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An integral i cannot solve

  1. Apr 12, 2004 #1


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    I have the attached graph, a part of a cosinusoidal curve (not sure if this is a correct translation).

    I don't have the function, but if I am correct:

    y = 3 cos (x*pi/2)

    The questions are:

    a) Calculate the area between that part of the curve and the x axis.
    b) Calculate the volume of the body created by rotating the part of the curve around the y axis.

    a) No problem here.

    P = 2 * integral from 0 to 1 of 3cos(xpi/2) dx

    I got the solution 12/pi.

    b) I got this far:

    To calculate the volume I need the area of the circles which constitute the body. To get that, I need to express x in terms of y.

    y = 3 cos (x*pi/2)
    arccos(y/3) = x*pi/2
    x = (2/pi)arccos(y/3)

    The area is: P = x^2*pi

    So the volume is:

    V = integral from 0 to 3 of [(2/pi)arccos(y/3)]^2*pi dy

    Let I = integral of [(2/pi)arccos(y/3)]^2*pi dy

    I = integral of (4/pi)arccos^2(y/3) dy
    I = (4/pi) integral of arccos^2(y/3) dy

    substitute t = y/3, dt = dy/3, dy=3dt

    I = (12/pi) integral of arccos^2(t) dt

    I don't know how to integrate this. Anyone able to help?

    Attached Files:

  2. jcsd
  3. Apr 12, 2004 #2
  4. Apr 12, 2004 #3


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    that was fast! thanks a lot :)

    by the way, is there a way for me to calculate it myself that i should have thought of?
    Last edited: Apr 12, 2004
  5. Apr 14, 2004 #4
    If you know how to differentiate arccos(x), you could try integration by parts and using some trigonometric identities?!
  6. Apr 14, 2004 #5


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    The whole problem was that I didn't know how to integrate arccos(x)dx. I didn't immediately think of dividing it into arccos(x)*1dx and using partial integration.. Stupid, I know :)
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