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Double integral boundaries.

  • Thread starter Kuma
  • Start date
  • #1
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Homework Statement



D = {x>0, x^2 < y < 10-x^2)

compute

integral (integral D of y^2 sqrt x)



Homework Equations





The Attempt at a Solution



I'm having trouble figuring out the bounds of the integral. y goes from x^2 to 10-x^2 but I think I have to split this integral up into two parts. I am not sure how to bound x. The parabolas intersect at the point sqrt 5
 

Answers and Replies

  • #2
2,571
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What do you mean probably at sqrt(5)? They intersect at two points
 
  • #3
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They intersect at (sqrt 5, 5) and (-sqrt 5, 5). x > 0 so we don't need the negative point.
 
  • #4
2,571
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Yes you got, it
 
  • #5
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So x goes from 0 to sqrt 5? thats the bound for x?
 
  • #6
2,571
1
So x goes from 0 to sqrt 5? thats the bound for x?
What do you think it should be? Let's put it this way, if it isn't x = 0, where would you start from? You mentioned splitting the integral, how do you plan to do that?
 
  • #7
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I guess I'm overthinking. It looks like x goes from 0 to x^2 and then stops when x reaches sqrt 5, then goes from 10-x^2 back to 0. This is just from the drawing I mean.
 
  • #8
Redbelly98
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So x goes from 0 to sqrt 5? thats the bound for x?
Yes, provided that you integrate over y first -- which, as you said, goes from x^2 to 10-x^2.
 

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