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Calculus

Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.

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Michael

  1. The region is a triangular patch with vertices at (1,0), (1,1), and (1,e)
    So the volume is a stack of washers of thickness dx:
    v = ∫[1,e] π(R^2-r^2) dx
    where R=(1+3) and r=(lnx+3)
    v = ∫[1,e] π(4^2-(lnx+3)^2) dx

    Or, you can set it up as a stack of nested shells of thickness dy
    v = ∫[0,1] 2πrh dy
    where r=y+3 and h=x-1=e^y-1
    v = ∫[0,1] 2π(y+3)(e^y-1) dy

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    oobleck

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