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Double integral

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data


    Evaluate the integral

    ∫∫R y√(x^2 + y^2) dA

    with R the region {(x, y) : 1 ≤ x^2 + y^2 ≤ 2, 0 ≤ y ≤ x.}

    2. Relevant equations



    3. The attempt at a solution

    Solution:

    ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx

    ∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0) dx

    1/3 ∫(2 to 1) [(x^2 + y^2)^(3/2) - (x^2)^(3/2)] dx

    1/3 x 2/5 [(x^2 + y^2)^(5/2) - (x^2)^(5/2)] |(x=2 to x=1)

    final answer: 4(124√2 - 33)/15

    Is this the correct answer??
     
  2. jcsd
  3. May 18, 2008 #2

    Dick

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    No, I don't think that's right. For one thing the number is far too large to fit in the annulus 1<=r<=2. And for another, why don't you use the polar coordinates the problem is so obviously screaming for?
     
  4. May 19, 2008 #3
    so

    ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx
    ∫(2 to 1)∫(x to 0) ((r sin(theta)(r^2)^1/2)r dr dtheta
    ∫(2 to 1)∫(x to 0) ((r^2 sin(theta)(r^3)^1/2) dr dtheta

    Like this?
     
  5. May 19, 2008 #4

    Dick

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    Sort of like that, but without the errors. Where did (r^3)^(1/2) come from? And you are integrating dtheta. What are the theta limits?
     
  6. May 19, 2008 #5
    I got the (r^3)^(1/2) from multiplying the r on the outside to the inside.. the (1/2) is the sqrt and the r^3 is from multiplying it with the r..

    also the theta limit would be r to 0, rather than x to 0??
     
  7. May 19, 2008 #6

    Dick

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    The r power is wrong. You started with (r^2)^(1/2). Not (r)^(1/2). Why do you think the theta limits are 0 to r? Is that just a random guess? Did you draw a picture of the region?
     
  8. May 19, 2008 #7
    oh its a circle that would go from 2pie to 0.. but iam still confused about the other part
     
  9. May 19, 2008 #8

    Dick

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    ????? What do you think it looks like??? 0<=y<=x. Start picking some values of x and sketching the region.
     
  10. May 19, 2008 #9
    What if i tried not to use polar and did this..

    from above we have:

    ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx

    ∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0) dx

    1/3 ∫(2 to 1) x^3 dx

    1/3 x 1/4 x^4 | (x = 2 and x =1)

    final answer: 5/4
     
  11. May 19, 2008 #10

    Dick

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    Just because you aren't using polar coordinates doesn't mean you can skip understanding the problem. Sketch the region!!!! If you do this you will realize you can't do it dx*dy using a single set of limits. You'd have to split the region up. Polar is easier.
     
  12. May 19, 2008 #11
    iam not so sure i know how to sketch it.
     
  13. May 19, 2008 #12

    Defennder

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    It's kind of tedious if you don't use polar coordinates. If you do by polar, you won't have to worry about finding the limits of y as a function of x. The limits in polar coordinates are just numbers.
     
  14. May 19, 2008 #13

    Dick

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    Start with 1<=r<=2. Any ideas what that might look like? Then try 0<=y<=x. As I said, pick a value of x, say x=1. What values of y work? Draw them. Try x=2. Draw them. Etc. Until you get the general picture.
     
  15. May 19, 2008 #14

    Defennder

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    Well, you have x^2 + y^2 ≤ 2. Consider the case x^2 + y^2 = 2. This represents a circle of radius sqrt(2). Consider also x^2 + y^2 = 1. This is circle of radius 1. The region you want is in between these 2 in the inequality. How does this translate graphically? For y<=x, draw the line y=x and determine the locus of points where y<x. The intersection of these 2 regions is the region over which the double integration is performed.
     
  16. May 19, 2008 #15
    hm what I came up with, which is probably wrong, is almost like a triangle in the first quad...

    so therefore you would go from (pie/2)/2 = pie/4 to 0??
     
  17. May 19, 2008 #16

    Defennder

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    Yes, in the first quadrant it is a triangular region extending to infinity. I'm assuming you haven't found the other region corresponding to 1 ≤ x^2 + y^2 ≤ 2. But what about the other quadrants?
     
  18. May 19, 2008 #17

    Dick

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    Yessss! The region is 1<=r<=2 and 0<=theta<=pi/4. It is a wedge shaped triangular region intersected with a ring shaped region.
     
  19. May 19, 2008 #18

    Dick

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    Yessss! The region is 1<=r<=2 and 0<=theta<=pi/4. It is a wedge shaped triangular region. See how helpful sketching the region is?
     
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