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Volume of a torus

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    http://faculty.tcu.edu/richardson/Calc2/H20090323torusVolume.htm [Broken] This is a link to the homework.

    2. The attempt at a solution
    I did number 1 by doing 2 * integral from -2 to 2 of sqrt(4-x^2) = 4pi

    The second problem is where im completely confused and dont know how to do it or even conceptualize it. Trying to look at an example of a similar problem on another site, I tried to do 4pi * integral from -2 to 2 of xsqrt(4-x^2) but it equals 0.

    I think if I understand this one ill be able to do the rest of the homework.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 21, 2009 #2

    Dick

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    You are close. But the x in the second integral is supposed to be the radius of rotation of the slice located at x and rotated around x=3. How far is x from the center of rotation at x=3?
     
  4. Mar 21, 2009 #3
    So it should be 4pi * integral from -2 to 2 of 3sqrt(4-x^2) ?
     
  5. Mar 21, 2009 #4

    Dick

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    No. If x is -2 the radius of rotation is 5. If x is +2 the radius of rotation is 1. Right? The distance of those points from x=3. What is the radius for a general point x between -2 and 2?
     
  6. Mar 21, 2009 #5
    r= 3-x
     
  7. Mar 21, 2009 #6

    Dick

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    You've got it. Put that into your integral instead of x.
     
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