Asked by Kait

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

I got the integral from 1 to 2.718 of pi(1)^2-pi(ln(x))^2

Is this correct or is there a trick I'm missing because that's often the case with calculus

Answers

Answered by Steve
what's this 2.718 stuff? That is e. Use it, just like π, without some sloppy decimal approximation.

since the axis is the line y = -3, the radius of each disc is y+3. So,

v = ∫[1,e] π((1+3)^2-(lnx + 3)^2) dx

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