Surface Integral (Divergence Theorem?)

[jegues]

Homework Statement

See figure attached for problem statement.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

What I decided to do was add a surface z=0 so that S became a closed surface.

Then I preformed the integration using divergence theorem and obtained a value of 972pi

Now all I have to do is remove the portion that the surface z=0 contributes.

How do I figure this out? I'm looking at the given F and I can see that the normal for z=0 will simply be, $$\hat{k}$$ and that will be dotted with the z component of F which is,

$$z^{2}$$ but z=0 so it's going to be 0.

Is what have done the right approach?

Thanks again!

 

[LCKurtz]

Is this a take-home exam problem?

 

[jegues]

Is this a take-home exam problem?

No, I am studying for my upcoming final exam using old final exams. It was an exam problem from 2009.

 

[LCKurtz]

Homework Statement

See figure attached for problem statement.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

What I decided to do was add a surface z=0 so that S became a closed surface.

Then I preformed the integration using divergence theorem and obtained a value of 972pi

Now all I have to do is remove the portion that the surface z=0 contributes.

How do I figure this out? I'm looking at the given F and I can see that the normal for z=0 will simply be, $$\hat{k}$$ and that will be dotted with the z component of F which is,

$$z^{2}$$ but z=0 so it's going to be 0.

Is what have done the right approach?

Thanks again!

Is this a take-home exam problem?

No, I am studying for my upcoming final exam using old final exams. It was an exam problem from 2009.

OK, that's good. I will try to delete another post where I asked the same question.

As for your solution. First, your surface isn't just in the first quadrant. Second, when you set up your triple integral, you don't plug the z on the surface in the integrand. The inner integral will have limits z going from z =0 to z on the surface.

 

[jegues]

OK, that's good. I will try to delete another post where I asked the same question.

As for your solution. First, your surface isn't just in the first quadrant. Second, when you set up your triple integral, you don't plug the z on the surface in the integrand. The inner integral will have limits z going from z =0 to z on the surface.

I only drew the surface in the first quadrant to get an idea of what it looked like, I knew it was in all quadrants but perhaps I should do a full drawing next time.

Alrighty here's my second crack at it. (See figure attached)

Any problems?

 

[LCKurtz]

Your setup is OK. I didn't check all your steps but at most there would only be algebra errors, if any. Looks like you're good to go...

 

[jegues]

Your setup is OK. I didn't check all your steps but at most there would only be algebra errors, if any. Looks like you're good to go...

Also do I have to think about removing the portion that the surface z=0 contributes?

I think if you were to calculate that it would be 0 anyways.

We know that the unit normal to this surface will simply be $$\hat{k}$$ dotting this with the k-component of F when z=0 will simply provide us with the integral of 0.

Is this correct?

 

[LCKurtz]

Also do I have to think about removing the portion that the surface z=0 contributes?

I think if you were to calculate that it would be 0 anyways.

We know that the unit normal to this surface will simply be $$\hat{k}$$ dotting this with the k-component of F when z=0 will simply provide us with the integral of 0.

Is this correct?

Yes. I didn't comment about that part because you were thinking correctly. If it wasn't zero you would have to account for it.