Parametric Integral: Understanding Math Problem

Click For Summary
SUMMARY

The discussion focuses on the integration of the function ƒ: ℝ² → ℝ defined by (x,y) ↦ z = ƒ(x,y) = y² over the top half of the unit circle, specifically in the context of a double integral. The user seeks clarification on the correct setup of the double integral, which is expressed as ∫∫_M f(x,y) dA = ∫∫_M y² dy dx, and questions the bounds for this integral. The conversation reveals that the author of the referenced material employs a single variable integral along the curved top arc, transitioning to higher dimensions later in the text.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with parametric equations and parameterization
  • Knowledge of integration bounds in polar coordinates
  • Basic concepts of differential forms
NEXT STEPS
  • Study the derivation of double integrals in polar coordinates
  • Learn about parameterization techniques for curves and surfaces
  • Explore the concept of differential forms in multivariable calculus
  • Review examples of single variable integrals along curves
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable integration and differential forms, as well as anyone seeking to deepen their understanding of parameterization in mathematical contexts.

sponsoredwalk
Messages
531
Reaction score
5
Hi, I'm having some trouble understanding what's going on when integrating
the region M on page 10 of http://www.math.boun.edu.tr/instructors/ozturk/eskiders/fall04math488/bachman.pdf" , It may just be the language.

ƒ : ℝ² → ℝ defined by (x,y) ↦ z = ƒ(x,y) = y² is the function we're
integrating over the top half of the unit circle.

1: I think what he's trying to communicate in this derivation is the
standard double integral, [itex]\int \ \int_M \ f(x,y) \ dA \ = \ \int \ \int_M \ y^2 \ dy \ dx[/itex].
Is that correct? You'll notice he jumps straight into his paramaterization
but would what I've just done here be right?

2: If so then would the bounds on the integral become:

[itex]\int \ \int_M \ f(x,y) \ dA \ = \ \int_{-1}^1 \ \int_0^{( \sqrt{1 - x^2})} \ y^2 \ dy \ dx[/itex] ?

3: If that is correct then I think it would explain why the author chose to
set up a paramaterization of the region M. When he goes on to show that
the unit circle can be paramaterized in different ways it reduces a double
integral to a single integral & is just easier. Is that why?

4: I've never seen anyone paramaterize double integrals in the way he
does, could you recommend some reading material that explains what he
is doing as I can't seem to find any myself.

I have more questions, mainly to do with pages 11-14 where, I think, he is
deriving differential forms (in my meagre estimation) but I'll hold off for
now, thanks for any assistance! :biggrin:
 
Last edited by a moderator:
Physics news on Phys.org
sponsoredwalk said:
Hi, I'm having some trouble understanding what's going on when integrating
the region M on page 10 of http://www.math.boun.edu.tr/instructors/ozturk/eskiders/fall04math488/bachman.pdf" , It may just be the language.

ƒ : ℝ² → ℝ defined by (x,y) ↦ z = ƒ(x,y) = y² is the function we're
integrating over the top half of the unit circle.

1: I think what he's trying to communicate in this derivation is the
standard double integral, [itex]\int \ \int_M \ f(x,y) \ dA \ = \ \int \ \int_M \ y^2 \ dy \ dx[/itex].
Is that correct? You'll notice he jumps straight into his paramaterization
but would what I've just done here be right?

2: If so then would the bounds on the integral become:

[itex]\int \ \int_M \ f(x,y) \ dA \ = \ \int_{-1}^1 \ \int_0^{( \sqrt{1 - x^2})} \ y^2 \ dy \ dx[/itex] ?

No, that isn't what he is doing on page 10. He is doing a single variable integral along the curved top arc. He doesn't up the dimension until page 14.

Disclaimer: Don't expect too much more from me on this topic; I'm not a differential forms kind of guy.:cool:
 
Last edited by a moderator:

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
20
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K