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April 19, 2014

April 19, 2014

Posted by **bobby** on Friday, September 21, 2012 at 8:33pm.

y=e^(5x) , \ y=0, \ x=0, \ x=0.8

around the y-axis

Please help, i have been attempting these problems for a couple of days

- calculus 2 -
**Reiny**, Friday, September 21, 2012 at 11:44pmthe y-intercept of y = e^(5x) is (0,1)

and y = e^(5x) intersects x = 0.8 at (.8 , 54.598)

so I made a sketch , and I am looking at a region made up of a rectangle topped by a another region resembling a triangle.

the volume of the cylinder with radius .8 and height 54.598 is

V = π(.8)^2 ( 54.598) = appr. 109.776

we now have to hollow-out the part which is not part of the solid, and we have to solve y = e^(5x) for x

y = e^(5x)

ln y = 5x

x = (1/5)lny

so the volume of the triangular - looking region

= π∫(1/25) (lny)^2 dy from 1 to 54.598

At this point I must admit that I "cheated" and used Wolfram to find the integral

http://integrals.wolfram.com/index.jsp?expr=%28ln%28x%29%29%5E2&random=false

= (1/25)π [ y(ln y)^2 - 2y ln y + 2y ] from 1 to 54.598

= ...

I got appr. 68.358

so that would give us a volume of 109.776 - 68.358 = 41.42

You better check my arithmetic, I would suggest you differentiate my integral to make sure you get (1/25) (ln y)^2

(just do the y(ln y)^2 - 2y ln y + 2y part)

- calculus 2 -
**Steve**, Saturday, September 22, 2012 at 12:29amor, if you like to use shells, you have

v = integral [0,0.8] 2πrh dx

where r = x h = y = e^(5x)

v = 2π integral[0,0.8] xe^)5x) dx

= 1/25 e^(5x) (5x-1)

= 41.417

Hmmm. using washers, I get

v = π*.64*1 + integral[1,e^4] π (R^2-r^2) dy

where R = .8 and r = 1/5 ln y

v = .64π + π integral[1,e^4] (.64 - 1/25 ln^2(y)) dy

= .64π + 12.544π = 13.184π = 41.418

Agrees with shells, and Reiny. :-)

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