Asked by Nathan

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:

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.

Answers

Answered by Steve
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.
Answered by Andrew
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?
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