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Homework Help: Double Integrals Bounded by Cylinders

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Bounded by the cylinders x2 + y2 = r2 and y2 + z2 = r2

    We're supposed to stick to double integrals as triple integrals are taught in a later section.

    3. The attempt at a solution
    Edit:
    Alright, I think I go to the right answer.
    x = sqrt(r2 - y2)
    z = sqrt(r[SUP2[/SUP] - y2)

    The bounds are 0 < y < r and 0 < x < sqrt(r2 - y2)

    So I get:
    integrate integrate sqrt(r2 - y2) dx from 0 to sqrt(r2 - y2) dy from 0 to r
    The first integral goes to:
    sqrt(r2 - y2) * x from 0 to sqrt(r2 - y2) =
    (r2 - y2)

    The second integral then makes it:
    r2 * y - 1/3 * y3 from 0 to r which gives us:
    2/3 * r3

    However, you're supposed to throw in an 8 at the beginning to times everything by (why?) so you get:
    16/3 * r3

    So, my questions:
    Was my edited solution correct and why are you supposed to multiple the entire double integral by 8.
     
    Last edited: Mar 24, 2010
  2. jcsd
  3. Mar 24, 2010 #2

    LCKurtz

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    I presume you are to calculate a volume, but you never actually said so. Have you drawn a picture of the required volume? Do you understand what portion, if not all, the volume you have calculated using those limits?
     
  4. Mar 24, 2010 #3
    This object is in all the octants of 3 space and your integral bounds go from 0 to the radius of the cylinders. So you are only finding the volume of the 1st octant which is fine since this is symmetrical across the origin.

    When I did the problem I only obtained (8/3)*r3. I could have made an error some where but here is what how I solved it.

    I convert my coordinates to cylindrical which makes the differential r dr d(theta). Since you are using r in the given problem, I simply called it "a" so it wouldn't cause a problem with the r dr of the differential.

    Looking at just the first octant we see the angle runs from 0 to pi/2 this will be one of the bounds of integration. Now we have x2+y2=a2. x2+y2=r2 and now we have r=a; therefore, our other bounds are 0 to a.

    Then I solved for z2 which yields z=sqrt(r2-y2). But remember y=rsin(theta); thus, we get z=sqrt(r2-(r*sin)2)=r*sqrt(1-sin2)=r*cos.

    Now we have all the data for the integral:

    8*[tex]\int^{pi/2}_{0}\int^{a}_{0}(r*r*cos)drd(theta)[/tex]
     
  5. Mar 24, 2010 #4
    The volumes are cylinders around the Z axis and the X axis.
    2*2 is for symmetry of the cylinder (so you're only using 1/4 of the circle about the xy axis instead of the entire circle. However, I'm unsure as to where the final 2 comes into play.
     
  6. Mar 24, 2010 #5

    LCKurtz

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    I will repeat my question. Do you understand from your picture what portion of the volume you have calculated? Describe it to me. Once you do that correctly you will see where the 8 comes from.
     
  7. Mar 25, 2010 #6
    Does the 8 come from:
    2 (take the upper part of the symmetrical circle on the z & y axis)
    2 (take the upper part of the symmetrical circle on the x & y axis)
    2 (take the positive part of the half-circle on the x & y axis)

    The shape would be slightly larger than a sphere.

    I'm having a bit of trouble trying to visualize it still though.
     
  8. Mar 25, 2010 #7
    Look what I wrote and you will see where the 8 comes from.
     
  9. Mar 25, 2010 #8

    LCKurtz

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    It's a miracle you can get the correct limits, which you did, without having the picture first. You should always draw a sketch and use it to get the limits.

    What you are describing above can be better phrased as "the volume in the first octant". There are 8 octants and lots of symmetry....
     
  10. Mar 25, 2010 #9
    Thank you for the explanation. I got the limits from trying to find the answer on the web and what have you. I knew the answer, just not how to get it which is rather important.

    Thanks for the help.
     
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