## QRAwarrior Group Title How would you draw a cylinder for the shell method for the region that is enclosed by x = pi/2, x = -pi/2 and y = cosx? This is rotated about y = -1. 2 years ago 2 years ago

1. QRAwarrior

@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

2. QRAwarrior

@jim_thompson5910

3. akamushi

the definite integrand of f(x)=cosx from -pi/2 to pi/2 sinx from -pi/2 to pi/2 = 1-(-1)=2

4. QRAwarrior

I do not want the answer. @akamushi

5. QRAwarrior

@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

6. QRAwarrior

@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

7. lgbasallote

@lgbasallote lol jk...idk shell using trig functions

8. PROSS

|dw:1337557715702:dw|

9. QRAwarrior

The height would be 2cosx, and the radius would be cosx - (-1)?

10. QRAwarrior

ie. h = 2y r = y - (-1) ?

11. QRAwarrior

|dw:1337558085638:dw|

12. QRAwarrior

The height (which are the horizontal lines of the cylinder) would be given by h = 2cosx? And the radius would be given by r = cos(x) - (-1)?

13. QRAwarrior

Wait, I think the height would be: h = 2x, due to symmetry

14. QRAwarrior

|dw:1337558202025:dw|

15. QRAwarrior

So then, radius is r = cos(x) - (-1) for sure, right?

16. QRAwarrior

And I can restate that cos(x) as y So then, r = y +1, h = 2x = 2arccos(y), I believe.

17. PROSS

now you are on the right track.

18. PROSS

if you are familiar with winplot, you can make a nice 3d representation.

19. QRAwarrior

@shivam_bhalla! Yes you finally came.

20. QRAwarrior

|dw:1337558589994:dw|

21. QRAwarrior

I have one question for either of you: I drew a vertical line from the x-axis to the function cos(x). Can that generically be represented as y?

22. shivam_bhalla

23. amistre64

x = pi/2, x = -pi/2 and y = cosx|dw:1337561871090:dw|

24. amistre64

|dw:1337562804283:dw|

25. QRAwarrior

Ok so it is: h = x y = cosx - (-1)

26. amistre64

we only need half of this, then double the results height is a measure of arccos(y) radius is from from 1 to 2 by perspective $2*2pi\int_{0}^{1}(y+1)*\cos^{-1}(y)\ dy$

27. QRAwarrior

I think I came up with this a few posts back

28. amistre64

height is a function of x; y=cos(x) arccos(y) = x = height

29. QRAwarrior

-------------------------------------------------------- Ok consider this question: Find the length of y^2 = x^3 from (0,0) to (4,8) I know the formula for arclength, but I end up getting this: |dw:1337563218324:dw|

30. QRAwarrior

|dw:1337563244465:dw|

31. QRAwarrior

How the heck can you resolve the radical when you have: |dw:1337563293983:dw| As your argument?

32. amistre64

have you tried looking at it form both forms? $\sqrt{1+(x')^2}\ or\ \sqrt{1+(y')^2}$

33. QRAwarrior

No. I think I will try the other form, BUT the problem is that I am not sure whether to pick the POSITIVE root or the NEGATIVE root..

34. amistre64

for the most part, i think you stick with the positive root

35. QRAwarrior

Why do you say "most part"

36. amistre64

distance tends to be a measure in absolute value, so the positive root seems most apt

37. QRAwarrior

WOW! I just realized how EASY it is doing it wrt x.

38. QRAwarrior

|dw:1337563549037:dw|

39. QRAwarrior

Easy U-sub. Thanks a lot amistre64. What level of education is your mathematics?

40. QRAwarrior

I was just curious

41. amistre64

i dunno what level it is, im guessing its at least on a 3rd grade level ;)

42. QRAwarrior

Ok anyways, thanks for your help

43. amistre64

youre welcome. Levelwise is difficult since im self taught for the most part of this stuff