Calc 2

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.

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  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.

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  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?

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