Double Integral

  • Thread starter tnutty
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  • #1
327
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Homework Statement



[tex]\int_{D}\int y^2[/tex]
where D = {(x,y) | -1 [tex]\leq[/tex] y [tex]\leq1[/tex], -y-2[tex]\leq x\leq y[/tex]

The integral I set up is below :

[tex]\int^{1}_{-1} \int^{y}_{-y-2} y^2 dx dy [/tex]

From that I get the answer 0, but the book says its 4/3.

I get 0 because It reduces to this integral :

[tex]\int^{-1}_{1} 2y^3 + 2y[/tex]

Any idea where I could be wrong?
 

Answers and Replies

  • #2
149
1

Homework Statement



[tex]\int_{D}\int y^2[/tex]
where D = {(x,y) | -1 [tex]\leq[/tex] y [tex]\leq1[/tex], -y-2[tex]\leq x\leq y[/tex]

The integral I set up is below :

[tex]\int^{1}_{-1} \int^{y}_{-y-2} y^2 dx dy [/tex]

From that I get the answer 0, but the book says its 4/3.

I get 0 because It reduces to this integral :

[tex]\int^{-1}_{1} 2y^3 + 2y[/tex]

Any idea where I could be wrong?

In your last step you reversed your bounds, it should be (-1,1) not (1,-1) as you wrote. Also the final step should be [tex]\int[/tex]2y3 +2y2 dy with the bounds (-1 to 1).
 
Last edited:
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi tnutty! :smile:
Any idea where I could be wrong?

erm :redface:

2y3 + 2y2 ? :wink:
 
  • #4
327
1
I mean to write 2y^3 + 2y^2. But its was the bounds. Thanks guys.
 

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