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
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.
4 answers
I have these ans. chocies:
pi(e-1)
2pi(e^(2)-1)
pi(e^(2)-1)
pi e
Is it pi(e^(2)-1)
pi(e-1)
2pi(e^(2)-1)
pi(e^(2)-1)
pi e
Is it pi(e^(2)-1)
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.
but, I don't see why it should not be
π(e^2 + 1)
check my arithmetic.
First, it's pi, not pie! 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)
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)
4 answers