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Homework Help: Calculating a volume through cylindrical coordinates

  1. May 30, 2010 #1
    Just a question.
    Say you have a function, which in cylindrical coordinates it gives that
    [tex]\int\int\int \sqrt{x^2 + y^2} dx dy dz[/tex]

    which is

    [tex]\int\int\int r^2 dr d/theta dz[/tex]

    i want to find in cylindrical coordinates, in the area limited by the functions :

    [tex]x^2 + y^2 = z^2[/tex]
    z is greater or equal than -1 and less than equal to 1.

    so, i solve it, but then i realize:
    why is the value of solving the inner integral in order to dr (with limits of integration from 0 to z) different that the one in order to dz (with limits from 0 to r), with the middle one being dz or dr, respectively, with limits 0 to 1, and the outside one (d theta) 0 to 2pi?
    (all multiplied by two)

    the order shouldnt matter, i think.

    please explain what the order should be and why, im really confused on this one... thanks
     
  2. jcsd
  3. May 30, 2010 #2

    lanedance

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    you're correct the value of the integral shouldn't change - though exactly what you;re doing isn't very clear - try tex below to show your limits

    [tex]\int_{z_1}^{z_2} \int_{\theta_1(z)}^{\theta_2(z)}\int_{r_1(z,\theta)}^{r_2(z,\theta)} r^2 dr d\theta dz[/tex]
     
  4. May 30, 2010 #3

    LCKurtz

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    You haven't stated the original problem clearly enough to even tell whether the first integral you have written down is correct, regardless of your limits.

    What are you trying to find? A volume? A mass given a variable density? Are you interested in the region (not area) inside or outside of the surface? Do you know what the surface looks like?

    If you draw a picture you will see that z going from 0 to r doesn't describe the same region as r going from 0 to z.
     
    Last edited: May 30, 2010
  5. May 31, 2010 #4
    The problem doesn't state what exactly i'm trying to calculate.

    anyway, the integral would be

    [tex]\int_{0}^{2\pi} \int_{0}^{1}\int_{0}^{r} r^2 dz dr d\theta [/tex]

    or

    [tex]\int_{0}^{2\pi} \int_{0}^{1}\int_{0}^{z} r^2 dr dz d\theta [/tex]

    yes, i think i know what it looks like.

    Basically, since the limits are

    [tex]
    x^2 + y^2 = z^2
    [/tex]

    it should be a cone, with symetry about the xy plane

    [tex]
    z = r
    [/tex]

    the limits of z are
    [tex]
    -1 <= z <= 1
    [/tex]

    so i can integrate the triangle that makes the cone, considering only the xz plane, plotting z/r.
    Then I integrate the volume, since it's a solid of revolution and multiply it by two, because of the symmetry.

    So ultimately, I know:

    [tex]2 \cdot 2\pi \int_{0}^{1}\int_{0}^{r} r^2 dz dr d\theta [/tex] or [tex]2 \cdot 2\pi \int_{0}^{1}\int_{0}^{z} r^2 dr dz d\theta [/tex]

    and yes, i want the region. sorry, if i said area before
     
    Last edited: May 31, 2010
  6. May 31, 2010 #5

    LCKurtz

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    No, you don't get to choose. Those integrals don't describe the same region. One gives the region inside (above) the cone and the other outside (below) the cone but inside the cylinder r = 1.
     
  7. May 31, 2010 #6
    i got it. thanks
     
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