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What am I doing wrong with this integral?

  • Thread starter duki
  • Start date
  • #1
264
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Homework Statement



Homework Equations



[tex]\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy[/tex]

The Attempt at a Solution



I've gotten [tex] \int_{-1}^3 \frac{x^3}{3}[/tex] = [tex] \int_{-1}^{3}y^3 dy[/tex] for the first part, but this can't be right?
 

Answers and Replies

  • #2
166
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How are you getting that equals sign in there?
Aren't you solving:
[tex]\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy[/tex] = ?

Try splitting it up to help you see it more easily, maybe:
[tex]\int_{-1}^3 \int_0^{3y}x^2dxdy+\int_{-1}^3 \int_0^{3y}y^2dxdy[/tex] = ?
 
  • #3
264
0
For the first integral I get:

[tex] \int_{-1}^3 y^3 dy[/tex]
Is that right so far?
 
  • #4
166
0
Well, no... because you know:

[tex]\int_{-1}^3 \int_0^{3y}x^2dxdy=\int_{-1}^3 \frac{(3y)^3}{3}dy[/tex]
 
  • #5
264
0
Oh. Oops. :)
Thanks
 

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