anonymous
  • anonymous
Find volume of solid generated
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
1 Attachment
welshfella
  • welshfella
|dw:1436654820728:dw|
anonymous
  • anonymous
I don't understand the infinity part on the left side if it isn't enclosed from there

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oops sorry i was looking at the wrong part XD
welshfella
  • welshfella
|dw:1436654961888:dw|
anonymous
  • anonymous
I'm not sure what that is, but that is not even close. Thank you though
anonymous
  • anonymous
so it is ln15-e^(ln15)
anonymous
  • anonymous
so you have the area of the box ln15*1 - pi integral(e^(x/2)^2)dx
anonymous
  • anonymous
oops forgot pi
anonymous
  • anonymous
the integral is just pi*integra(e^x)=pi*e^x
anonymous
  • anonymous
Thank you as well, but no
anonymous
  • anonymous
|dw:1436655396490:dw|
anonymous
  • anonymous
Your drawing is incorrect as well. I do appreciate the help, but its not helping
anonymous
  • anonymous
whoops...
anonymous
  • anonymous
forgot the triangle bit thing with the curvy roof
anonymous
  • anonymous
No, it is bounded bounded by 1 you have it bounded by y=0
1 Attachment
anonymous
  • anonymous
1 Attachment
welshfella
  • welshfella
V = pi(e^(ln15) - e^0) - ln15 pi = 14pi - ln15pi = 11.29 pi
anonymous
  • anonymous
\[V=\ln15-\int\limits_{0}^{\ln15}y^2dx\]
anonymous
  • anonymous
YES How did you get that I couldn't make out what yo had up top
anonymous
  • anonymous
mathway.com for graphing
anonymous
  • anonymous
Thta was to welshfella
welshfella
  • welshfella
the diagram you mean?
anonymous
  • anonymous
no the answer its pi(14-ln15)
welshfella
  • welshfella
yes
anonymous
  • 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
  • welshfella
|dw:1436655815791:dw|
welshfella
  • 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
  • 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
  • welshfella
|dw:1436655967976:dw|
welshfella
  • welshfella
that gives volume of revolution of the shaded area below |dw:1436656237239:dw|
welshfella
  • welshfella
- then you need to subtract the volume of the cylinder radius 1 and length ln15
anonymous
  • anonymous
ok, thank you
welshfella
  • 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
  • anonymous
its 2pi integral (shell height)(shell radius)
welshfella
  • welshfella
ah i see - the method i used was integrating infinitely thin discs
anonymous
  • anonymous
yeah the shell is integrating "coffee cans"
anonymous
  • anonymous
cylinders
welshfella
  • welshfella
ok lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.