Question

Solve the problem.

Find the volume of the solid generated by revolving the region bounded by the curve y=lnx, the x-axis, and the vertical line x=e^(2) about the x-axis.

Answer 1

One of the integrals you should have in your repertoire of common integrals is
∫lnx = xlnx - x

so volume
= π∫ lnx dx from x-1 to e^2
= xln - x | from 1 to e^2
= e^2(lne^2) - e^2 - (1ln1 - 1)
= e^2(2) - e^2 - 0 + 1
= e^2 + 1

Answer 2

I have these ans. chocies:

pi(e-1)

2pi(e^(2)-1)

pi(e^(2)-1)

pi e

Is it pi(e^(2)-1)

Answer 3

oops, sorry about dropping that π in the 4th last line

but, I don't see why it should not be
π(e^2 + 1)

check my arithmetic.

Answer 4

First, it's <b>pi</b>, not <b>pie</b>! Grrr! Does no math teacher ever teach that any more?

You can always check your answer using shells, instead of discs. As youi recall, the volume of a thin shell is essentially just the circumference times the height. So, since

y = lnx,
x = e^y
dx = e^y dy
and we have

v = ∫[0,2] 2πrh dy
v = 2π∫[0,2] y(e^2 - e^y) dy
= 2π(e^2/2 y^2 - ye^y + e^y) [0,2]
= 2π(e^2 - 1)

So, where did discs go wrong?

v = ∫[1,e^2] πr^2 dx
= π∫[1,e^2] (lnx)^2 dx
πx(ln^2(x)-2lnx+2) [1,e^2]
= πe^2(4-4+2) - π(0-0+2)
= 2π(e^2 - 1)