Surface Integral: Evaluating with Spherical Coordinates

[TheFerruccio]

Homework Statement

Find
\iint\limits_S \mathbf{F}\cdot \hat n\, dA

Homework Equations

\mathbf{F} = [1, 1, a]

S: s^2+y^2+4z^2 = 4, z \geq 0

The Attempt at a Solution

I parameterized in spherical coordinates

x=4\sin{\phi}\cos{\theta}
y=4\sin{\phi}\sin{\theta}
z=\cos{\phi}

Then, I found the surface normal vector, and finding the normal vector is what exploded into something that I couldn't simplify very well. I have a feeling that, because it exploded, that there is a simpler way for me to go about doing this. I thought about using the divergence theorem, but I didn't see how I could use it with an open surface.

 

[vela]

If you have a surface defined implicitly by F(x,y,z)=0, the normal is given by ∇F.

 

[HallsofIvy]

Well, what did you get for that "normal vector"? I, taking the cross product of the the derivatives of \vec{r}= 4 sin(\phi)cos(\theta)\vec{i}+ 4 sin(\phi)sin(\theta)\vec{j}+ cos(\phi)\vec{k}, got, giving the "vector differential of surface area",
4sin^2(\phi)cos(\theta)\vec{i}+ 4sin^2(\phi)sin(\theta)\vec{j}+ 16 sin(\phi)cos(\phi)\vec{k}

With \vec{F}= \vec{i}+ \vec{j}+ a\vec{k} then the integrand is
(4 sin^2(\phi)cos(\theta)+ 4sin^2(\phi)sin(\theta)+ 16a sin(\phi)cos(\phi))d\theta d\phi

I don't see anything terribly difficult to integrate in that- use the standard identity sin^2(\phi)= (1/2)(1- cos(2\phi).

Or, avoid the surface integral entirely by using the divergence theorem: integrate \nabla\cdot \vec{F} over the interior of the ellipsoid.

 

[TheFerruccio]

Well, what did you get for that "normal vector"? I, taking the cross product of the the derivatives of \vec{r}= 4 sin(\phi)cos(\theta)\vec{i}+ 4 sin(\phi)sin(\theta)\vec{j}+ cos(\phi)\vec{k}, got, giving the "vector differential of surface area",
4sin^2(\phi)cos(\theta)\vec{i}+ 4sin^2(\phi)sin(\theta)\vec{j}+ 16 sin(\phi)cos(\phi)\vec{k}

With \vec{F}= \vec{i}+ \vec{j}+ a\vec{k} then the integrand is
(4 sin^2(\phi)cos(\theta)+ 4sin^2(\phi)sin(\theta)+ 16a sin(\phi)cos(\phi))d\theta d\phi

I don't see anything terribly difficult to integrate in that- use the standard identity sin^2(\phi)= (1/2)(1- cos(2\phi).

Or, avoid the surface integral entirely by using the divergence theorem: integrate \nabla\cdot \vec{F} over the interior of the ellipsoid.

Yes, that's what I got. However, you putting "normal vector" in quotes hints at me entirely borking up the terminology. I was basing "normal vector" off of: http://en.wikipedia.org/wiki/Surface_normal The whole reason I got stuck was because I thought I had to divide, then, by the magnitude, so I can get a pure direction. Is this not the case?

My confusion over using the divergence theorem was addressed in the first post. The surface isn't closed (since it's only the top half), and I think I'd be getting a net 0 divergence over the entire volume (the vector field uniformly travels in and out of the object). It doesn't seem like it would be helpful.

Maybe my confusion is from mixing up the concept of calculating flux (which is what I'm doing here) and outright finding the surface area. Is finding the surface area itself where I should be focusing on finding the surface normal?

*edit* also, since the integrand is now in spherical coordinates, isn't the _______ (forgot the name) now \sin{\theta}d\theta d\phi?

 

[vela]

Yes, that's what I got. However, you putting "normal vector" in quotes hints at me entirely borking up the terminology. I was basing "normal vector" off of: http://en.wikipedia.org/wiki/Surface_normal The whole reason I got stuck was because I thought I had to divide, then, by the magnitude, so I can get a pure direction. Is this not the case?

Yes and no. The vector \hat{n} in the integral indeed refers to the unit vector normal to the surface, but to actually evaluate the integral, you also need dA. It turns out, the cross product method of finding the normal automatically gives you dA as well. In other words, using the notation from the wiki article, you have

\hat{n}\,dA = \frac{\partial\mathbf{x}}{\partial s}\times\frac{\partial\mathbf{x}}{\partial t}\,ds\,dt

*edit* also, since the integrand is now in spherical coordinates, isn't the _______ (forgot the name) now \sin{\theta}d\theta d\phi?

In the normal vector you found, there's a common factor of \sin\phi. That's exactly the factor you're referring to here. (Note you switched the roles of theta and phi from how you were using them in the original post.)