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Evaluate the double integral?

  • Thread starter Math10
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  • #1
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Homework Statement


Evaluate the double integral (x+2y)dA, where R is the region in the first quadrant bounded by the circle x^2+y^2=9.

Homework Equations


None.

The Attempt at a Solution


I know how to evaluate the double integral but I just don't know how to find the limits of integration. I know that the radius of the circle is 3.
 

Answers and Replies

  • #2
PeroK
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First, what are the limits for x?
 
  • #3
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I have no idea. Is it 0 to 3?
 
  • #4
PeroK
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It looks like you do have an idea. That's right.

For each x, what are the limits of y? You may need to draw the area.
 
  • #5
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From 0 to 3?
 
  • #6
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Homework Statement


Evaluate the double integral (x+2y)dA, where R is the region in the first quadrant bounded by the circle x^2+y^2=9.

Homework Equations


None.

The Attempt at a Solution


I know how to evaluate the double integral but I just don't know how to find the limits of integration. I know that the radius of the circle is 3.
How would you describe the region R as a set? IOW, R = {(x, y) | <inequality involving x> and <inequality involving y>}.

If you are tempted to write the inequalities as ##0 \le x \le 3, \text{and } 0 \le y \le 3##, resist that temptation! That region would be a square, which the one in this problem is not.

As I said in another of your threads, many times carrying out the integration is the easy part, requiring very little thinking. Figuring out the limits of integration is often the more difficult task. Your textbook should have a number of examples. Have you looked at them?
 
  • #7
PeroK
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From 0 to 3?
That would be a square in the first quadrant.
 
  • #8
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So y=sqrt(9-x^2) and y= - sqrt(9-x^2)?
 
  • #9
PeroK
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So y=sqrt(9-x^2) and y= - sqrt(9-x^2)?
Why would y be negative?
 
  • #10
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So y=0 to sqrt(9-x^2)?
 
  • #11
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So y=0 to sqrt(9-x^2)?
Yes. Now what are your limits for x?
 
  • #12
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0 to 3?
 
  • #13
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Yes. Now you need to figure the order in which you're going to integrate.
 
  • #14
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dy dx?
 
  • #15
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  • #16
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dy dx.
 
  • #17
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OK. Now you're ready to set up the iterated integral and evaluate it...
 
  • #18
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Thank you!
 
  • #19
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You're welcome!
 

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