Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y = x + 2 and the parabola y = x^2 about the following lines:

Question

a) The line x=2
b) The line x=-1
c) The x axis
d) The line y=26

I know what the formulas are, but I can't seem to apply them in this question. Please help me.

Answer 1

the graphs intersect at (-1,1) and (2,4)

Recall that the volume of a shell of thickness dx is

v = 2πrh dx

So, to rotate around the line x=2,

v = ∫[-1,2] 2πrh dx
where r = 2-x and h=(x+2)-x^2
v = ∫[-1,2] 2π(2-x)(x+2-x^2) dx = 27π/2

Just to verify, let's do it with discs (washers) as well. This one is bit more complicated, since the left boundary changes from the parabola to the line.

v = ∫π(R^2-r^2) dy
we have to break it up into two intervals.
#1: v = ∫[0,1] π(R^2-r^2) dy
where R=2+√y, r=2-√y
v = ∫[0,1] π((2+√y)^2-(2-√y)^2) dy = 16π/3

#2:
v = ∫[1,4] π(R^2-r^2) dy
where R=2-(y-2), r=2-√y
v = ∫[1,4] π((2-(y-2))^2-(2-√y)^2) dy = 49π/6

16π/3 + 49π/6 = 81π/6 = 27π/2
Looks like the shells were ok

Now do the others similarly.

If you type your integrals as I did, you can enter them in at wolframalpha.com and it will evaluate them, as well as show the indefinite integrals involved.

Answer 2

I'm trying to do this with just the x-axis but I keep getting the wrong answer to it. What am I supposed to do?