Calc III Double Integral Question

Click For Summary

Discussion Overview

The discussion revolves around a problem related to setting up a double integral for evaluation, specifically focusing on determining the correct bounds for the integral. The context includes aspects of calculus, particularly in the area of multiple integrals and their representation in different coordinate systems.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in setting the bounds for a double integral and believes the region is a Type 2 region (dxdy), initially proposing y bounds from 0 to √2 and x bounds from x=sqrt(1-x^2) to x=sqrt(4-x^2).
  • The same participant realizes that these bounds may include extra space outside the intended shaded region and suggests that the line y=x must be involved in the bounds.
  • Another participant suggests that the integral should be written in polar form rather than Cartesian form, indicating a belief that a single integral cannot be properly set up in Cartesian coordinates for this problem.
  • A later reply indicates that the participant successfully completed the problem using polar coordinates and shares the result of the integral as 15/16, while also acknowledging the need to post in the appropriate section in the future.

Areas of Agreement / Disagreement

There is no clear consensus on the initial bounds for the integral, as participants express differing views on how to approach the problem. The discussion includes a mix of suggestions and realizations without resolving the initial confusion about the bounds.

Contextual Notes

Participants do not clarify the assumptions underlying their proposed bounds or the specific characteristics of the region in question, which may affect the setup of the integral.

dropoutofschool
Messages
3
Reaction score
0
View attachment 96628
96628.gif


This is the problem I'm trying to solve. The directions require me to rewrite as a single integral and evaluate. But I'm having trouble setting the bounds for a complete compounded integral. The graph of the region would look something like this...
View attachment 96629
96629.png

Where the shaded area is the region. I would think its a Type 2 region (dxdy). The y bounds would be 0 to√2 and the x bounds would be from x=sqrt(1-x^2) to x=sqrt(4-x^2)... so so I thought. I then realized that these bounds would extra space outside of the intended shaded region, so the y=x linemust be involved in the bounds somehow.

I just need help setting the integral up, the actual integrating should be easy. Thanks ! Any help is appreciated!
 
Last edited by a moderator:
Physics news on Phys.org
Ugh Pictures didn't post... please refer to this...
Hw.GIF


pic.png
 
dropoutofschool said:
View attachment 96628
This is the problem I'm trying to solve. The directions require me to rewrite as a single integral and evaluate. But I'm having trouble setting the bounds for a complete compounded integral. The graph of the region would look something like this...
View attachment 96629
Where the shaded area is the region. I would think its a Type 2 region (dxdy). The y bounds would be 0 to√2 and the x bounds would be from x=sqrt(1-x^2) to x=sqrt(4-x^2)... so so I thought. I then realized that these bounds would extra space outside of the intended shaded region, so the y=x linemust be involved in the bounds somehow.

I just need help setting the integral up, the actual integrating should be easy. Thanks ! Any help is appreciated!

I'm pretty sure the idea is to write the integral in polar form, not in Cartesian form... There's no way you can write this integral in Cartesian form with only one integral.
 
Last edited:
In the future, please post questions like this in the Homework & Coursework sections (under Calculus), not here in the technical math sections.
 
WOW! Completely forgot about polar, just completed the problem----for future reference, the answer is 15/16. Thank you sir and will change where I post next time!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K