Polar Integral Conversion Correct?

  • Thread starter Thread starter bigedd1227
  • Start date Start date
  • Tags Tags
    Integral Polar
Click For Summary
SUMMARY

The discussion focuses on converting a double integral from Cartesian to polar coordinates. The original integral is represented as Int(Int((x^2+y^2))dy)dx with bounds from -1 to 1 for the outer integral and from -sqrt(1-y^2) to sqrt(1-y^2) for the inner integral. The correct conversion to polar coordinates results in Int(Int(r^3)dr)dtheta, with outer bounds of 0 to 2pi and inner bounds of 0 to 1. A key correction noted is that the inner bounds should be -sqrt(1-x^2) to sqrt(1-x^2) in the original integral.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with polar coordinate transformations
  • Knowledge of integration techniques
  • Basic proficiency in mathematical notation and formatting
NEXT STEPS
  • Study the process of converting Cartesian coordinates to polar coordinates in integrals
  • Learn about the properties of double integrals and their applications
  • Explore advanced integration techniques, such as using Jacobians for coordinate transformations
  • Review examples of polar integrals in calculus textbooks or online resources
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and integral transformations, as well as professionals working with mathematical modeling and analysis.

bigedd1227
Messages
1
Reaction score
0
I'm not sure if I'm doing this conversion correctly. I have to convert the following double integral into polar. The integration part I can do, I just want to make sure I converted correctly. And sorry about the formatting bc I don't kno how to do the math formatting.


Int(Int((x^2+y^2))dy)dx
the bounds on the outer integral are -1 to 1
the bounds on the inner integral are -sqrt(1-y^2) to sqrt (1-y^2)

now here is what I got converting it to polar

Int(Int(r^3)dr)dtheta
outer bounds = 0 to 2pi
inner bounds = 0 to 1

Did I do the conversion correctly?
Thanks for the help
 
Physics news on Phys.org
Lookin' good, but in the first integral the bounds on the inner integral should be -sqrt(1-x^2) to sqrt (1-x^2).
 
Last edited:

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
2K
Replies
3
Views
3K
Replies
20
Views
2K
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
13K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K