Flux of a vector field through the boundary of a closed surface

[LSS2303]

I am supposed to calculate the flux of the vector field [MATH] \vec G(r) = x\hat e_x + y \hat e_y + z \hat e_z[/MATH]through the boundary of the volume ? defined by [MATH]x^2 +y^2 +z^2 \leq 1, z \geq 0[/MATH] (the northern hemisphere of the sphere of radius 1 centred at the origin. The formula would be

[MATH] \iint \vec G(r) \cdot d\vec S[/MATH]
for a sphere, the vectorial surface area element is: [MATH] R^2 \sin \theta d \theta d \phi [/MATH]. I don't understand how I can compute this, as [MATH]d \vec S[/MATH] is given in spherical polar coordinates but [MATH]\vec G(r) [/MATH] is given in cartesian coordinates. I am also given the following hint: 'first answer the question why the disc [MATH] x^2+y^2 \leq 1[/MATH]does not contribute, then compute the flux through the curved part of the sphere. Any help would be great!

 

[HallsofIvy]

Well, first, the unit normal to the xy-plane is <0, 0, 1> so that any flux of \(\displaystyle u(x,y,z)dx+ v(x,yz)dy+ w(x,y,z)dz= \int u(x,y,z)(0)+ v(x,y,z)(0)+ w(x,y,z)dz=\int w(x,y,z)dz\). But here w(x,y,z)= z which is 0 on the xy--plane!

Now, the "vectorial surface area element" for a sphere is NOT "\(\displaystyle R^2 sin(\theta)d\theta d\phi\)" because that is not a vector! In polar coordinates, we can write the "position" vector as \(\displaystyle \left< x, y, z\right>= \left< Rcos(\theta)sin(\phi), Rsin(\theta)sin(\phi). Rcos(\phi)\right>\). The derivative with respect to \(\displaystyle \theta\) (so a vector in the \(\displaystyle \theta\) direction and tangent to the sphere) is \(\displaystyle \left<-Rsin(\theta)sin(\phi), Rcos(\theta)sin(\phi), 0\right>\) and the derivative with respect to \(\displaystyle \phi\) (so a vector in the \(\displaystyle \phi\) direction and also tangent to the sphere) is \(\displaystyle \left< Rcos(\theta)cos(\phi), Rsin(\theta)cos(\phi), -Rsin(\phi)\right>\).

The cross product of those two vectors (so normal to the surface) is \(\displaystyle \left<R^2cos(\theta)sin^2(\phi), R^2sin(\theta)sin^2(\phi), R^2sin(\phi)cos(\phi)\right>\) and the "vector surface area element" is \(\displaystyle \left<R^2cos(\theta)sin^2(\phi), R^2sin(\theta)sin^2(\phi), R^2sin(\phi)cos(\phi)\right>d\theta d\phi\).
(\(\displaystyle R^2sin(\theta)d\theta d\phi\) is the length of that, the scalar surface area element.)

Your function G (it isn't really a function of "r") is \(\displaystyle x\vec{i}+ y\vec{j}+ z\vec{k}= (Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}+ Rcos(\phi)\vec{k}\). Take the dot product of that with the vector surface area element and integrate.

 

[LCKurtz]

Hey Halls, were you out partying too long last night? You forgot to use backslash like \cos for your trig functions and \langle and \rangle for your vector brackets. Or did I just happen onto your post before you had a chance to edit it?

 

[HallsofIvy]

I've always thought "sin" and "cos" looked perfectly good without the "\" and I never use "langle" and "rangle"!

 

[LCKurtz]

[MATH]<\rho sin(\phi)cos(\theta), \rho sin(\phi)sin(\theta), \rho cos(\phi) >[/MATH]
versus

[MATH]\langle \rho \sin(\phi)\cos(\theta), \rho \sin(\phi)\sin(\theta), \rho \cos(\phi) \rangle[/MATH]
I guess beauty is in the eye of the beholder.

 

[pka]

[MATH]<\rho sin(\phi)cos(\theta), \rho sin(\phi)sin(\theta), \rho cos(\phi) >[/MATH]versus
[MATH]\langle \rho \sin(\phi)\cos(\theta), \rho \sin(\phi)\sin(\theta), \rho \cos(\phi) \rangle[/MATH]I guess beauty is in the eye of the beholder.

Versus \(\left< \rho \sin(\phi)\cos(\theta), \rho \sin(\phi)\sin(\theta), \rho \cos(\phi) \right>\) using \left< \(\cdots\) \right>.

 

[HallsofIvy]

"Versus"
(Unless we are singing a song.)

 

[LSS2303]

Thank you Hallsoflvy.
There is a typo in my last post. I meant to say that the vectorial surface area element is [MATH] R^2 \sin \theta d \theta d \phi { \hat e_r} [/MATH]. I think my problem is that I don't understand how this is equivalent to the vector you gave me.