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I don't understand the infinity part on the left side if it isn't enclosed from there
oops sorry i was looking at the wrong part XD
I'm not sure what that is, but that is not even close. Thank you though
so it is ln15-e^(ln15)
so you have the area of the box ln15*1 - pi integral(e^(x/2)^2)dx
oops forgot pi
the integral is just pi*integra(e^x)=pi*e^x
Thank you as well, but no
Your drawing is incorrect as well. I do appreciate the help, but its not helping
forgot the triangle bit thing with the curvy roof
V = pi(e^(ln15) - e^0) - ln15 pi = 14pi - ln15pi = 11.29 pi
YES How did you get that I couldn't make out what yo had up top
mathway.com for graphing
Thta was to welshfella
the diagram you mean?
no the answer its pi(14-ln15)
how did you derive that i forgot to add that its supposed to be the shell method, but I'm not too concerned with that
the shaded area is rotated about x axis so we need the volume of the area down to the x axis minus the area of the cylinder below
I'm not doing very well... i kept thinking it was a square and forgot it had to be revolved into a cylinder...
that gives volume of revolution of the shaded area below |dw:1436656237239:dw|
- then you need to subtract the volume of the cylinder radius 1 and length ln15
ok, thank you
yw as for the 'shell method' i dont know what that is ( i cant remember that in school - mind that was a long time ago)
its 2pi integral (shell height)(shell radius)
ah i see - the method i used was integrating infinitely thin discs
yeah the shell is integrating "coffee cans"