Flux Field Across the Boundary of the Sphere

**Lafexlos** · April 4th 2011, 10:52 PM

i'm kinda stuck with this question and dont know where to begin solving.

Work out the outward flux of the field $\vec{F}=xz\hat{i}-zy\hat{j}+yz\hat{k}$ across the boundary of the sphere $x^2+y^2+z^2=1$ . Verify your result using Divergence Theorem.

Since it's the vector calculus, i posted this thread here. hope, didn't made it wrong.

**HallsofIvy** · April 5th 2011, 06:08 AM

Originally Posted by Lafexlos

i'm kinda stuck with this question and dont know where to begin solving.

Work out the outward flux of the field $\vec{F}=xz\hat{i}-zy\hat{j}+yz\hat{k}$ across the boundary of the sphere $x^2+y^2+z^2=1$ . Verify your result using Divergence Theorem.

Since it's the vector calculus, i posted this thread here. hope, didn't made it wrong.

It would have been better to post what you have tried so we could see where you were stuck. There are many ways of doing a problem like this and we cannot be sure which is appropriate for you. In any case, here is how I would do this problem.

Using spherical coordinates, with $\rho$ equal to the radius of the sphere, 1, we can write the "position vector" of any point on the surface as
$\vec{r}(\theta, \phi)= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}$

Differentiating with respect to $\theta$ and $\phi$ gives two vectors in the tangent plane.
$\vec{r}_\theta= -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}$
$\vec{r}_\phi= cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{k}- sin(\phi)\vec{k}$

The cross product of those two tangent vectors, the "fundamental vector product" for this surface, gives the "vector differential of surface area",
$d\vec{S}= \left(cos(\theta)sin^2(\phi)\vec{i}+ sin(\theta)sin^2(\phi)\vec{j}+ sin(\phi)cos(\phi)\right)d\theta d\phi$
(The order in which you take the cross product determines the sign. I chose the way that gives an outward pointing normal [all components positive] because the problem said "the outward flux".)

You want to integrate $\int\int \vec{F}\cdot d\vec{S}$ over the surface of the sphere: $\theta$ from 0 to $2\pi$ , $\phi$ from 0 to $\pi$ . Of course, you will need to change $\vec{F}$ to these coordinates.

**Lafexlos** · April 5th 2011, 08:52 AM

i was stuck at finding $d\vec{S}$ and now i understand the way and see why i couldn't solve.
i put $\phi$ at wrong place hence i got $\vec{r}(\theta, \phi)= cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}+ sin(\phi)\vec{k}$ as position vector.
Thanks for help.

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