Calculus

Consider the solid obtained by rotating the region bounded by the following curves about the line x=1.
y=x,y=0,x=4,x=6
Find the volume
So it would be
pi (integral from 3 to 6) of ((1-y)^2 -(1-0)^2) right?
so then you integrate it and get
pi(Y^3/3-y^2) from 3 to 6.
?

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  1. If you are going to integrate over y, the solid has two parts: a plain old cylinder of height 4 and thickness 2, and a variable-thickness shape of height 2.

    So,

    v = π(5^2-3^2)(4) + ∫[4,6] π(R^2-r^2) dy
    where R=5 and r=x-1=y-1
    v = 64π + π∫[4,6] 25-(y-1)^2 dy
    = 64π + π∫[4,6] -y^2+2y+24 dy
    = 64π + 52/3 π
    = 244/3 π

    I think shells are easier in this case. SO, since each shell has thickness dx, we have

    v = ∫[4,6] 2πrh dx
    where r = x-1 and h = y = x
    v = 2π∫[4,6] (x-1)(x) dx
    = 2π∫[4,6] x^2-x dx
    = 244/3 π

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    posted by Steve

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