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Surface Integral over Half Shell

  1. Jul 28, 2011 #1

    mzh

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    From Div,Grad, Rot and all that
    Disclaimer: Sorry, why do the Latex tags not work?

    1. The problem statement, all variables and given/known data
    "An electrostatic field is given by [itex]\vec{E} = \lambda (\hat{\vec{i}}yz + \hat{\vec{j}}xz + \hat{\vec{k}}xy)[/itex], where [itex]\lambda[/itex] is a constant. Use Gauss' law to find the total charge enclosed by a surface [itex]S_1[/itex], the hemisphere [itex] z=(R^2 - x^2 - y^2)^{1/2}[/itex] and [itex]S_2[/itex], its circular base in the xy-plane."


    2. Relevant equations
    So I understand I'm supposed to solve the integral [itex]\iint_S \vec{E} \cdot \hat{\vec{n}} dS[/itex], since this equals the enclosed charge divided by the dielectric constant.



    3. The attempt at a solution
    I am trying to find a way of making it clear that the electric field is independent of angle for a given radius, so I can discard the angular dependence and it essentially becomes a constant at a given radius. What should I do first? Express the field in spherical coordinates? I also calculated the divergence and it is, obviously, zero, so technically there can be no charge enclosed. But I would also like to know how to calculate the integral/apply Gauss law.

    Any hints are very appreciated (working on this problem for a week now, and I'm still not successful).
     
    Last edited: Jul 28, 2011
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  3. Jul 28, 2011 #2

    Disconnected

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    Gold Member

    My brain isn't working completely awesomely today (headache :/ ) but have you tried taking a look as stokes/greens/divergence?
     
  4. Jul 28, 2011 #3
    I am not familiar with the physical side so far as I will only start with physics this winter term, but mathematically speaking, if you know that divergence is zero, and you know from the physical side that the formula you mentioned above, is indeed valid for that type of exercise, then you have solved it: There's no charge included. If you would like to make the technical calculation, you first have to find a normal vector on the upper half-sphere of dimension 2. this can easily be achieved, as you simply use he vector pointing from the origin to an abitrary point on the sphere, normalize it and be fine. For the base part of the surface, I'd recommend regarding the entire sphere and calculate the value of the integral for a new potential which takes the same values on points symmetric with respect to the base part. Once you have calculated the integral, you know that the solution must be half, but since you only want to apply gauss' law, you can drop that part.
    One advice I can give you is that you should introduce spherical coordinates as it would simplify the problem for the case that the surface is the sphere of dimension 2 a lot. However, not having performed this calculation myself, I may a priori assure that it involves a lot of calculations most of which are really technical (and boring ;) ).
     
  5. Jul 28, 2011 #4
    This electric field isn't constant over any radius. If this electric field were the same magnitude at every point on the sphere, then you can use the argument that the flux of the top half of the sphere is taken by using the magnitude of the E field, times the magnitude of the outward area vector, and the cosine of the angle theta. In this case though, we have a E field that isn't constant and changes at every point in space, so you will have to use Stokes or the Divergence Theorem.

    *edit to add: Sorry wasn't paying attention. XD

    You are on the right track with calculating it. So the question is that you want to know how to calculate using Stokes theorem?
     
  6. Jul 29, 2011 #5

    mzh

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    @thepatient: Thanks for the feedback. I know I can apply the divergence theorem and find that the enclosed charge is zero, since [itex]div(E) = 0[\itex]. *But*, I want to carry out surface integration for the sake of it.
    Another dude suggested to introduce spherical coordinates by the common transformation laws: [itex]x=r*sin(theta)*cos(phi)[\itex], [itex]y=r*sin(theta)*sin(phi)[\itex], [itex]z=r*cos(theta)[\itex]. But, what does [itex]dS[\itex] become here?

    @dongo: sure, boring like snit no question, thats why I would use Maple or something. still, I need to know what to calculate.

    Once again, thanks for the feedback so far, really appreciating it.
     
