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Volume of Revolution question

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the volume of 2sin(x) and -sin(x) from 0 to pi revolving around the y-axis


    3. The attempt at a solution
    My problem is with the geometry of this problem, 2sin(x) is above the x-axis and -sin(x) is below the x-axis. My belief was that I should be adding on the extra area of -sin(x) because it lies below the x-axis. I don't understand why I should be subtracting in this case. Thanks
    edit: the axis of rotation is the y-axis.

    there are two parts, the first asks me to show, through differentiation, that the integral of x(sin(x))dx = sin(x) - x(cos(x)) + C. Done. For the second part they say use the result of part (a) to find the volume of the solid generated by revolving each plane region, the area between 2sin(x) and -sin(x), about the y-axis. I said x-axis, sorry my mistake.
     
    Last edited: Mar 1, 2013
  2. jcsd
  3. Mar 1, 2013 #2

    SammyS

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    This problem doesn't make a lot of sense.

    Please state the whole problem word for word as it was given to you.
     
  4. Mar 1, 2013 #3
    post edited
     
  5. Mar 1, 2013 #4

    SammyS

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    Yes, it makes much more sense for this region to be revolved about the y-axis, rather than about the x-axis .

    Also, the extra information is very helpful in regards to performing an integration which will give the volume.

    If you set up the integral(s) corresponding each method of finding this volume,
    The washer method

    The shell method​
    you will find the one of these methods is consistent with the given hint (the additional info in your edit).

    Start by doing the set-up for each method.



    Furthermore, it's helpful to graph the region to be revolved. WolframAlpha gives:
    attachment.php?attachmentid=56239&stc=1&d=1362158108.gif



    Added in Edit:

    As for your question regarding subtracting: You are subtracting a negative quantity.
    2sin(x) - ( -sin(x) ) = 2sin(x) + sin(x) = 3sin(x) .​
     

    Attached Files:

    Last edited: Mar 1, 2013
  6. Mar 1, 2013 #5
    let me ask you this, if you were just doing a plain integral of -sin(x) what area would you shade as the region of integration? The area between the purple curve and the x-axis or everything below that purple curve?
     
  7. Mar 2, 2013 #6

    Mark44

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    If the question asked you to find the area between the graph of y = -sin(x) and the x-axis, on the interval [0, ##\pi##], it's pretty clear what region you would shade, right? If for some reason you shaded everything below the curve, that would be an infinite area.

    The integral to represent the area I described (i.e., between the curve and the x-axis) would be

    $$ \int_0^{\pi} 0 - (-sin(x))~dx$$

    This would yield a positive number, as areas should be.
     
  8. Mar 4, 2013 #7
    got it now, thank you.
     
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