Evaluate Area of R: Change of Variables

Click For Summary

Homework Help Overview

The problem involves evaluating the area of an elliptical region defined by the equation x²/3² + y²/2² = 1 in the first quadrant, using a change of variables to u and v where x = 3u and y = 2v. Participants are discussing how to set up the double integral for this transformation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to establish the correct form of the double integral and the appropriate functions to use for the transformation. Questions arise regarding the setup of the Jacobian and the integrand, as well as the limits of integration in the new uv plane.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the integral setup. Some have provided insights into the transformation process and the nature of the new region in the uv plane, while others are questioning the specific forms of the integrand and the limits of integration.

Contextual Notes

There is some uncertainty regarding the correct interpretation of the transformed region and the implications of the substitution on the integral setup. Participants are also considering the graphical representation of the region in the uv plane.

Dustinsfl
Messages
2,217
Reaction score
5
Let R be the elliptical regin in the first quadrant by x^2/3^2+y^2/2^2=1. x=3u, y=2v, evaluate the area of R.

How do I setup the double integral? By this I mean what do I put in for f(g(u,v),h(u,v))jacobian dudv?

I know the bounds of u and v and the Jacobian but I not sure what to do for g and h. Would that be just pluging 3u and 2v into x^2 and y^2, respectively? But what happens to the 1 in the eqaution?

So by solving the (u,v) cordinates, one comes up with (0,0); (1,0); and (0,1).
The Jacobian is 6.
 
Physics news on Phys.org
When you substitute your u's and v's for the x's and y's of the equation, for your region A in the xy plane you get a new region B in the uv plane. The point of the substitution is that instead of integrating

[tex]Area = \int \int_A 1\ dydx[/tex]

over the region A in the xy plane, which is "ugly", you can instead calculate

[tex]Area = \int \int_B 1 J\ dudv[/tex]

using u-v limits in the uv plane. If your new region B is nice, you may easily work out the new integral, or even avoid working it at all.
 
So the integral is just 0<=u<=1 and 0<=v<=-u+1 dvdu with no equation in the the integral?
 
Dustinsfl said:
So the integral is just 0<=u<=1 and 0<=v<=-u+1 dvdu with no equation in the the integral?

No. You need to draw a graph of your uv equation in the uv plane and use it to figure out correct limits.
 
The drawing is a triangle and those are the bounds. I am not concerned about them. I am trying to figure out what f(g(u,v),h(u,v)) are inside the integrals. Is it 1? Is it u+v not considering the 1 or is it u+v-1?
 
Last edited:
Dustinsfl said:
The drawing is a triangle and those are the bounds. I am concerned about them. I am trying to figure out what f(g(u,v),h(u,v)) are inside the integrals. Is it 1? Is it u+v not considering the 1 or is it u+v-1?

Are you trying to tell me that when you substitute x=3u, y=2v into the equation:

[tex]\frac {x^2}{3^2}+\frac {y^2}{2^2}=1[/tex]

that the graph you get in the uv plane is a triangle?

Your integrand, once you figure out the proper area, will be 1 times the Jacobian.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K