Surface Integrals of Vector Fields question

In summary, the student is having difficulty understanding where they are stuck in their solution to a surface integral. They parameterize their surface and compute the cross product of the vector field with the normal vector to get the solution. However, when they look at the answer that they get for the cross product, they realize that it is different from the answer that they get from the text. The book uses sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> for their cross product, while the student gets sin(phi)*<cos(theta)sin(phi),sin(theta)sin(
  • #1
physstudent1
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1

Homework Statement


F=<0,3,x^2> computer the surface integral over the hemisphere x^2 + y^2 + z^2 = 9
z greater than or equal to 0, outward pointing normal.


Homework Equations





The Attempt at a Solution



I don't know why I keep getting this problem wrong. The general formula for surface integrals of vector fields is [tex]\int[/tex][tex]\int[/tex]F([tex]\Phi[/tex]([tex]\theta[/tex],[tex]\phi[/tex])*(dot product) n([tex]\theta[/tex],[tex]\phi[/tex]) For a sphere the normal is defined in the chapter of my text as R^2 sin([tex]\phi[/tex])<cos([tex]\theta[/tex])sin([tex]\phi[/tex]),sin([tex]\theta[/tex])sin([tex]\phi[/tex]),cos([tex]\phi[/tex])> However in the solutions manual when they are setting up the Integral they just have the sin([tex]\phi[/tex])<cos([tex]\theta[/tex])sin([tex]\phi[/tex]),sin([tex]\theta[/tex])sin([tex]\phi[/tex]),cos([tex]\phi[/tex])> but no R^2 ? I was looking over a sample problem on Pauls online notes and he did include the R^2 when setting up the integral so I am a little confused They end up getting 9pi/4. Can anyone help this isn't a homework problem I am just studying for my final which is in a few days thanks a lot.
 
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  • #2
I will try to help, but I am by no means a pro at surface integrals. I a having a little difficulty understanding where you are stuck, so let's go through it quickly.

I assume you have
1.) parametrized your surface S as [itex]\vec{r}(\phi,\thete)=<x(\phi,\theta),y(\phi,\theta),z(\phi,\theta)>[/itex]

2.) Computed the cross product [tex]\frac{\partial{r}}{\partial{\phi}}\times\frac{\partial{r}}{\partial{\theta}}[/tex]

and

3.) Written the vector Field restricted to the parametrization of S=[itex]\vec{F}(\vec{r}(\phi,\theta))[/itex]

Yes?
 
  • #3
hey thanks, my surface I have parameterized as (3cos(theta)sin(phi),3sin(theta)sin(phi),3cos(phi)) 0<theta<2pi 0<phi<pi/2

the cross product came out to 9sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> I didn't actually do this cross product but in the chapter of my text the normal vector for a spherical surface is defined as R^2*sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)>

and I am given that the R=3. And the vector field restricted to the parameterization is 27sin(theta)(sin(phi))^2 + 81(cos(theta))^2 * (sin(phi))^3 * cos(phi) However my solution manual has this as 3sin(theta)(sin(phi))^2 + 9(cos(theta))^2 * (sin(phi))^3 * cos(phi) this is because what they are using for their normal is just sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> I don't see why they wouldn't have the R^2 in it it really doesn't make sense to me so that is where I am stuck! Any help would be greatly appreciated.
 
  • #4
Your parameterization is correct.

Your cross-product is also correct.

You are saying though that value that YOU GET for [itex]\vec{F}(\vec{r}(\phi,\theta))[/itex] is different from that of the text because of what they use for the cross-product?

Either you have made a typo, or you should see why that statement does not make sense. The value that they use for n has nothing to do with F(r) does it?
 
  • #5
Sorry if I caused confusion I typed that mess up fast cause I didn't want you to think I wasn't going to respond, I am saying mine and the text's answer vary because of what we got for the cross products. I have the same exact thing as the text for the Parametrization and the Vector field restricted to it, where my answer and the text's stray are when it does the dot product because their cross product answer has no 9 in front of it but mine does.
 
  • #6
Well in that case I would need to see what the book got for a cross-product, but I have a feeling that they either made a typo or what they have written is equivalent and you may just not be seeing how they wrote it. Solutions manuals are very prone to errors and they like to write things in ways that are tricky.

But you and I both got the same thing for the cross-product though.
 
  • #7
ah ok well that makes me feel better than because my prof won't care about simplification, I had the same feeling as you about me not seeing it so I just went ahead and calculated the integral out but I got a different answer. And I got the other problems in the set correct, I just wanted to be sure I wasn't doing the cross product wrong because my finals coming up and this is my first time learning this chapter...heh..skipped a few classes towards the end ...thanks a lot !
 
  • #8
oh by the way the book got sin(phi)*<cos(theta)sin(phi),sin(theta)sin(phi),cos(phi)> for the cross product.
 
  • #9
Ah yes, of course. It is a vector, so they simply removed the factor of 9. You can do that since the information that a vector provides is not lost by doing so.
 
  • #10
ooo, sigh you are correct thanks a lot for the help I feel really stupid now hahah
 

1. What is a surface integral of a vector field?

A surface integral of a vector field is a mathematical operation that calculates the total amount of the vector field that passes through a given surface. It is a way to measure the flux or flow of a vector field across a surface.

2. How is a surface integral of a vector field different from a regular integral?

A surface integral of a vector field is different from a regular integral in that it takes into account not only the magnitude but also the direction of the vector field. It also calculates the integral over a two-dimensional surface instead of a one-dimensional curve.

3. What is the significance of surface integrals of vector fields in physics?

Surface integrals of vector fields are important in physics because they can be used to calculate physical quantities such as electric flux, magnetic flux, and fluid flow through a surface. They are also used in the study of fluid dynamics and electromagnetism.

4. How is the surface chosen for a surface integral of a vector field?

The surface for a surface integral of a vector field is typically chosen based on the problem at hand. It can be any two-dimensional surface, such as a plane, a sphere, or a curved surface. The surface must be closed and have a defined orientation.

5. What are some real-life applications of surface integrals of vector fields?

Surface integrals of vector fields have many practical applications, including calculating the flow of air over an airplane wing, determining the strength of an electric field in a given region, and analyzing the flow of fluids in a pipe. They are also used in computer graphics to render 3D images and in geology to study the movement of fluids in the Earth's crust.

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