  7. Jul 29, 2011 #6
    O.K., for instance you could use an isomorphism which relates for instance the upper halfsphere (equator not inculded) diffeomorphically (i.e. isomorphically as well as differntiably) to the 2-dimensional Euclidean space. It involves something that is called Gram's determinant or, more generally speaking, the volume form of the sphere (I am saying this determinedly, as the volume of a twodimensional manifolfd has to be regarded as its surface area, but this is just messing with definitions). If you integrate this over the upper sphere, you can, by a suitable change of coordinates (i.e. spherical coordinates where the radius is a constant), after a tedious calculation , be yielded the area of the 2-sphere. (Note that the calculation so far does only include the upper halfsphere. You will have to multiply your result with two. Note also, that we have not included the equator in our calculation. To include it would not change the result as a 1-dimensional submanifold is, in this case, a zero-set of the sphere.
    Actually, you could avoid the whole half-sphere thingy if you simply integrate the sphercal coordinates (that is in this case, the angles) over the suitable interval(s) (you can look that up in any advanced calculus textbook!). The only thing that remains is calculating the Jacobian of the corrdinate change. This is actually simple as you already have the relation between the usual three dimensional coordinates x,y,z, and the angles phi1,phi2. If you have calculated the Jacobian, you will notice that is 2,3-matrix. But to take the determinant you need a 3,3 matrix. By matrix transpoition you can calculate a 3,2-matrix. Then multiply and take the determinant, then take the square root of the determinant. This is the function you will have to include in the intergral to account for the coordinate change.
     
  8. Jul 29, 2011 #7
    ^ You'll have to parametricize the surface in order words into a vector function of two variables instead of 3. Once you parametricize the surface to [itex]\overline{r}[/itex](u,v), you change the integral to the form:

    ∬E⃗(x,y) ⋅n⃗(x,y) ˆdS(x,y) to
    ∬E⃗(x(u),u(v))⋅n⃗(x(u),y(v))dS(u,v)

    Change all the x and y variables to their parametric functions in terms of the parametric variables.

    n is the outward unit vector on the surface, which can be calculated as:
    n =[itex]r[itex]_{u}[/itex] X r[itex]_{v}[/itex]}/|r[itex]_{u}[/itex] X r[itex]_{v}[/itex]|[/itex]

    dS =| r[itex]_{u}[/itex] X r[itex]_{v}[/itex]| dudv

    This will do only one surface, so you will have to add the surface integrals over the half sphere and the plane together.
     
  9. Jul 29, 2011 #8

    mzh

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    I know, yes. phi from 0 to 2pi ("around the equator"), theta from 0 to pi ("over the poles").

    Sorry, how do you mean, i can calculate a determinant from a 3x2 matrix? What do you mean 'multiply'?
    Thanks for the reply. But still, I have trouble figuring out the integration element dS. Please help me with this, it would be very appreciated.
     
  10. Jul 29, 2011 #9

    mzh

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    This is my problem, how do I do the parametrization for the surface in 3D?
     
  11. Jul 29, 2011 #10
    r(u,v) = <x(u), y(v), z(u,v)>

    For the sphere, there are only two things that change while the radius is constant: The angle between the positive z-axis and the radius, and the angle on the x,y plane between the positive x axis and the reflection of the radius onto the x,y plane. You can parametricize using theta and phi.
     
  12. Jul 29, 2011 #11
    In the sphere's case, you will have:
    r(theta, phi) = <x(theta, phi), y(theta, phi), z(phi)>, the cross product will get pretty messy though. There are many other ways you can parametricize using only two variables as well, not just theta phi.
     
  13. Jul 29, 2011 #12

    mzh

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    just for definitions sake, theta is the angle from the positive z-axis to the radius-vector, ranging from 0 to pi ("around the poles"), and phi is the angle "around the equator", ranging from 0 to 2pi, right?
    What do you mean with the angle brackets? And, can I thus write x(theta,phi)=sin(theta)*cos(phi), y(theta,phi)=sin(theta)*sin(phi), z=cos(theta), correct? I mean this is just the parametrization of x,y,z for spherical coordinates where I omitted the 'r', since it is constant. As I understand, i would need to calculate the surface element over which to integrate now... bit unsure how to do that i have to say.
    Thanks for the help again.
     
  14. Jul 29, 2011 #13
    That works. Usually they call phi the angle between the positive z axis and the position vector, but you can call it w if you wanted to.

    The only thing your missing is the radius of the sphere in the parametricization. You can remove roe from the usual parametrization but you must replace it by constant "R".
     
