Integration Help: Changing Order & Limits for Problem Solving - Need Assistance

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Homework Help Overview

The discussion revolves around a problem involving changing the order of integration in a double integral. Participants are focused on ensuring the limits of integration are correctly established before proceeding with the integration process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the original bounds of the integral and the implications of changing the order of integration. There is an emphasis on verifying the new limits based on the geometric interpretation of the integration region.

Discussion Status

Some participants have provided insights into the bounds of the integral and clarified the geometric region of integration. There is an ongoing exploration of the implications of these bounds, with no explicit consensus reached yet.

Contextual Notes

The original poster is seeking confirmation on the limits of integration and has referenced a specific image for context, which is not visible in the discussion. The problem is situated within the constraints of a homework assignment.

elle
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Hi, I've got a few integration problems that I would like some help with:

The one I'm currently working on is this:

http://i26.photobucket.com/albums/c109/mathsnerd/4c37af04.jpg"

I'm supposed to change the order of integration but before I start integrating, I want to make sure I've changed the limits correctly. Can someone please check for me? Thank you! :smile:
 
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The bounds on the first integral read:

0\leq x \leq 4 \mbox{ and } \sqrt{x}\leq y \leq 2

notice that y is bound below by \sqrt{x}

and so the integral you want has bounds:

0\leq y \leq 2 \mbox{ and } 0\leq x \leq y^2
 
To be clear, the domain of integration is the smaller roughly triangular region in the upper left corner of the box [0,4]X[0,2].
 
Afer changing the bounds, try applying the substitution

u=x+y^2\Rightarrow du=dx \mbox{ so that }0\leq x \leq y^2 \Rightarrow y^2\leq u\leq 2y^2
 

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