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
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
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