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Supposedly simple double integral

by raynoodles
Tags: double, integral, simple, supposedly
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raynoodles
#1
Nov3-07, 02:56 PM
P: 4
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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#2
Nov3-07, 03:25 PM
Sci Advisor
P: 1,232
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.)
raynoodles
#3
Nov3-07, 08:44 PM
P: 4
so then the parameters for y would be: 0 to x/3?

HallsofIvy
#4
Nov4-07, 06:33 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
Supposedly simple double integral

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.
raynoodles
#5
Nov4-07, 04:22 PM
P: 4
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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