  15. Jul 29, 2011 #14
    As in x=Rsin(theta)cos(phi) and so on.
     
  16. Jul 29, 2011 #15

    mzh

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    yeah, i was thinking something like that. Also, in the beginning i was thinking "hey man, dude. so R is constant, but then what about dr? Ain't 'dr' then equal zero?" But i guess I'm realizing now that we are not integrating over dr, and therefore there is essentially no 'dr'.

    Seriously appreciating the support here on physics forums.
     
  17. Jul 30, 2011 #16
    As a mathematician specializing in differential geometry, I usually work with this http://en.wikipedia.org/wiki/Gramian_matrix when it comes to subamnifolds embedded in the n-dimensional Euclidean space.

    EDIT: Especially this:
    "Given a real matrix A, the matrix ATA is a Gram matrix (of the columns of A), while the matrix AAT is the Gram matrix of the rows of A."
     
    Last edited: Jul 30, 2011
  18. Jul 30, 2011 #17

    mzh

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    thanks. unfortunately, it's bit above my head right now. I believe it should be possible to solve the exercise with the knowledge from the book up to the chapter.
     
  19. Jul 30, 2011 #18

    mzh

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    I tried the following (Maple). This is the transformation [itex]\Psi(\theta,\phi)[/itex]:
    Code (Text):

    x:=(theta,phi)-> R*sin(phi)*cos(theta):
    y:=(theta,phi)-> R*sin(phi)*sin(theta):
    z:=(phi)-> R*cos(phi):
     
    In Maple, I issue the following command (copy/paste if you like):
    Code (Text):

    Matrix(3, 3, {(1, 1) = diff(x(theta, phi), R), (1, 2) = diff(x(theta, phi), theta), (1, 3) = diff(x(theta, phi), phi), (2, 1) = diff(y(theta, phi), R), (2, 2) = diff(y(theta, phi), theta), (2, 3) = diff(y(theta, phi), phi), (3, 1) = diff(z(phi), R), (3, 2) = diff(z(phi), theta), (3, 3) = diff(z(phi), phi)});
     
    Evaluating the differentials gives the below determinant.
    [itex]
    \left[ \begin {array}{ccc} \sin \left( \phi \right) \cos \left(
    \theta \right) &-R\sin \left( \phi \right) \sin \left( \theta \right)
    &R\cos \left( \phi \right) \cos \left( \theta \right)
    \\ \sin \left( \phi \right) \sin \left( \theta
    \right) &R\sin \left( \phi \right) \cos \left( \theta \right) &R\cos
    \left( \phi \right) \sin \left( \theta \right) \\
    \cos \left( \phi \right) &0&-R\sin \left( \phi \right) \end {array}
    \right]
    [/itex]

    I obtain the following value for the determinant:
    [itex]
    - \left( \sin \left( \phi \right) \right) ^{3} \left( \cos \left(
    \theta \right) \right) ^{2}{R}^{2}- \left( \sin \left( \phi \right)
    \right) ^{3} \left( \sin \left( \theta \right) \right) ^{2}{R}^{2}-{
    R}^{2}\sin \left( \phi \right) \left( \sin \left( \theta \right)
    \right) ^{2} \left( \cos \left( \phi \right) \right) ^{2}-{R}^{2}
    \sin \left( \phi \right) \left( \cos \left( \theta \right) \right) ^
    {2} \left( \cos \left( \phi \right) \right) ^{2}
    [/itex]

    Man, does that look correct?
     
    Last edited: Jul 30, 2011
  20. Jul 30, 2011 #19
    Looks about right. You can do a bit of algebra to reduce it a bit.
     
  21. Jul 30, 2011 #20

    mzh

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    [itex]
    simplify(- \left( \sin \left( \phi \right) \right) ^{3} \left( \cos \left(
    \theta \right) \right) ^{2}{R}^{2}- \left( \sin \left( \phi \right)
    \right) ^{3} \left( \sin \left( \theta \right) \right) ^{2}{R}^{2}-{
    R}^{2}\sin \left( \phi \right) \left( \sin \left( \theta \right)
    \right) ^{2} \left( \cos \left( \phi \right) \right) ^{2}-{R}^{2}
    \sin \left( \phi \right) \left( \cos \left( \theta \right) \right) ^
    {2} \left( \cos \left( \phi \right) \right) ^{2});
    [/itex]
    [itex]
    -R^2\sin(\phi)[/itex]
     
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