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Double integrals over a region

  1. Jun 21, 2004 #1
    I'm kinda lost on this problem

    "Find the volume of the give solid:
    Bounded by the cylinder y^2+z^2=4 and the planes x=2y, x=0, z=0 in the first octant."

    Any help would be appreciated.
  2. jcsd
  3. Jun 21, 2004 #2
    First thing that helps is to draw a picture. Then, you'll see that it's helpful here to work in cylindrical coordinates. Are you stuck on the limits of integration or where?
  4. Jun 21, 2004 #3
    Volume of region

    In this case it doesn't seem like you need to use double integrals. you have y^2+z^2 = 4 and y = x/2 for one of the planes. If you draw the xy plane you will just have the line y=x/2. If you visualize any cross section of the cylinder made by placing a plane perpindicular to the xy and perpindicular to the xz plane, that the height of the cylinder at that point (the value of z) will be constant along the plane. So what I would do is, Integrate along the y-axis by breaking down the cylinder into shells.

    so you get V = Integral(2*y*sqrt(4-y^2)dy, 0, 2). Which is solved by a simple substitution.
  5. Jun 21, 2004 #4
    That's the thing, I dont really understand how to do it with cylindrical coordinates.
  6. Jun 22, 2004 #5
    I think my teacher wants me to use double integrals...
  7. Jun 22, 2004 #6
    ok, well first you should set it up so V=int(function) dxdydz. Then we'll change the function into cylindrical coordinates. To do this, remember x=rcostheta and y=rsintheta and z=z. You also need the Jacobian here.
  8. Jun 22, 2004 #7
    Why is there a dz in there, shouldnt it only be a double integral, since it's in that particular section?
  9. Jun 22, 2004 #8


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    Note that we have:
    Also, we have: [tex]0\leq{z}\leq\sqrt{4-y^{2}}[/tex]
    While [tex]0\leq{y}\leq{2}[/tex]
    Hence, we have the double integral V:
    Last edited: Jun 22, 2004
  10. Jun 23, 2004 #9
    Always recall that when finding volumes, these double integrals are simply "simplified" triple integrals.

    The definition of volume is just the triple integral of a volume element dV.


    [tex]V=\int\int\int_E dV[/tex]

    when the first integral is carried out, you are left with the integrand that arildno pointed out above this post.
    Last edited: Jun 24, 2004
  11. Jun 24, 2004 #10
    A dumbed down version of the definition I should point out is:

    Volume is the triple integral of 1.
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