Double Integral - Polar Coordinates

Click For Summary

Homework Help Overview

The problem involves evaluating a double integral by changing to polar coordinates. The original integral is defined over a specific region in the xy-plane, which is suggested to be related to a circular sector.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to express the differential area element in polar coordinates and the necessity of changing the limits of integration. There are questions about how to determine the appropriate values for r and φ based on the shape of the domain.

Discussion Status

Some guidance has been offered regarding the need to visualize the domain in the xy-plane, and there is an acknowledgment that the substitution for dxdy will depend on the specific shape of the region. Multiple interpretations of the region's shape are being explored.

Contextual Notes

Participants note the challenge of determining the limits of integration when converting to polar coordinates, as well as the potential complexity of the region defined by the original integral.

duki
Messages
264
Reaction score
0

Homework Statement



Evaluate by changing to polar coordinates

Homework Equations



Can't figure out how to make the integral stop after the sqrt(9-x^2)
[tex]\int_0^\frac{3}{\sqrt(2)} \int_x^{\sqrt(9-x^2)} e^-(x^2+y^2) dy dx[/tex]

The Attempt at a Solution



I'm not sure where to really start on this one. I know it will end up being e^-r^2 but beyond that I'm not sure.
 
Last edited:
Physics news on Phys.org
You'll have to express dxdy in other variables, and the intervals have to be changed.
 
How can I change them to polar coordinates?
 
Last edited:
First of all: Draw a figure in the xy-plane with to see what shape the domain yields (probably something like a circle sector). Then you should be able to figure out what values you should give [tex]r[/tex] and [tex]\phi[/tex]. You usually substitute [tex]r*drd\phi[/tex] for [tex]dxdy[/tex] when using polar coordinates. However, the exact expression depends on what shape the domain yields.
 

Similar threads

Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
5K
Replies
25
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K