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## anonymous one year ago Find volume of solid generated

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1. anonymous

2. welshfella

|dw:1436654820728:dw|

3. anonymous

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

4. anonymous

oops sorry i was looking at the wrong part XD

5. welshfella

|dw:1436654961888:dw|

6. anonymous

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

7. anonymous

so it is ln15-e^(ln15)

8. anonymous

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

9. anonymous

oops forgot pi

10. anonymous

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

11. anonymous

Thank you as well, but no

12. anonymous

|dw:1436655396490:dw|

13. anonymous

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

14. anonymous

whoops...

15. anonymous

forgot the triangle bit thing with the curvy roof

16. anonymous

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

17. anonymous

18. welshfella

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

19. anonymous

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

20. anonymous

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

21. anonymous

mathway.com for graphing

22. anonymous

Thta was to welshfella

23. welshfella

the diagram you mean?

24. anonymous

no the answer its pi(14-ln15)

25. welshfella

yes

26. 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

27. welshfella

|dw:1436655815791:dw|

28. 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

29. anonymous

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

30. welshfella

|dw:1436655967976:dw|

31. welshfella

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

32. welshfella

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

33. anonymous

ok, thank you

34. 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)

35. anonymous

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

36. welshfella

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

37. anonymous

yeah the shell is integrating "coffee cans"

38. anonymous

cylinders

39. welshfella

ok lol

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