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Supposedly simple double integral 
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#1
Nov307, 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. 


#2
Nov307, 03:25 PM

Sci Advisor
P: 1,205

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.)



#3
Nov307, 08:44 PM

P: 4

so then the parameters for y would be: 0 to x/3?



#4
Nov407, 06:33 AM

Math
Emeritus
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Thanks
PF Gold
P: 39,569

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. 


#5
Nov407, 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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