Calculate the given surface integral [Mathematical physics]

  • Thread starter Mulz
  • Start date
  • #1
113
3

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


[/B]
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.
 
Last edited:

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,979
3,133
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.
 
  • Like
Likes PeroK
  • #3
113
3
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
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,979
3,133
... 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
113
3
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 eachother so it should be 1?
 
  • #6
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,979
3,133
I repeat
Is that equal to ##1~##everywhere?
By "everywhere" I mean all points in space.
 
Top