Change of variables in double integrals textbook problem

Click For Summary
SUMMARY

The discussion focuses on a textbook problem involving the change of variables in double integrals, specifically finding the area of a region bounded by the parabolas \(y = x^2\), \(y = 2x^2\), \(x = y^2\), and \(x = 3y^2\). The variables \(u\) and \(v\) are defined as \(u = \frac{y}{x^2}\) and \(v = \frac{x}{y^2}\), transforming the region \(D\) into a rectangle \(R\) in the \(uv\)-plane defined by \(1 \leq u \leq 2\) and \(1 \leq v \leq 3\). The choice of these variables simplifies the integration process, allowing for easier evaluation of the area.

PREREQUISITES
  • Understanding of double integrals and area calculation
  • Familiarity with parabolic equations and their graphs
  • Knowledge of Jacobian determinants in coordinate transformations
  • Basic algebraic manipulation of equations
NEXT STEPS
  • Study the concept of Jacobian determinants in detail
  • Explore examples of variable transformations in double integrals
  • Learn about different coordinate systems such as polar and cylindrical coordinates
  • Practice solving double integrals with various boundary conditions
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus, as well as anyone interested in mastering the technique of variable substitution in double integrals.

Swatch
Messages
88
Reaction score
0
There is an example in my textbook which I´m having trouble with.

The example is like this.
"
Find the area of the finite plane region bounded by the four parabolas, y=x^2 , y=2x^2 , x=y^2 , and x=3y^2
The region is called D.

Let u=y/x^2 and v=x/y^2

The the region D corresponds to the rectangle R in the uv - plane given by
1≤ u ≤2 and 1≤ v ≤3
"
The rest with the Jacobian determinant and so on I understand.

What I don't understand in this example is the expression
u=y/x^2 and v=x/y^2
I don't see why u and v are chosen this way.
It would be great if someone could explain this to me.
 
Physics news on Phys.org
When u = y/x^2, y = u*x^2 and when v = x/y^2, x = v*y^2. So for example when y ranges from x^2 to 2x^2, u ranges from 1 to 2.
 
I understand this better now.
Thanks
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
21
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K