Find volume of solid generated

- anonymous

Find volume of solid generated

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- anonymous

##### 1 Attachment

- welshfella

|dw:1436654820728:dw|

- anonymous

I don't understand the infinity part on the left side if it isn't enclosed from there

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- anonymous

oops sorry i was looking at the wrong part XD

- welshfella

|dw:1436654961888:dw|

- anonymous

I'm not sure what that is, but that is not even close. Thank you though

- anonymous

so it is ln15-e^(ln15)

- anonymous

so you have the area of the box ln15*1 - pi integral(e^(x/2)^2)dx

- anonymous

oops forgot pi

- anonymous

the integral is just pi*integra(e^x)=pi*e^x

- anonymous

Thank you as well, but no

- anonymous

|dw:1436655396490:dw|

- anonymous

Your drawing is incorrect as well. I do appreciate the help, but its not helping

- anonymous

whoops...

- anonymous

forgot the triangle bit thing with the curvy roof

- anonymous

No, it is bounded bounded by 1 you have it bounded by y=0

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- anonymous

##### 1 Attachment

- welshfella

V = pi(e^(ln15) - e^0) - ln15 pi
= 14pi - ln15pi
= 11.29 pi

- anonymous

\[V=\ln15-\int\limits_{0}^{\ln15}y^2dx\]

- anonymous

YES How did you get that I couldn't make out what yo had up top

- anonymous

mathway.com for graphing

- anonymous

Thta was to welshfella

- welshfella

the diagram you mean?

- anonymous

no the answer its pi(14-ln15)

- welshfella

yes

- anonymous

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

- welshfella

|dw:1436655815791:dw|

- welshfella

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

- anonymous

I'm not doing very well... i kept thinking it was a square and forgot it had to be revolved into a cylinder...

- welshfella

|dw:1436655967976:dw|

- welshfella

that gives volume of revolution of
the shaded area below
|dw:1436656237239:dw|

- welshfella

- then you need to subtract the volume of the cylinder radius 1 and length ln15

- anonymous

ok, thank you

- welshfella

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)

- anonymous

its 2pi integral (shell height)(shell radius)

- welshfella

ah i see - the method i used was integrating infinitely thin discs

- anonymous

yeah the shell is integrating "coffee cans"

- anonymous

cylinders

- welshfella

ok lol

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