## anonymous 3 years ago have to do a shell volume; can do it easy with disc but that's not the question asked. bounded by y=1/x^2; y=0; x=1; x=4.about the x axis.... I have gotten this: $v=2\pi \int\limits_{0}^{1 }(\sqrt{y}/y)(y-1) dy$

1. anonymous

answer with disc is 21/64 pi...which is correct..but I cannot get it with shell. Help please!

2. anonymous

3. amistre64

shell is simple enough, so long as you remember how to find the area of a rectangle ...

4. amistre64

y=1/x^2; y=0; x=1; x=4 |dw:1365100001202:dw|

5. amistre64

you might wanna go ahead and shift the graph by x-a to get it looking somewhat normal

6. anonymous

ok, I had the r and h the right direction. I guess I am having trouble finding the lengths.

7. amistre64

h is from a to b, radius is from f(a) to f(b)

8. anonymous

since the equation is in terms of x; I resolved in terms of y since shell with the rotation about x axis means I have to be in terms of y and use dy. correct?

9. amistre64

$\int_{a}^{b}2\pi~r ~h$ $\int_{a}^{b}2\pi~f(x) ~x~dx$

10. amistre64

one issue i see that i alluded to but dint take care of

11. amistre64

the height at a is not x=a, its x-a $\int_{a}^{b}2\pi~f(x) ~(x-a)~dx$

12. anonymous

you have dx...thought it had to be dy for x axis rotation for shell method?

13. amistre64

no, we are integrating aling the height, which in this case runs along the x axis. one final remark is this. We have a missing core when we get to b that is just a solid cylindar to fill in |dw:1365100474873:dw|

14. amistre64

Volume of a cylindar is: pi r^2 h pi f(b)^2 (b-a)

15. amistre64

we could integrate it with respect to the radius .... but that really doesnt seem necessary to me looking at the picture

16. anonymous

the teacher said if you have shell and revolve area about x axis it will make r be the vertial length and h be the horizontal length like you have; but said the integrand will be y values and will be dy....because is shell method not disc....?

17. amistre64

forget about calling them x and y, it just throws in extra stuff that is really not related to height and radius for a shell ..... there are 2 ways we can approach this shell, drdh, or dhdr, all that changes is the order in which we run it. |dw:1365100962229:dw| |dw:1365101040649:dw|

18. amistre64

lets fall f(x) ... g(y) in the bottom case $\int_{r=0}^{r=f(a)}\int_{h=a}^{h=g(y)}~2\pi rh~dhdr$ OR $\int_{h=a}^{h=b}\int_{r=0}^{r=f(x)}~2\pi rh~drdh$

19. amistre64

i prolly messed that up some putting it into double integration ... but the point is, it can be "run" along the x axis, framing the radius and height in terms of x, thereby having a dx

20. amistre64

or, it can be framed in terms of radius and height in terms of y, thereby creating an inegration of dy

21. amistre64

we can see that the area of the shell at h=(x-1) , and radius = f(x) from x=1 to 4 gives us 2pi f(1) * 0 2pi f(3/2) * 1/2 2pi f(2) * 1 2pi f(5/2) * 3/2 2pi f(3) * 2

22. anonymous

i'm looking...so confused. I thought depending on which way you revolved it had to be a set thing like if revolved about x; use dy and y value....if revolved about y; use dx and use x values for shell method. is this statement not correct?

23. amistre64

that statement is a general rule of thumb but is contingent on how you are moving and what is variable along the way

24. amistre64

if you stick to trying to figure these things out by remembering certain formula per se, your not going to get the overall picture and be stuck on a test going .... what was the rule i should use? We know the basic set up for a shell has these components, and how you define them will depend on dx or dy $\int2\pi~r~h$

25. amistre64

now, one way or the other may be simpler to work out .... so knowing both ways to address it is good

26. anonymous

ok; that makes sense to not rely on formula. if I were to revolve about x axis and want my parameters on the integral to be y values; would you say the one length would be 1/(sqrt of y) ?

27. anonymous

I see the x set up fine; i cannot see the y setup and need to see both as you say ...sometimes need one or the other. for some reason the y setup gets me in this case.

28. amistre64

i agree :) ill address the y setup in a minute, i just want to run thru this x setup .... helps me remember the things ive forgoten :) r = f(x), h=(x-1), along the interval of 1 to 4, and remember to addon the hollow core $2pi\int_{1}^{4}rh$ $2pi\int_{1}^{4}\frac{x-1}{x^2}dx$ $2pi\int_{1}^{4}\frac{x}{x^2}-\frac{1}{x^2}dx$ $2pi\int_{1}^{4}\frac{1}{x}-{x^{-2}}dx+3pi(\frac{1}{16})^2$ this is very doable yes?

29. anonymous

if I am spinning a shell about the x axis; the radius would be the length from the function to y=0. would I write that as (y) or as (1/sqrt y). i know the ordered pair on the function is (x,y) but y is the function. i get confused when to use y and when to use the function equation.

30. anonymous

why 3pi(1/16)^2?

31. amistre64

1/x^2 at x=4 is 1/16 for a radius of the core volume of a cylindar is pi r^2 h: h=1 to 4 = 3 3pi (1/16)^2 to fill in the hollow core that was left over

32. anonymous

oh your intuitively doing the math not by technique..is that true?

33. anonymous

like logically figuring shape;etc.

34. amistre64

yes, im applying stuff from geometry way back when that is still applicable :) we could run a integration on it but it would be a little silly to me in terms of y, r = f(a) to f(b); h = g(y) = x = 1/y^2

35. anonymous

problem is the teacher is going to grade on technique of washer method...and it has to be exactly documented as that or it's 100 percent wrong.

36. amistre64

x = 1/y^2 y^2 = 1/x y = +- 1/sqrt(x) for the height

37. anonymous

Ha; ok...well I agree with you totally!!! wish you were the teacher :)

38. anonymous

however; i follow your logic...much like mine...but it won't please her to do this. that is why I am stuck. it's not like if I get the right answer it is ok...has to be her format.

39. amistre64

|dw:1365102773341:dw|

40. amistre64

i just orientated it in a more normal looking fashion

41. anonymous

x goes from 1 to 4..?

42. amistre64

x goes from 1 to 4 hasnt changed any, its an inherent part of the problem

43. anonymous

so in this picture can you help me see what a vertical line from bottom to top of shape would be for length? Is it simply y or is it 1/(sqrt y) ?

44. amistre64

h = 1 to 1/sqrt(y)

45. amistre64

might have to subtract 1 again , to get a zero there

46. anonymous

why did you shift up?

47. amistre64

i didnt shift up, thats just the way the problem looks when orientated in this manner $h=0 \to (\frac{1}{\sqrt{y}}-1)$$y = \frac12\to1$

48. amistre64

r = y

49. anonymous

oh dear...feeling lost. I think I will take a break and look at your comments in a little. maybe it will sink in better. thanks so very much for all your help!!

50. amistre64

$\int 2pi~rh$ $\int_{\frac12}^{1} 2pi~y(\frac{1}{\sqrt y}-1)~dy$

51. amistre64

notice that we still have an empty core

52. amistre64

y/sqrty = sqrty $2pi~\int_{\frac12}^{1} \sqrt y-y~~dy+(core)$

53. anonymous

got it! thanks . the core adding helped. later...

54. amistre64

dbl chk my limits on y tho

55. amistre64

yeah, 1/16 not 1/2 .. :)