Surface Integral in Spherical Coordinates for Arbitrary Vector Field

[Pengwuino]

Homework Statement

I'm looking to do the surface integral of \oint {\vec v \cdot d\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over a} } where v is arbitrary and in spherical coordinates and the surface is the triangle enclosed by the points (0,0,0) -> (0,1,0) -> (0,1,1) -> (0,0,0).

The Attempt at a Solution

Now the obvious method is to convert the differential area to cartesian giving \int_0^y {\int_0^1 {\vec v \cdot dydz} } \hat x. However, I want to know how this would be done in spherical. The limits are what confuse me, would they be something like <br /> \int_{\pi /2}^{\pi /4} {\int_0^{r\sin \theta } { - v \cdot r\sin \theta drd\theta } } \hat \theta? That can't make sense because obviously the integral is all screwed up then... Come to think of it, that area element doesn't even make sense...

 

[HallsofIvy]

What makes you think there is going to be any simple way to do it in spherical coordinates? There is no spherical symmetry so I doubt that it will look at all reasonable in spherical coordinates. You would have to do this "piecewise" taking different limits of integration between the different vertices of the triangle.

 

[Pengwuino]

Yah I am having one of those "I know there's an easy way to do this but that doesn't mean its the only way" moments. I don't expect any thing reasonable. What do you mean taking it "piecewise"? I'm trying to solidify my understanding of integration. I know something might become disgusting to solve, but I want to at least be able to be confident about the setup.

 

[Pengwuino]

Ok after some thought, I'm cool with the area element still being d\vec a = - rdrd\vec \theta. I'm guessing the problem lies with the fact that r and \theta can't be integrated, in a sense, independently in this problem and that's what you mean by the symmetry? It seems the trick with cartesian is that each coordinate can be written as a function of the others or you can hold coordinates constant while you perform integrations. With spherical, are you saying I'd need to do the integration over an infinitesimal integrated "slice" of that triangle? Am i getting warmer?