# Surface integral use stokes/divergence/whatever is convenient

1. Feb 18, 2012

### ArcanaNoir

1. The problem statement, all variables and given/known data
Consider the closed surface S consisting of the graph $z=1-x^2-y^2$ with $z \ge 0$ and also the unit disc in the xy plane. Give this surface an outer normal. Compute: $\int \int_s \mathbf{F} \cdot d \mathbf{S}$

2. Relevant equations

Stokes theorem, divergence theorem

3. The attempt at a solution

Well the divergence of F is 5.
So I should calculate $$\int \int \int_S 5 dV$$

I'm not really sure where to go with this.

2. Feb 18, 2012

### I like Serena

That expression already looks a lot simpler then the double integral! ;)

So how to calculate a 3 dimensional integral?
Perhaps you should switch to cylindrical coordinates to make it easier?

3. Feb 18, 2012

### ArcanaNoir

So I really just calculate $$\int \int \int_S 5 dV$$ ?

so.. would it be $$5 \int \int \int r \; dz \; dr \; d \theta$$ ?

How do I bound z?

4. Feb 18, 2012

### I like Serena

Yes and yes. :)

What is the highest value of z where it still has an actual value (and does not become imaginary)?

5. Feb 18, 2012

### ArcanaNoir

oh, 1.
so I get 5 pi ?

6. Feb 18, 2012

### I like Serena

Almost.
I get a slightly different result.
Perhaps one of us made a calculation error?

7. Feb 18, 2012

### ArcanaNoir

I calculated $$5 \int_0^{2\pi } \int_0^1 \int_0^1 r \; dz \; dr \; d \theta$$

8. Feb 18, 2012

### I like Serena

That's not quite right.
The bounds of r depend on z.

9. Feb 18, 2012

### ArcanaNoir

Are we integrating over a cylinder with height 1 and radius 1? If we are, we are allowed to use volume formulas. We aren't supposed to actually calculate stuff.

10. Feb 18, 2012

### I like Serena

No, it's not a cylinder.

The graph of $z=1−x^2−y^2$ is a paraboloid that extends downward.
You can think of it as a kind of rounded cap of which you need the volume.

Can you convert the equation of that graph to cylindrical coordinates?

11. Feb 18, 2012

### ArcanaNoir

z=1-r^2?

12. Feb 18, 2012

### I like Serena

Yep.
The r in this equation is the upper bound for the r in your integral.

Can you write this upperbound of r as a function of z?

13. Feb 18, 2012

### ArcanaNoir

I shouldn't have to. I'm doing something wrong. If I actually calculate an integral I've done it wrong and missed the point here. Perhaps some other theorem would be a better approach?

14. Feb 18, 2012

### I like Serena

Seems unlikely.

You would not supposed to be calculating the double integral with the dot product.
That is indeed a lot of work.

But you would need to calculate the volume integral.
Do you have a formula handy for the volume of a paraboloid?

Otherwise, you will have to calculate it.
There is an easier way however if you consider the volume to consist of a number of stacked circle disks.
Each circle disk has volume $\pi r^2 dz$ with $r^2$ being equal to $1-z$.
This should be easy to integrate.

15. Feb 18, 2012

### ArcanaNoir

Okay, I'm going to let this problem go for a while. I'm not getting any of the other problems either, so I'm going to go back to the drawing board and study the theorems some more. thanks for the help so far :)