Question

Find the volume of the solid of revolution if the region bounded by the x-axis, y=(1-lnx)/x^3 and x>e, is revolved around the x-axis

Answer 1

\[\pi \int\limits_{e}^{\infty}(0-\frac{ 1-lnx }{ x^3 })^2\]

Answer 2

is what i came up with

Answer 3

does y cross the x axis any?

Answer 4

looks okay

Answer 5

no but it converges as it goes to infinity

Answer 6

it should converge ...

Answer 7

so \[\pi \int\limits_{e}^{\infty}\frac{ (1-lnx)^2 }{ x^6 }dx \]

Answer 8

ln(x) < sqrt(x) ... if you use this inequality, you find that it converges.

Answer 9

it crosses at x=e and stays negative so yeah

Answer 10

then u=1-lnx du=-1/x dx

Answer 11

\[\pi \int\limits_{-\infty}^{0}\frac{ u^5 }{ e ^{(5-5u)} }\]

Answer 12

am i doing this right?

Answer 13

u^2 on top

Answer 14

according to wolf answer is 2/(125 E^5)
http://www.wolframalpha.com/input/?i=+Integrate%5B%281+-+Log%5Bx%5D%29%5E2%2Fx%5E6%2C+%7Bx%2C+E%2C+Infinity%7D%5D
just use do it by parts.

Answer 15

or .. I think you can make use of Gamma function.

Answer 16

i have not learned that yet

Answer 17

then do it by parts ... 5 times.

Answer 18

or do you mean the tabular method??

Answer 19

The numerator is supposed to be \(\large u^2\), right? So you'll only need to apply integration by parts twice.

Integration by parts says that \(\large\displaystyle \int u\,dv = uv-\int v\,du\)

Answer 20

huh? what tabular method?

Answer 21

u=(1-lnx)^2 du=-2/x (1-lnx) dv=1/x^6 dx v=-5/x^5

Answer 22

No, we meant to apply IBP to \(\large\displaystyle \pi\int_{-\infty}^0 \frac{u^2}{e^{5-5u}}\,du = \frac{\pi}{e^5}\int_{-\infty}^0 u^2e^{5u}\,du\)

Answer 23

It's much easier to work with that instead of the logarithm stuff. :-)

Answer 24

also without by parts
\[ \large\displaystyle \pi\int_{-\infty}^0 \frac{u^2}{e^{5-5u}}\,du = \frac{\pi}{e^5}\int_{-\infty}^0 u^2e^{5u}\,du \\ = \frac{\pi}{e^5 5^3}\int_{0}^\infty (5u)^{3-1}e^{5u}\,d(5u) =\frac{\pi \Gamma(3)}{e^5 5^3} = \frac{2\pi}{e^5 5^3}\]

Answer 25

my answer sheet says the answer is pi/4e^2...