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find the volume of the solid generated by revolving around the x-axis: y=e^(x-8), y=0, x=0, x=10

So I know you can use the washer method, but the shell method can also be used and it should be better. The issue is I'm not certain how to actually do it. Any help is appreciated!

  • calc -


    v = ∫[0,10] πr^2 dx
    where r = y = e^(x-8)
    v = π∫[0,10] (e^(x-8))^2 dx
    = π∫[0,10] e^(2x-16) dx
    = π/2 e^(2x-16) [0,10]
    = π/2 (e^4 - 1/e^16)


    y = e^(x-8), so x = 8+lny

    v = ∫[e^-8,e^2] 2πrh dy
    where r = y and h = 10-x = 2-lny
    v = 2π∫[e^-8,e^2] y(2-lny) dy
    = π/2 (y^2(5-2lny)) [e^-8,e^2]
    = π/2 (e^4 - 21/e^16)

    Hmmm. better double-check my math. Anyway, that's the method.

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