Double integral

  • Thread starter simba_
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  • #1
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Main Question or Discussion Point

Evaluate

[tex]\int[/tex][tex]\int[/tex] xy dxdy

where D is the triangular region {(x,y) element of R2| x+y <= 3, x >=0, y >= 0}


( You have to work out the limits of the integrals from the region D)



the bit i get confused about in these questions are the limits of the integrals

So i just want to check my answer. I got the limits of the y integral to be (3,0) and the limits of the x integral to be (3,0)

so my answer was 20.25, is that right??
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
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That's not right. The region is a triangle bounded by the x-axis, the y-axis and the line y=3-x. You evaluated the region as a square.

What you need to do is put the outer integral with the overall limits (from 0 to 3) and the inner integral with limits that depend on the other parameter (if dx is inside, this would be x goes from 0 to 3-y)
 
  • #3
19
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right i see now... thanks
 
  • #4
19
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so i put (0,3) on the outer integral and i got 27/8 as my answer then...
 
  • #5
LeonhardEuler
Gold Member
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I would check again, I think you missed a factor of 2 somewhere.
 
  • #6
19
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SS xy dxdy

1/2x2y (sub in 3-y, 0)

= y(9 - 6y + y2)1/2

= (9y - 6y2 + y3)1/2


S 1/2(9y - 6y*y + y*y*y)dy

= 1/2((9/2)y*y - 2y*y*y + (1/4)y*y*y*y) sub in (3, 0)

= 27/8 ?

thats how i got it... im very tired so prone to small mistakes atm, but i cannot find any here. is there something wrong with my process
 
Last edited:
  • #7
LeonhardEuler
Gold Member
859
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:redface: My mistake.
 

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