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Surface Integration of a Cone (Sloped Surface)

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Given Parameterization:
    x = u cos [tex]\phi[/tex]
    y = sin [tex]\phi[/tex]
    z = u cot [tex]\Omega[/tex]

    Find the sloping surface of a right cone with semi-angle [tex]\Omega[/tex] with a base radius of a.

    2. Relevant equations

    Surface area of a cone = [tex]\pi r\sqrt{r^2 + h^2}[/tex]



    3. The attempt at a solution

    Solid angle:
    [tex]\Omega = \int(r dS)/ (r^3)[/tex]

    semi angle = (1/2) [tex]\Omega[/tex]

    Cartesian Equation of a cone:

    (x2 + y2) / (r / h)2 = z2

    I understand the concepts of surface integration but I have not run across a problem where F was not given. I have a feeling that I am much more likely to run into this issue in the future and I would like to know what the process of determining F is.

    Should I start by taking the div of the cartesian equation and then plugging in the given parameters (x,y,z) and integrating?
     
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2

    HallsofIvy

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    It doesn't make sense to tell us that "F" is not given, when you don't tell us what "F" is!

    In any case, if [itex] x= u cos(\phi)[/itex], [itex]y= u sin(\phi)[/itex], and [itex]z= u cot(\Omega), where u and [itex]\phi[/itex], and [itex]\Omega[/itex]is a constant, [itex]\vec{r}= u cos(\phi)\vec{i}+ u sin(\phi)\vec{j}+ u cot(\Omega)\vec{k}[/itex] so [itex]\vec{r}_u= cos(\phi)\vec{i}+ sin(\phi)\vec{j}+ cot(\Omega)\vec{k}[/itex] and [itex]\vec{r}_\phi=- -u sin(\phi)\vec{i}+ u cos(\phi)\vec{j}[/itex] are tangent vectors to the surface in the direction of "coordinate lines". The "fundamental vector product" [itex]\vec{r}_u\times \vec{r}_\phi[/itex] is [itex]-ucos(\phi)cot(\Omega)\vec{i}- usin(\phi)cot(\Omega)\vec{j}+ u\vec{k}[/itex]. The "differential of surface area" is the length of that vector times [itex]du\d\phi[/itex]: [itex]u\sqrt{cot^2(\Omega)+ 1}dud\phi[/itex]
     
  4. Nov 12, 2008 #3
    The "F" I was referring to is the function inside a double or triple integral when doing a surface integration

    ie

    [tex]\int\int\int F dV[/tex]

    With your help I believe that I have come up with the answer, however I am having a little trouble showing that it is correct.

    A = (2[tex]\Pi[/tex])(Slope Height) / 2

    after my integration I end up with the following:

    [ ([tex]\Pi[/tex]a2)/2 ] (cot2[tex]\Omega[/tex] + 1).5

    Which simplifies down to

    [([tex]\Pi[/tex]a2)/2] csc[tex]\Omega[/tex]

    Can someone give me a push (or shove) in the right direction to prove this equals A from above?
     
  5. Nov 12, 2008 #4

    HallsofIvy

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    That's a volume integral, not a surface integral! The "F" you want to find the surface area, when dS is the differential of surface area is just "1":
    [tex]\int\int 1 dS= S[/tex].

    [/quote]With your help I believe that I have come up with the answer, however I am having a little trouble showing that it is correct.

    A = (2[tex]\Pi[/tex])(Slope Height) / 2

    after my integration I end up with the following:

    [ ([tex]\Pi[/tex]a2)/2 ] (cot2[tex]\Omega[/tex] + 1).5

    Which simplifies down to

    [([tex]\Pi[/tex]a2)/2] csc[tex]\Omega[/tex]

    Can someone give me a push (or shove) in the right direction to prove this equals A from above?[/QUOTE]
     
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