Surface integral use stokes/divergence/whatever is convenient

[ArcanaNoir]

Homework Statement

Consider the closed surface S consisting of the graph [itex] z=1-x^2-y^2 [/itex] with [itex] z \ge 0 [/itex] and also the unit disc in the xy plane. Give this surface an outer normal. Compute: [itex] \int \int_s \mathbf{F} \cdot d \mathbf{S} [/itex]

Homework Equations

Stokes theorem, divergence theorem

The Attempt at a Solution

Well the divergence of F is 5.
So I should calculate [tex] \int \int \int_S 5 dV [/tex]

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

 

[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?

 

[ArcanaNoir]

So I really just calculate [tex] \int \int \int_S 5 dV [/tex] ?

so.. would it be [tex] 5 \int \int \int r \; dz \; dr \; d \theta [/tex] ?

How do I bound z?

 

[I like Serena]

Yes and yes. :)

Your problem statement already gives a lower bound for z.
What is the highest value of z where it still has an actual value (and does not become imaginary)?

 

[ArcanaNoir]

oh, 1.
so I get 5 pi ?

 

[I like Serena]

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

 

[ArcanaNoir]

I calculated [tex] 5 \int_0^{2\pi } \int_0^1 \int_0^1 r \; dz \; dr \; d \theta [/tex]

 

[I like Serena]

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

 

[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.

 

[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?

 

[ArcanaNoir]

z=1-r^2?

 

[I like Serena]

z=1-r^2?

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?

 

[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?

 

[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.

 

[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 :)