I am going to do this in a bit of a tricky way.
I can foresee having to integrate (ln(x))^2 which could be rather messy
so let's take the inverse of the whole mess
If we had:
Give the volume of the solid generated by revolving the region bounded by the graph of y=e^x, the y-axis, the lines y=1 and y=e, about the x-axis,
we would have exactly the same solid.
Volume = pi[integral](e^2 - (e^x)^2)dx from x=0 to x=1
= pi[(e^2)x - (1/2)e^(2x)] from 0 to 1
= pi[e^2 - (1/2)e^2 - (0 - (1/2)e^0))]
= (1/2)_e^2 - 1/2 or (e^2 - 1)/2
check my arithmetic.
Give the volume of the solid generated by revolving the region bounded by the graph of y=ln(x), the x-axis, the lines x=1 and x=e, about the y-axis
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