1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Vector calculus - Divergence Theorem

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\int_{s} \vec{A} \cdot d\vec{a}[/tex] given [tex]\vec{A} = ( x\hat{i} + y\hat{j} + z\hat{k} ) ( x^2 + y^2 + z^2 )[/tex] and the surface S is defined by the sphere [tex]R^2 = x^2 + y^2 + z^2[/tex] directly and by Gauss's theorem.

    2. Relevant equations

    [tex]\int_{s} \vec{A} \cdot d\vec{a} = \int_{V} \nabla \cdot \vec{A} da = \int\int\int \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 A_{r}) r^2 \sin\theta dr d\theta d\phi = \int\int\int \frac{\partial}{\partial r}(r^2 A_{r}) \sin\theta dr d\theta d\phi = 4 \pi R^5[/tex]

    3. The attempt at a solution

    Given the answer, I was able to work backwards using limits of integration of 0 to 2pi for phi, 0 to pi for theta and 0 to R on r to find out that A = r^3. Now, the question I have is how can I show that A = r^3 given the A in cartesian coordinates? Since [tex]r=\sqrt{x^2 + y^2 + z^2}[/tex] its clear that [tex]A=r^2(x\hat{i} + y\hat{j} + z\hat{k})[/tex] but to call the vector valued term r is not agreeing with me. Obviously, since the surface is a sphere, I thought it would be easier to use spherical coordinates...

    Edit: Using cartesian coordinates I can find [tex]\nabla \cdot \vec{A} = 5(x^2 + y^2 + z^2) = 5r^2[/tex] but using the formula given for spherical coordinates, I would need to show A = r^3 to do a full solution in spherical.
    Last edited: Feb 17, 2007
  2. jcsd
  3. Feb 17, 2007 #2


    User Avatar
    Science Advisor

    What do you mean "A= r3"? What is A? There is no A in the statement of the problem.
  4. Feb 17, 2007 #3
    A is given, it is the vector. It's in the first line... Of course, to solve for A in spherical I worked backwards from the answer (I needed an R^5 in the final answer, and the divergence in spherical gives me an integral of a partial of r^2 * A, so obviously r^2 * A needs to have a 5th power, thus A = r^3, working it through, it does come out correctly).

    When I try to convert the given A into spherical directly, I can only get as far as [tex]\vec{A} = r^2(x\hat{i}+y\hat{j}+z\hat{k})[/tex]

    Another question I have is whether it would be easier to show it directly in cartesian or spherical...I'm leaning towards cartesian because A is given in cartesian.
    Last edited: Feb 17, 2007
  5. Feb 18, 2007 #4


    User Avatar
    Homework Helper
    Gold Member

    As Halls pointed out, that's not right.

    The conversion from cartesian to spherical, and vice versa, should be found in any vector calculus text.

    (or try, http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/body.htm [Broken])

    The integral [itex]\int_{s} \vec{A} \cdot d\vec{a}[/itex] over [itex]x^2 + y^2 + z^2=R^2[/itex] should be easier to evaluate in spherical coordinates because of the symmetry (ie, you're integrating over a sphere).
    Last edited by a moderator: May 2, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook