Surface Integral (Divergence Theorem?)

In summary, the conversation discussed a problem involving the integration of a closed surface using the divergence theorem. The individual had added a surface z=0 to make the surface closed and performed the integration, obtaining a value of 972pi. The question then arose about removing the portion that the surface z=0 contributes, but it was determined that it would be 0 due to the unit normal being \hat{k} and the k-component of F being 0 when z=0. The individual also clarified that this was not a take-home exam problem, but rather a problem from an old final exam.
  • #1
jegues
1,097
3

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, [tex]\hat{k}[/tex] and that will be dotted with the z component of F which is,

[tex]z^{2}[/tex] but z=0 so it's going to be 0.

Is what have done the right approach?

Thanks again!
 

Attachments

  • 2009Q3.jpg
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  • #2
Is this a take-home exam problem?
 
  • #3
LCKurtz said:
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.
 
  • #4
jegues said:

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, [tex]\hat{k}[/tex] and that will be dotted with the z component of F which is,

[tex]z^{2}[/tex] but z=0 so it's going to be 0.

Is what have done the right approach?

Thanks again!

LCKurtz said:
Is this a take-home exam problem?

jegues said:
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.
 
Last edited:
  • #5
LCKurtz said:
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?
 

Attachments

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  • #6
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...
 
  • #7
LCKurtz said:
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 [tex]\hat{k}[/tex] dotting this with the k-component of F when z=0 will simply provide us with the integral of 0.

Is this correct?
 
  • #8
jegues said:
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 [tex]\hat{k}[/tex] 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.
 

What is a surface integral?

A surface integral is a mathematical concept used in vector calculus to calculate the flux of a vector field through a surface. It involves integrating a scalar or vector function over a surface in three-dimensional space.

What is the divergence theorem?

The divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem, is a fundamental theorem in vector calculus that relates the flux of a vector field through a closed surface to the volume integral of the field's divergence over the enclosed volume.

How is the divergence theorem related to surface integrals?

The divergence theorem states that the surface integral of a vector field over a closed surface is equal to the volume integral of the field's divergence over the enclosed volume. In other words, it provides a way to calculate surface integrals using volume integrals, which can often be easier to evaluate.

What are some practical applications of surface integrals and the divergence theorem?

Surface integrals and the divergence theorem are used in many fields of science and engineering, including fluid dynamics, electromagnetism, and heat transfer. They can be used to calculate volumes, fluxes, and other physical quantities in three-dimensional systems.

Are there any limitations or special cases of the divergence theorem?

While the divergence theorem is a powerful tool for calculating surface integrals, it only applies to certain types of surfaces and vector fields. In particular, the surface must be closed, and the vector field must be continuously differentiable in the region enclosed by the surface. Additionally, the divergence theorem does not apply to surfaces that intersect themselves or have sharp corners.

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