Alright, so I forgot how to do Surface Integrals

[Poop-Loops]

Homework Statement

I have a vector function, and I need to take the surface integral of it over a hemisphere, top half only. I'm "confirming" the divergence theorem by doing a volume integral and surface integral. Already did the volume one so I have something to compare to already.

The Attempt at a Solution

Yeah yeah, "look at your book!"

I've checked my calc book, my mathematical physics book, and my E&M book (which assigned the problem) and for some reason I just can't get it.

EDIT: Also, I should mention that the vector isn't given in "i,j,k" form, but "r, theta, phi", which is even more confusing for me.

The books either use a simple cube (ya thanks) or cylindrical coordinates, which still makes it easier for me to grasp.

Let me see if I am thinking of this correctly:

dS will be d(theta)d(phi) for the top half, and should be d(r)d(phi) for the disk (My phi goes from 0 to 2pi). However I think I am missing something (Jacobian?). Besides that, I need to dot the function vector with the normal vector, yes?

The book says I need to take the gradient of the surface function (and normalize it) to make it a normal vector. So for the top half it would just be <r,0,0>? And I can't figure it out for the disk at the bottom. I guess I could use -k and then just transformed into spherical coordinates?

 

[Hurkyl]

dS will be d(theta)d(phi) for the top half, and should be d(r)d(phi) for the disk (My phi goes from 0 to 2pi). However I think I am missing something (Jacobian?).

Yes -- you're missing the area. dS is an area element, so if you parametrize it with, say, $d\theta d\phi$, then you need to include the area of the sphere element spanned by that differential.

Besides that, I need to dot the function vector with the normal vector, yes?

and $d\vec{S}$ is $\bf{\hat{n}}dS$, so yes.

The book says I need to take the gradient of the surface function (and normalize it) to make it a normal vector. So for the top half it would just be <r,0,0>?/

That points in the right direction (radially outward), but that's not a unit vector.

And I can't figure it out for the disk at the bottom. I guess I could use -k and then just transformed into spherical coordinates?

Sounds good.

 

[Poop-Loops]

Yes -- you're missing the area. dS is an area element, so if you parametrize it with, say, $d\theta d\phi$, then you need to include the area of the sphere element spanned by that differential.

Limits of integration? Or are you talking about something else?

and $d\vec{S}$ is $\bf{\hat{n}}dS$, so yes.

That points in the right direction (radially outward), but that's not a unit vector.

Ok, so if I had <1,0,0>, that would work, right? Since it would be 1 unit in the "r" direction, so radially outward.

 

[Poop-Loops]

I hate myself and I want to die.

I started reading different sections of the book (going over curvilinear coordinates) and at the bottom of one of the pages, it gives me exactly what I needed... what da is when I'm integrating over a sphere, and when it's a disk.

Should I switch my major?

EDIT: Oh, and 5 minutes after my discovery, I am done with the problem. I spent over an hour on it before that.

I really want to cry...

 

[Hurkyl]

Just an hour? Pfft. I've spent days, maybe even weeks wondering about something before I discovered it was in one of the books I have lying around!

Let this be a lesson in research -- knowing where to find information is almost as good as knowing that information.

"Look at your book" isn't idle advice: it's a vital habit for a mathematician!

 

[Poop-Loops]

Yeah. Every time I think I know something, it turns out I don't. I was looking under the "divergence" and "divergence theorem" sections, when I should have been looking at the curvilinear section. I mean, I already know that stuff, and I didn't think the author would just idly throw in "Oh yeah, use *this* for da using these coordinates in these 2 circumstances."

And actually I lost track of time. 'twas already two hours and I had also spent another 2 at school. :(

 

[Mindscrape]

Don't worry, I don't think I would be able to do a surface integral without symmetry. I would either have to think really hard about it for a day and derive it myself, or look it up. While looking something up is a time saver, there is always something to be said about deriving math.

 

[Poop-Loops]

To sort of go against my thread on professors only doing proofs: yes, I find I learn something better when I do it the hard way.