Surface Integral Limits: Solving for u and v

Woolyabyss
Messages
142
Reaction score
1

Homework Statement


Problem is in image uploaded

Homework Equations


n/a

The Attempt at a Solution


x = u, y = v and z = 1 - u - v

∂r/∂u × ∂r/∂v = i + j + k
F dot N = u^2 + 3v^2

∫∫(u^2 + 3v^2 )dudv

My problem is I'm not sure what I should take as the limits?
Should I flip around the order of integration (dvdu) and have v go from 0 to 1-u and have u go from 0 to 1?
 

Attachments

  • Surface Integral.jpg
    Surface Integral.jpg
    30.8 KB · Views: 401
Physics news on Phys.org
Woolyabyss said:

Homework Statement


Problem is in image uploaded

Homework Equations


n/a

The Attempt at a Solution


x = u, y = v and z = 1 - u - v

∂r/∂u × ∂r/∂v = i + j + k
F dot N = u^2 + 3v^2

∫∫(u^2 + 3v^2 )dudv

My problem is I'm not sure what I should take as the limits?
Should I flip around the order of integration (dvdu) and have v go from 0 to 1-u and have u go from 0 to 1?

It doesn't matter which order you use for the integration. You have a triangle in the uv plane and you can do it either way. It isn't clear why your author changes the names of the variables from xy to uv instead of letting x and y just represent themselves. Set it up like you would any double integral.
 
They have given you the limits in the problem, where they write ##D=\{(u,v)\ :\ 0\leq u\ ...##. Have you tried using those? What did you get?
 
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...
Back
Top