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Triple-integrating a region with no volume

  1. Apr 25, 2005 #1
    From Shaum's: Compute the triple integral of [tex]f(r,\theta ,z)=r^2[/tex] over the region [tex]R[/tex] bounded by the paraboloid [tex]r^2=9-z[/tex] and the plane [tex]z=0[/tex]

    This has me stumped. The volume bounded by [tex]r^2=9-z[/tex] and [tex]z=0[/tex] is not closed in 3-space. But if they really meant region, triple-integrating a region with no volume gives 0. What should I do?
     
  2. jcsd
  3. Apr 25, 2005 #2

    matt grime

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    Isn't th region just the region given by rotating the area bound by

    x^2 = 9-z and the x axis about the z axis. Why isn't that closed?

    If we imagine the x-y plane on the floor with the positive z axis coming upwards, it's like integrating over the volume given by putting a cap on the floor.
     
  4. Apr 25, 2005 #3

    dextercioby

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    I tried to attach a graph,but the OS wouldn't let me...:yuck: Matt's right,it is a cap.

    Daniel.
     
  5. Apr 25, 2005 #4
    Why would we rotate it? r^2 = 9 - z describes a parabolic cylinder.
     
  6. Apr 25, 2005 #5

    dextercioby

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    Nope,a revolution paraboloid around the "Oz" axis...

    Daniel.
     
  7. Apr 25, 2005 #6

    HallsofIvy

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    All I can do is echo the others: z= 9- r2 is a paraboloid, with vertex at (0,0,9), opening downward, and "closing" it at z= 0 certainly does form a closed figure. When z= 0, r2= 9 so r= 3. The integration (in polar coordinates) will be over [itex]0\le \theta\le 2\pi[/itex],[itex] 0\le r \le 3[/itex].
     
    Last edited: Apr 25, 2005
  8. Apr 25, 2005 #7
    So I must watch out for the "special" variables rho, r, theta and phi, such that the inclusion of one of them will automatically mean polar, spherical or cylindrical coordinates?
     
  9. Apr 25, 2005 #8
    Indeed, it should have been specified that cylindrical coordinates were being used.
     
  10. Apr 25, 2005 #9
    Icebreaker, use [tex] 9-r^2[/tex] as the upper limit in your integration of f over z.


    [Edit: Wait, do you already get the idea? Sorry, if I'm being redundant.]
     
    Last edited by a moderator: Apr 25, 2005
  11. Apr 25, 2005 #10
    Thanks to everyone. Yes, I know the idea, but because the question itself did not specify cylindrical coordinates, and that it was amid a bunch of other integrals, I thought they simply replaced the x,y,z axes with r, theta and z. Like one of those "trick" questions where they replace f(x) with x(f).
     
  12. Apr 25, 2005 #11

    HallsofIvy

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    No, it is not a "trick" question and it does specify that cylindrical coordinates are to be used- your original post said "Compute the triple integral of f(r,θ,z), over the region bounded by the paraboloid and the plane"

    What do you think f(r,θ,z) means?
     
    Last edited: Apr 26, 2005
  13. Apr 25, 2005 #12
    HallsofIvy, perhaps he hasn't had enough exposure to cylindrical coordinates to realize that immediately. :rolleyes:
     
  14. Apr 26, 2005 #13

    dextercioby

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    If "he hasn't had enough exposure" (i like how it sounds),then what is he doing solving integrals with them...??

    The way i know these variables,it should have been [itex]f\left(\rho,\varphi,z\right) [/itex].

    Daniel.
     
  15. Apr 26, 2005 #14
    Why should it necessarily mean that they're using cylindrical coordinates? Just because the variables are called [itex]r[/itex], [itex]\theta[/itex], and [itex]z[/itex] should not imply anything about how they're parameterized with respect to the "real" coordinate axes. The problem, of course, is that the question doesn't give you bounds on [itex]\theta[/itex] unless you know implicitly that you're dealing with cylindrical coordinates (or some similar parameterization).

    Now, the fact that the question said that the region was bounded by a paraboloid might be grounds for a guess as to the parameterization...
     
    Last edited: Apr 26, 2005
  16. Apr 26, 2005 #15

    matt grime

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    Sadly lots of us are used to the abuse of mathematics that goes on in undergrad calc (in the US sense) and the courses and books all assume that the second you see r, theta, z, or rho phi theta that you know its polars of some flavour. Just as when they say "find the domain of sqrt(xy)" every undergrad should be able to write the answer down but anyone with an exposure to higher mathematics should be tearing their hair out at the sheer awfulness of it.
     
  17. Apr 26, 2005 #16
    f(r,θ,z) is simply a 3 variable function. There's no reason why I must treat it differently than f(a,b,c) or x(s,r,t). The point is that I must GUESS what the question is trying to ask, and that's the problem.
     
  18. Apr 27, 2005 #17

    matt grime

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    Well, the point is that you're using a book that presupposes you can deduce that it is talking about polar coordinates. As soon as you see these labels you ought to think that's what they mean. There is a reason to suppose it is different from other variables, because in lots of maths people assign explicit meaning to letters, particularly engineering courses: t is time, r is radial distance, theta and phi are angles in the xy plane and from the azimuth.

    These are commonly accepted practices, even if some of us agree that it is not necessarily a good thing.
     
  19. Apr 27, 2005 #18

    saltydog

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    What about the answer? Can I say or no? I've attached a plot. Suppose it's a . . . cone-head cap, whatever. Anyway, I get:

    [tex]\int_0^{2\pi}\int_0^3\int_0^{9-r^2} r^3dzdrd\theta=\frac{243\pi}2[/tex]
     

    Attached Files:

  20. Apr 27, 2005 #19
    The function was r^2.

    Is that just a typo, or did you calculate the wrong integral?
     
  21. Apr 27, 2005 #20

    matt grime

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    An infinitesimal volume element is rdr(dtheta)dz in cylindrical polars.
     
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