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im trying to calculate the current running through a sphere while given a current density $$J=Jcos(2\theta)e^{-t/r} \hat r $$ J is given in spherical coordinates. I know that: $$I=\int J\,da $$ But I can't seem to understand how to "build" my integral. I was thinking that i need to do something like $$ \int r\,dr\,sin\theta\,d\theta*J $$ but this gets very messy because of the exponent containing expression with "r". Any tips/help please? Thanks

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  • $\begingroup$ A sphere has a constant radius $R$, therefore you don't have to integrate over that. OTOH, your integral is missing $\phi$. And BTW, your integrals get much more readable if you prefix each $d$ with a thin space which you get using \,. $\endgroup$
    – celtschk
    Commented May 5, 2014 at 9:57

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I am not sure what is exactly your problem with evaluating the integral. Here is what I would do. We have hat $J$ points radially, so the total current entering or exciting a sphere of radius $R$ is simply given by $$ I= \int |J| |da|=\int |J| R^2 d\Omega = \int |J| R^2 d\varphi \sin\theta d\theta.$$ as the normal of the area element $da$ is always parallel to the current. The resulting integrals are not at all difficult to evaluate. We have (evaluating first the integral over $\varphi$) $I = JR^2 \int \cos(2\theta) d\varphi \sin\theta d\theta = 2\pi J R^2 e^{-t/R} \int_{0}^\pi \cos(2\theta) \sin\theta d\theta = - \frac{4}{3} J R^2 e^{-t/R}$$

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  • $\begingroup$ I was given an expresion of J which is affected by the radius "r", therefore I don't understand why there is no need to integrate over r dr also.? the way i see it, the current will run through "many" spheres with radiuses getting bigger and bigger until the current reaches the shpere with the "original" R radius $\endgroup$
    – user3921
    Commented May 5, 2014 at 10:01
  • $\begingroup$ To get from the current density to the current, you should integrate over an area. Your area is the surface of a sphere (as given in the task you have to follow). You should know how to integrate over the surface of the sphere?! Does this involve an integration over $R$???? The answer is no, in fact $R$ is a fixed parameter indicating the size of the sphere (that is why I gave it a different letter from $r$ which was the radial index of the vector field $J$.) $\endgroup$
    – Fabian
    Commented May 5, 2014 at 10:04

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