Supposedly simple double integral

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Homework Help Overview

The discussion revolves around evaluating a double integral of the function xy over a triangular region defined by the vertices (0,0), (3,0), and (0,1). Participants are focused on determining the appropriate limits of integration for both dx and dy.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to establish the limits for the integration but is uncertain about the correct bounds for dy. Some participants suggest visualizing the region to clarify the limits based on the geometry of the triangle. There is also a discussion about swapping the order of integration and how that affects the limits.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to determine the limits of integration. The original poster expresses difficulty in arriving at the correct answer despite trying different parameter setups. There is no explicit consensus on the correct approach yet, but various interpretations and methods are being explored.

Contextual Notes

There are indications of confusion regarding the limits of integration, particularly for dy, and the original poster has attempted multiple configurations without success. The problem is framed within the constraints of a homework assignment, which may impose specific methods or formats for the solution.

raynoodles
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double integral of xy dA
in the triangular region of (0,0), (3,0), (0,1).
my problem that I am having is finding the limits I am suposed to find dx or dy in. I figure I should use 0 to 3 for dx, but then i do dy from 0 to what? Help appreciated.
 
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Try drawing a picture of the region. Then, for a given value of x, what values of y lie within the region? This gives you the limits of integration for y, given x. (Of course, you must then do the y integral before you do the x integral.)
 
so then the parameters for y would be: 0 to x/3?
 
Yes, because the upper boundary is the line y= x/3.

It is a very good exercise to "swap" the limits of integration. Suppose you wanted to integrate with respect to x first and then y? Clearly to cover the entire triangle, you must take y going from 0 to 1. For each y, then, x must go from the left boundary, x= 0, to the "right" boundary which is still that line y= x/3. That is, x must go from x= 0 to x= what? Do the integral of xy both ways and see if you get the same thing.
 
still not getting the right answer.

I used the parameters dy= 0 to 1 and dx= 0 to -3y+3 and got 2.375.
the answer was wrong.
I did it the other way with dy=0 to x/3+1 and dx= 0 to 3 and got another wrong answer.
what am I doing wrong?
 

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