Double integral limits?

  • #1

Homework Statement



∫∫ydxdy over the triangle with vertices (-1,0), (0,2), (2,0)

Homework Equations




I did it like this and got the right answer:

∫dy ∫ydx

this first:
∫ydx from x = (y-2)/2 to x = 2-y

then ∫dy from y = 0 to y = 2

I got 2 which is correct


but when I tried another method, splitting them up across the y axis i got 8/3:

I did I = A+B

A = ∫dx∫ydy

∫ydy from y = 0 to y = 2x+2

then I integrated from x = -1 to x =0 (I got 2/3)

B = ∫dx∫ydy

∫ydy from y =0 to y = -x+2

then I integrated wrt x from x = 0 to x = 2 (I got 2)

so the second answer was 8/3


Why was the second one wrong? I keep getting questions wrong because of the limits I choose! Could someone help me out please?




The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



∫∫ydxdy over the triangle with vertices (-1,0), (0,2), (2,0)

Homework Equations




I did it like this and got the right answer:

∫dy ∫ydx

this first:
∫ydx from x = (y-2)/2 to x = 2-y

then ∫dy from y = 0 to y = 2

I got 2 which is correct


but when I tried another method, splitting them up across the y axis i got 8/3:

I did I = A+B

A = ∫dx∫ydy

∫ydy from y = 0 to y = 2x+2

then I integrated from x = -1 to x =0 (I got 2/3)

B = ∫dx∫ydy

∫ydy from y =0 to y = -x+2

then I integrated wrt x from x = 0 to x = 2 (I got 2)

so the second answer was 8/3


Why was the second one wrong? I keep getting questions wrong because of the limits I choose! Could someone help me out please?




The Attempt at a Solution


Your limits look to be correct with both methods. Without you showing your calculations instead of just giving us your answers, we can't tell where your mistake is.
 

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