way

In order to find the area of the region you need to look at it with respect to y.

Your radius would then be y
Your height would be the difference of the two equations annie gave you (remember upper minus lower)
Your bounds would be from 0 to 1