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Volume generated by 2sin(x/2) & sin(x) about y=2 with shells

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the volume generated by
    x=0
    x=π
    y=2sin(x/2)
    y=sin(x)
    using the shell method.
    2. Relevant equations
    x=2arcsin(y/2)
    x=arcsin(y)

    3. The attempt at a solution

    So to setup my integral I got the following values,
    Outer Radius=(2-arcsin(y))
    Inner Radius=(2-2arcsin(y/2))
    I'm pretty sure outer/inner radius is for the shell method so I was thinking a way to get around this would be by setting up two different integrals to get the volumes and then subtracting the smaller one from the larger one.
    I'm stuck at finding out the length because for length I have
    L=π-x
    I'm confused as to which function I should to use here or is this similar to the outer/inner radius where I use them separately for their respective integrals?
     
  2. jcsd
  3. Mar 5, 2015 #2

    SammyS

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    (You should give the problem statement directly in your post without regard to what you put in the thread title.)

    Your length is incorrect. To help get it correct, sketch the region that is to be rotated about y = 2.

    What is the integration variable? If it's y, then x should be in terms of y .
     
  4. Mar 5, 2015 #3
    Okay after looking at my sketch I think I need to separate this up a lot.
    for 0->1 and x=0 to x=pi/2: It appears as though the length is 2arcsin(y/2)-arcsin(y) and then from x=pi/2 to pi length=pi-arcsin(y)
    Is this correct?
     
  5. Mar 5, 2015 #4
    Okay and from y=1 to 2 the length=pi-2arcsin(y/2)
    So to get the volume of the solid now I would integrate these lengths with the corresponding radius? What I mean is:
    The lengths L=2arcsin(y/2)-arcsin(y) & L=pi-arcsin(y) with Outer Radius=(2-arcsin(y)) and L=pi-2arcsin(y/2) with Inner Radius=(2-2arcsin(y/2))
     
  6. Mar 5, 2015 #5

    SammyS

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    Almost correct.
    arcsin(y) will not give a result greater than π/2, so, to get the x from the y value, for the for the y=sin(x) curve beyond x = π/2, you need x = π - arcsin(y).

    Don't forget y from 1 to 2 .
     
  7. Mar 5, 2015 #6

    SammyS

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    I deleted that post, but not in time.
     
  8. Mar 5, 2015 #7
    Okay I'm slightly confused, don't I already have x=π-arcsin(y) for the y=sin(x) curve?
     
  9. Mar 5, 2015 #8

    SammyS

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    What portion of the y = sin(x) curve, are you using the expression,x=π-arcsin(y) for the length?
     
  10. Mar 5, 2015 #9
    from x=pi/2 to x=pi and y=0 to 1
     
  11. Mar 5, 2015 #10

    SammyS

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    Right, but arcsin(y) only gives values from pi/2 to 0 for y from 1 to 0. So to get these x values (not the lengths), you need x = π - arcsin(y). Then to get the associated length, take π minus this x value. → length = π - ( π - arcsin(y) ) . The x values here will be between π/2 an π, The lengths will be between π/2 an 0 .
     
  12. Mar 5, 2015 #11
    Okay thanks I understand what you mean now although I should probably review inverse trigonometric functions. I have been looking at my graph and I think I mixed up the length L=2arcsin(y/2)-arcsin(y) from x=0 to pi/2, should this be L=arcsin(y)-2arcsin(y/2) since x=arcsin(y) is the curve furthest to the right?
     
  13. Mar 5, 2015 #12

    SammyS

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    Yes.

    To check that, try y = .5, for instance. The answer should be positive. (and ... It is.)
     
  14. Mar 5, 2015 #13
    Okay so now to calculate the volume I am thinking this is a method of doing it:

    Upper & Lower limits for the integrals in order: 0-1; 0-1; 1-2
    V=2π*∫(arcsin(y)-2arcsin(y/2))*(2-arcsin(y))*dy+2π*∫(π-(π-arcsin(y))*(2-arcsin(y))*dy+2π*∫(2π-2arcsin(y/2)*(2-2arcsin(y/2))*dy


    Sorry if this is really hard to read but I can't seem to get latex to work.
     
    Last edited: Mar 5, 2015
  15. Mar 6, 2015 #14

    SammyS

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    I see now that the expressions for radius were wrong. Each should be simply ##\ 2 - y\ ## .

    Also: Notice that π-(π-arcsin(y)) = arcsin(y) .
     
  16. Mar 6, 2015 #15

    BvU

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    Re TeX:

    ##\#\# ## V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy ##\#\# ##

    gives

    ##V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy##

    whereas

    $## ##$V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy$## ##$

    gives displaystyle (bigger int sign, bigger y/2):

    $$V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy$$
     
    Last edited: Mar 6, 2015
  17. Mar 6, 2015 #16
    Whoops, is the volume integral from my previous post correct though if the radiuses are (2-y) or am I setting it up improperly?
     
  18. Mar 6, 2015 #17

    SammyS

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    Right, (2-y) is what you should use for the radii.

    For example, the red in the following should be changed ##\displaystyle\ 2\pi \int_0^1 \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \color{red}{\left (2-\arcsin y\right )} dy\ ##
    to ##\displaystyle\ 2\pi \int_0^1 \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \color{red}{\left (2- y \right )} dy\ ##

    Likewise in the other integrals .
     
  19. Mar 7, 2015 #18
    Okay thanks for the help this question was a real hassle considering how much easier it would have been using disks.
     
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