Surface integral, grad, and stokes theorem

trelek2
Messages
86
Reaction score
0
Hi
I'm practicing for my exam but I totally suck at the vector fields stuff.
I have three questions:


1.
Compute the surface integral

\int_{}^{} F \cdot dS

F vector is=(x,y,z)
dS is the area differential
Calculate the integral over a hemispherical cap defined by x ^{2}+y ^{2}+z ^{2}=b ^{2} for z>0

2.
If C is a closed curve in the x-y plane use Stokes theorem to show that area enclosed by C is given by:
S= \frac{1}{2} \oint_{}^{C}(xdy-ydx)
I don't see how :(
And then use it to obtain the area S enclose by ellipse \frac{x ^{2} }{a ^{2} }+ \frac{y ^{2} }{b ^{2} } =1. Use the folowing parametrisation: vector r=acos \phi _{x}+bsin \phi _{y} I also can't get PI*a*b...

3. (probably the easiest one)
Calculate grad of scalar field
(a dot r)/r^2
Where a is a constant vector, r is simply the vector (x,y,z). Supposed to use the chain rule...

Thanks in advance for any help. Please also include the answers to 1 and 3 as my book has very few examples and i want to be sure that what I'm doing is correct:)
 
Physics news on Phys.org
Hi trelek2, we are not here to do your homework (or homework-like problems) for you.

Why not make an attempt at each problem and post your work so we can see where you are getting stuck?:smile:

As a hint for problem 1, the integration is easiest in spherical coordinates.
 
I already did 1 and 3, however I'm worried they might be wrong. I got 2Pi*b^2 in the first one. I did use spherical polar coordinates and computed dS by taking the corss product of dr/dphi and dr/dtheta. Then just integration.
And in 3 i get vector a divided by length r... I guess its just one step, I don't know if I'm using the chain rule correctly.
As for 2 I have no clue what to do. I guess curl sth?
 
trelek2 said:
I already did 1 and 3, however I'm worried they might be wrong. I got 2Pi*b^2 in the first one. I did use spherical polar coordinates and computed dS by taking the corss product of dr/dphi and dr/dtheta. Then just integration.

Your method looks good, but you somehow got the wrong answer...you'd b etter post you calculations so I can see where you went wrong...

And in 3 i get vector a divided by length r... I guess its just one step, I don't know if I'm using the chain rule correctly.

You should be getting \frac{\vec{a}-2(\vec{a}\cdot\hat{r})\hat{r}}{r^2}, so you must be applying the chain rule incorrectly...again, post your calculations and I'll be able to see where you are going wrong.

As for 2 I have no clue what to do. I guess curl sth?

As a hint, \vec{\nabla}\times(-y\hat{x}+x\hat{y})=____?
 
alright, great thanks. 2 works fine. The answer to 3 seems obvious now that I see it. So for ex:
grad r^n will be for example nr^(n-2)*r-vector, right?
If your saying the method is right maybe I did get the right answer in 1. As a matter of fact I got 2Pi*b^3, but copied into here incorrectly. If that's still incorrect I totally give up and I'll have to copy all the calculations in here. Thanks again
 
trelek2 said:
alright, great thanks. 2 works fine. The answer to 3 seems obvious now that I see it. So for ex:
grad r^n will be for example nr^(n-2)*r-vector, right?
Yep.:smile:

If your saying the method is right maybe I did get the right answer in 1. As a matter of fact I got 2Pi*b^3, but copied into here incorrectly. If that's still incorrect I totally give up and I'll have to copy all the calculations in here. Thanks again

Your answer is still incorrect; better post your calculation...

EDIT: Your answer is correct: when I first read the problem statement, I had thought that the surface was the entire sphere, re-reading it I see that it is only the z>0 hemisphere.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
7
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
26
Views
3K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
8
Views
2K
Back
Top