Calculate the given surface integral [Mathematical physics]

In summary: You can't get an integral to work by ignoring all the points where the integrand is not equal to 1.You have to be careful here. When you find the divergence in cartesian coordinates, you get$$\vec{\nabla} \cdot \vec{F} = 1-\frac{2(x^2+y^2)}{(x^2+y^2)^2}+\frac{2}{x^2+y^2}$$Is that equal to ##1~##everywhere?By "everywhere" I mean all points in space. You can't get an integral to work by ignoring all the points where the integrand is not equal to 1.In summary, the conversation
  • #1
Mulz
124
5

Homework Statement


Calculate

[tex] \int_{S} \vec{F} \cdot d\vec{S} [/tex] where

[tex] \vec{F} = z \hat{z} - \frac{x\hat{x} + y \hat{y} }{ x^2 + y^2 } [/tex]

And S is part of the Ellipsoid [tex] x^2 + y^2 + 2z^2 = 4 , z > 0 [/tex] and the normal directed such that

[tex] \vec{n} \cdot \hat{z} > 0 [/tex]

Homework Equations


All the above. There are the Gauss integrals

[tex] \int_{S} \vec{F} \cdot d\vec{S} = \int_{V} \vec{\nabla} \cdot \vec{F} dV [/tex]

The Attempt at a Solution


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I tried solving the problem by introducing

[tex] \rho \hat{\rho} = x \hat{x} + y \hat{y} [/tex]

Thus expressing the vectorfield in cylindrical coordinates. I also drew the figure which resulted in a half sphere with origo at (0,0.0), top at [tex] (0,0,\sqrt{2}) [/tex] and the rest in XY has 2.

In order for me to solve the problem with Gauss I had to enclose the bottom with another surface.

Because the field is now

[tex] \vec{F} = z\hat{z} - \frac{1}{\rho} \hat{\rho} [/tex]. One can easily see that it is undefined when [tex] \rho \rightarrow 0 [/tex] So I will have to introduce another figure, perhaps a cylinder through origo and the subract this part from the total integral. I don't know how to do this or if I can.

I can also integrate the problem by dividing up the force field but I don't know how to do then. I'm also not sure if it is wise to implement cylindrical coordinates when the figure is clearly more spherical. Makes using the surface integral impossible since it encompasses a cylinder surface and not sphere.
 
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  • #2
I have a few questions you might find helpful.
1. Must you do the surface integral or can you do the volume integral and then invoke the divergence theorem? If the former, then you can still do the volume integral and use the result as an independent check of your answer.
2. Have you found ##\hat n##? If so, what is it?
3. Are you sure you have the correct form for the vector field ##\vec F##? If so, then you have purely a math problem because the two terms in ##\vec F## are dimensionally incompatible (mathematicians pay little attention to dimensions.) However, it makes the algebra annoyingly cumbersome because one may not use dimensional analysis as a means of checking one's algebra at the different stages of development.
 
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  • #3
kuruman said:
I have a few questions you might find helpful.
1. Must you do the surface integral or can you do the volume integral and then invoke the divergence theorem? If the former, then you can still do the volume integral and use the result as an independent check of your answer.
2. Have you found ##\hat n##? If so, what is it?
3. Are you sure you have the correct form for the vector field ##\vec F##? If so, then you have purely a math problem because the two terms in ##\vec F## are dimensionally incompatible (mathematicians pay little attention to dimensions.) However, it makes the algebra annoyingly cumbersome because one may not use dimensional analysis as a means of checking one's algebra at the different stages of development.

I'm not sure how to do the volume integral. I try to use the [tex] \vec{\nabla} \cdot \vec{F} [/tex] and see that it yields 1. Thus the volume integral is simply the volume of the ellipsoid. This is a wrong answer as the answer sheet says that it should be [tex] \vec{\nabla} \cdot \vec{F} = 1-2\pi \delta^{2}(\vec{\rho}) [/tex] but I only get it to one when doing the calculations. The second term is the "line source density", that is [tex] kh = -2\pi \sqrt{2} [/tex] and doing the volume integral of the 1 and adding in this second term gives the right answer. I'm not sure why it is that way. The root 2 is the height.

2. I don't know how to get [tex] \hat{n} [/tex]. On the bottom surface it's obviously -z but I'm not sure what it is on the ellipsoid. Maybe I can use [tex] d\vec{S} = r^2d\theta d\phi \hat{n} [/tex]
 
  • #4
Mulz said:
... but I only get it to one when doing the calculations.
You have to be careful here. When you find the divergence in cartesian coordinates, you get
$$\vec{\nabla} \cdot \vec{F} = 1-\frac{2(x^2+y^2)}{(x^2+y^2)^2}+\frac{2}{x^2+y^2}$$Is that equal to ##1~##everywhere?
To get ##\hat n##: remember that on the surface of the ellipsoid you have ##\varphi (x,y,z)=x^2+y^2+2z^2=4=const##. In what direction is ##\vec{\nabla}\varphi(x,y,z)## on the surface of the ellipsoid? In other words, pretend that the ellipsoid is a charged conductor and you are looking for the direction of the electric field on its surface.
 
  • #5
kuruman said:
You have to be careful here. When you find the divergence in cartesian coordinates, you get
$$\vec{\nabla} \cdot \vec{F} = 1-\frac{2(x^2+y^2)}{(x^2+y^2)^2}+\frac{2}{x^2+y^2}$$Is that equal to ##1~##everywhere?
To get ##\hat n##: remember that on the surface of the ellipsoid you have ##\varphi (x,y,z)=x^2+y^2+2z^2=4=const##. In what direction is ##\vec{\nabla}\varphi(x,y,z)## on the surface of the ellipsoid? In other words, pretend that the ellipsoid is a charged conductor and you are looking for the direction of the electric field on its surface.
The two last terms in the divergence take out each other so it should be 1?
 
  • #6
I repeat
kuruman said:
Is that equal to ##1~##everywhere?
By "everywhere" I mean all points in space.
 

1. What is a surface integral?

A surface integral is a mathematical tool used to determine the flux, or flow, of a vector field through a surface. It involves calculating the dot product of the vector field and a small element of the surface, and then integrating over the entire surface.

2. How is a surface integral different from a regular integral?

A regular integral involves integrating over a one-dimensional interval, while a surface integral involves integrating over a two-dimensional surface. This means that in a surface integral, we are calculating the area under a curve on a surface, rather than just a line.

3. What is the importance of surface integrals in mathematical physics?

Surface integrals are used in mathematical physics to calculate various physical quantities, such as electric and magnetic flux, work done by a force, and fluid flow. They are essential in understanding and solving problems related to vector fields and surfaces in three-dimensional space.

4. How do you calculate a surface integral?

To calculate a surface integral, you first need to parameterize the surface using two variables, usually u and v. Then, you need to find the normal vector to the surface at each point and calculate the dot product with the vector field. Finally, you integrate the dot product over the surface using the limits of the parameterized variables.

5. What are some real-world applications of surface integrals?

Surface integrals have many real-world applications, such as calculating the flow of a fluid through a surface, determining the amount of heat transfer on a curved surface, and finding the electric flux through a closed surface. They are also used in computer graphics to render 3D objects and in engineering to analyze stress and strain on surfaces.

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