Surface integrals/Surface areas of arbitrary domain regions

  • #1

Main Question or Discussion Point

I'm having trouble evaluating this surface integral. This would be very simple to solve if the parameter domain of the variables u and u was a square region. However, that isn't the case here. I've tried using a change of variables and saying that u = r cos x, and v = r sin x. Where 0 < x < 2pi, and 0 < r < 2 for the limits of integration. However, computing the surface differential (absolute value of the cross product of the partial derivatives) using the new variables has become way to complicating and tedious. I end up with about 6 or 7 terms of sines and cosines under a square root that cannot be simplified. Was this the way to approach this problem? I also thought about parametrizing the region R, but I'm not sure how that would work.


Also, since I'm already here, how would I be able to do this for some general region R in the u-v plane? This is what lead me to think of a parametrization of a region R. So basically a parametrization within a parametrization. Mhmm, sounds interesting. Is that possible?

Thank you.

P.S. Tried looking around and haven't been able to find something that could answer my question.
 

Attachments

Last edited:

Answers and Replies

  • #2
18,037
7,391
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Given parametric equations for a surface, x= f(u, v), y= g(u, v), there are two ways of handling surface integrals. One is just to use the formula [itex]dxdy= \frac{\partial x, y}{\partial u,v} dudv[/itex] where [itex]\frac{\partial x,y}{\partial u, v}[/itex] is the Jacobian,
[tex]\left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|[/tex].

More fundamentally, you need the "algebra of differential forms" which is anti-symmetric (to really understand why requires a course in "differential geometry" and "differential forms").

If x= f(u,v) then [itex]dx= f_u du+ f_v dv[/itex]. If y= g(u. v) then [itex]dy= g_u du+ g_v dv[/itex].
Multiplying, [itex]dxdy= (f_u)(g_u)(du)(du)+ (f_u)(g_v)(du)(dv)+ (f_v)(g_u)(dv)(du)+ (f_v)(g_v)(dv)(dv)[/itex]. Since multiplication of differential forms is anti-symmetric, (du)(du)= (dv)(dv)= 0 and (dv)(du)= -(du)(dv) so that reduces to [itex]dxd7= (f_ug_v- g_vf_u)dudv[/itex] as given by the Jacobian.
 
  • #4
Zondrina
Homework Helper
2,065
136
Here's the problem in a latex form:

$$I = \frac{1}{4} \iint_S dS$$

You are given the surface ##\vec \phi(u,v) = <u+v, u^2 + v^2, u-v>## with ##u^2 + v^2 \leq 4##.

For the surface integral of a scalar function, ##dS = ||\vec \phi_u \times \vec \phi_v|| \space dA## or ##dS = \sqrt{z_u^2 + z_v^2 + 1} \space dA##.

The first form of ##dS## seems more appropriate as you are given ##\vec \phi##. So:

$$I = \frac{1}{4} \iint_S dS = \frac{1}{4} \iint_D ||\vec \phi_u \times \vec \phi_v|| \space dA$$

Simply switching to polar co-ordinates from here gives your limits for ##r## and ##\theta##. Then:

$$I = \frac{1}{4} \iint_S dS = \frac{1}{4} \iint_D ||\vec \phi_u \times \vec \phi_v|| \space dA = \frac{1}{4} \iint_{D'} ||\vec \phi_{u=rcos(\theta)} \times \vec \phi_{v=rsin(\theta)}|| \space rdrd\theta$$
 
  • #5
Hk, I solved it not to long after posting it. Sorry. Hopefully someone finds this useful though
 

Related Threads for: Surface integrals/Surface areas of arbitrary domain regions

  • Last Post
Replies
4
Views
2K
Replies
2
Views
4K
Replies
1
Views
2K
  • Last Post
Replies
7
Views
4K
Replies
13
Views
5K
  • Last Post
Replies
2
Views
651
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
Top