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Triple Integral Spherical Coordinates?

  1. Nov 5, 2008 #1
    I don't think so since it's not a sphere (disk). I have not learned about cylindrical coordinates and Cartesian is just a pain, so I am assuming I am supposed to use polar or something.

    Can someone clear up my confusion?

    [tex]\int\int\int_E y\,dV[/tex]

    where E lies above the plane z=0, under the plane z=y and inside the cylinder [itex]x^2+y^2=4[/itex]

    How would you proceed here if you DON'T want to use Cartesian Coordinates?

    Casey
     
  2. jcsd
  3. Nov 5, 2008 #2
    Duuuudddddeeeeeeee........... I don't get it.... I see a cirlcle (the domain of E) and I automatically think polar... but it is 3-D so tha doesn't male sense right? For 3-D you need spherical coordinates, but that doesn't make sense either.

    I am thinking I am not supposed to use Cartesians, cause that integrals a pita. But maybe I am wrong; however we usually don't get integrals that require the use of tables (for now).
     
  4. Nov 5, 2008 #3

    Dick

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    To find the volume you integrate the height y, over the area in the x-y plane. Use polar coordinates in the x-y plane. That's really cylindrical coordinates, but you don't have to call it that.
     
  5. Nov 5, 2008 #4
    Hey Dick! I don't have to much experience with this. Do I set up the integral like this:

    [tex]\int\int_D\int_0^y y\,dz\,dA[/tex]

    where D is the region bounded by x=0 and x^2+y^2=4

    thus the integral becomes

    [tex] \int_{\theta=\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{r=0}^2\,y^2\,dr\,d\theta[/tex]

    and then just plug in what y is in polar coordinates? Or did I F this one up royally?

    Thanks :smile:
     
    Last edited: Nov 5, 2008
  6. Nov 5, 2008 #5

    Dick

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    Not that royally, but it could use some work. Why y^2? If you are integrating y over the area in the x-y plane it's only a double integral over r and theta. And doesn't theta go from 0 to pi? And isn't the area measure in polar coordinates r*dr*dtheta?
     
  7. Nov 5, 2008 #6
    I am missing something... where did the third integral go then? The y^2 came from integrating y*dz wtr z--->y*z from z=0 to z=y.

    Then I though you integrate that over the x-y area?
     
  8. Nov 5, 2008 #7

    Dick

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    If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.
     
  9. Nov 5, 2008 #8
    Maybe I should start over here. If I take the Volume and project it onto the x-y plane, I get D, which takes the form of a semi-circle that follows a path from theta=0 to pi at a radial distance 2 from the origin.

    Now I want to integrate the function y wrt r and theta? Is that what you are saying?
     
  10. Nov 5, 2008 #9

    Dick

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    Yes, that's exactly what I'm saying. You integrate the height of the region over the area of the base.
     
  11. Nov 5, 2008 #10
    Why am I integrating 1 ? I am looking for its 4-D volume (or whatever it is called) aren't I.
     
  12. Nov 5, 2008 #11
    [tex]\int\int\int y\,dz\,dr\,d\theta[/tex]

    Once more. Is this my integral? I am just confused, now.

    I think it's time for bed! Thanks for the help Dick :smile: I will look at it in the a.m. I am getting too flustered right now. I understand every other triple integral assigned to us except this one. Moreover, I can solve this one in Cartesian Coordinates just fine, so long as I have an integral table handy. So I have no idea why this does not make sense to me in Cylindricals.

    In Cartesian I have

    [tex]\int_x\int_y\int_z\,y\,dz\,dy\,dx[/tex] .... Not sure what the big deal is now.
     
  13. Nov 5, 2008 #12

    Dick

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    I'm going to pretend you didn't say that. A volume integral is the integral of 1 over the 3d coordinate limits. Look it up.
     
  14. Nov 5, 2008 #13
    What is the definition of the [tex]\int\int\int_E f(x,y,z)\,dV[/tex]

    It is not the volume of E

    correct? If I wanted the volume of E I would integrate the function

    [tex]\int\int\int_E \,dV[/tex]

    Correct?

    But my problem does not ask me to find the volume of E. It asks me to evaluate

    [tex]\int\int\int_E f(x,y,z)\,dV[/tex]

    where f(x,y,z) is simply y. Is that not the case here? I am NOT looking for volume (in the three dimensional case)?
     
  15. Nov 5, 2008 #14

    Dick

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    Yes. Completely, utterly, right. I somehow thought you were looking for the volume and you're not. Sorry, sorry. Yes, the y^2 is correct. But shouldn't it be y^2/2?
     
  16. Nov 5, 2008 #15
    I don't think so: if I have [tex]\int\int_D\int_0^y y\,dz\,dA[/tex]

    Just taking care of that innermost integral with respect to z I have

    [tex]\int_0^yy\,dz=y\int_0^y\,dz=y*z|_0^y=y^2[/tex]

    yes?
     
  17. Nov 5, 2008 #16

    Dick

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    Yep, you win again. For a fixed x,y y is a constant. You might be better at this than I am.
     
  18. Nov 5, 2008 #17
    Yessss! I win! Just kidding.:biggrin: Okay so back to bed. I think I'll be able to sleep now :smile:

    So back to my proposal:

    Now I have integrated the function with respect to height and got y^2. Now I evaluate a double integral over D of y^2 (but in polar terms=(r*sintheta)^2). Something to the affect of:

    [tex]\int\int_D\,y^2\,dA=\int_0^\pi \int_0^2\, r^2*sin^2\theta\,* r\,dr\,d\theta[/tex]
     
  19. Nov 6, 2008 #18

    Dick

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    That looks fine to me.
     
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