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Homework Help: Spherical surface area element

  1. Mar 30, 2014 #1
    See image attached.

    (I've had a google but can't find anything).

    I am trying to understand the expression : Rdθ.2∏Rsinθ

    Here are my thoughts so far:

    Rdθ is the width of a strip, θ being the variable changing/to integrate over, giving arise to the elements.

    2∏Rsinθ must then be the length. However I dont understand this expression.

    I think I understand Rdθ - it is the arc length.

    Here are my length thoughts: From the triangle sin θ =(l/2)/R, where l is the length we are after, and l/2, because this triangle only gives half of the length we are after.

    Ofc this must be wrong, as I have a ∏ unexplained.

    Many thanks to anyone who can help shed some light on this, greatly appreciated !

    Attached Files:

  2. jcsd
  3. Mar 30, 2014 #2


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    It would help if you give us a statement of what problem is being solved. Also, in the figure it doesn't appear that R is the radius of the circle. Or maybe it is, just overwritten on the vertical strip.
  4. Mar 30, 2014 #3
    I am given a surface change densiity σ = P.[itex]\hat{n}[/itex] = P cos θ, where P is a constant, and am multiplying this by the area element to find the total charge.
    (I am then going to apply Coulomb's law to find the total electric field).

    Does this help?

    My apologies, R is the radius of the circle
  5. Mar 30, 2014 #4


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    It doesn't help me. I don't see what you are doing. I don't see any spherical dS element and I don't understand the problem. Maybe someone else will stop by.
  6. Apr 4, 2014 #5
  7. Apr 4, 2014 #6


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    [itex]r\,d\theta[/itex] is the arclength in the direction of increasing [itex]\theta[/itex] (with [itex]\phi[/itex] and [itex]r[/itex] fixed). Arclength in the direction of increasing [itex]\phi[/itex] (with [itex]\theta[/itex] and [itex]r[/itex] fixed) is [itex]r\sin\theta\,d\phi[/itex]. The area of the surface corresponding to [itex][\theta, \theta + \delta\theta]\times[\phi,\phi + \delta\phi][/itex] with [itex]r = R[/itex] constant is then approximately [itex](R\delta\theta)(R \sin (\theta)\delta\phi) = R^2 \sin (\theta)\delta\theta\delta\phi[/itex]. Thus the integral of [itex]\sigma[/itex] over the sphere [itex]r = R[/itex] is
    \int_{r = R} \sigma\,dS = \int_0^{2\pi} \int_0^{\pi} \sigma R^2 \sin \theta\,d\theta\,d\phi
    [/tex] and if [itex]\sigma[/itex] is independent of [itex]\phi[/itex] this reduces to
    R^2\left(\int_0^{2\pi}\,d\phi\right)\int_0^\pi \sigma\sin \theta\,d\theta = 2\pi R^2 \int_0^\pi \sigma\sin\theta\,d\theta.

    Are you given that [itex]\sigma = P \cos \theta[/itex] for constant [itex]P[/itex], or are you given [itex]\sigma = \mathbf{P} \cdot \hat{\mathbf{n}}[/itex] for constant [itex]\mathbf{P}[/itex]? I can't tell whether [itex]\mathbf{P} \cdot \hat{\mathbf{n}} = P\cos\theta[/itex] is something you were given or the reflex action of calling the angle in the definition of the dot product "[itex]\theta[/itex]". But that angle might not be equal to the coordinate [itex]\theta[/itex], so should be given a different name.

    Assuming [itex]\hat{\mathbf{n}}[/itex] is the outward normal to the sphere, you have [itex]\hat{\mathbf{n}} = \hat{\mathbf{r}}[/itex], so [itex]\mathbf{P} \cdot \hat{\mathbf{n}}[/itex] is the radial component of [itex]\mathbf{P}[/itex]. But if [itex]\mathbf{P}[/itex] is constant then its radial component is not constant but depends on both [itex]\theta[/itex] and [itex]\phi[/itex].
    Last edited: Apr 4, 2014
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