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Double integral

  1. May 11, 2010 #1
    can someone explain to me why my teacher divided the area into two?
    I=[tex]\int[/tex]1,0[tex]\int[/tex](2x-x^2)^.5,0 1/(x^2+y^2)^.5dydx



    ugggggh i tried to use the latex.....
    anyway...

    he used the polar coordinate to do this. once he turned it into polar coordinate, he divided the area into 2 bounded by (pi/4 - o d[tex]\theta[/tex]) (1/cos[tex]\theta[/tex] - 0 for dr) and then + (d[tex]\theta[/tex] bounded by pi/2 - pi/4) (2cos bounded by 2 - 0)


    sorry i don't know how to make this more clear...ask me if you need to clarify
     
    Last edited: May 11, 2010
  2. jcsd
  3. May 11, 2010 #2

    lanedance

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    Homework Helper

    see the correct (hopefully) tex below, click on it to see script, note you can put the whole equation in one set of tags
    [tex]\int_0^1 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx[/tex]

    is this what you meant?
     
  4. May 11, 2010 #3

    lanedance

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    Homework Helper

    so what was the question?
     
  5. May 11, 2010 #4
    I think it might actually be

    [tex]\int_0^1 \int_0^{\sqrt{2x-x^2}} \frac{1}{\sqrt{x^2+y^2}} dy dx[/tex]
     
  6. May 11, 2010 #5

    lanedance

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    Homework Helper

    yeah i think you're right
     
  7. May 11, 2010 #6
    my teacher said you have to separate the area into to parts in order to solve it and i don't understand why


    sorry about my crappy drawing again
     

    Attached Files:

    Last edited: May 11, 2010
  8. May 11, 2010 #7
    What exactly was the problem statement?
     
  9. May 11, 2010 #8

    Mark44

    Staff: Mentor

    The region over which integration takes place is the upper left quarter of a circle with radius 1 and centered at (1, 0). When you switch to a polar integral, the description for r changes, and this is why you need to write the polar form of your given integral with two integrals. For one integral, r ranges from 0 to the appropriate value on the line x = 1. For the other integral, r ranges from 0 to the appropriate value on the circle.
     
  10. May 11, 2010 #9
    then how you suppose to figure out the boundaries? and can you expand on the description for r changes??
     
  11. May 11, 2010 #10
    Enters Mark! You're in much better hands than mine now xstetsonx so I'm definitely stepping back here. Have fun! :smile:
     
  12. May 11, 2010 #11
    thx phyzmatrix anyways
     
  13. May 11, 2010 #12

    Mark44

    Staff: Mentor

    Have you drawn a sketch of the region? I have described the boundaries. All you need to do is change the Cartesian equation that represents each boundary to its polar form.
     
  14. May 11, 2010 #13
    yup under the attachment above. well you said appropriate i am not clear. my teacher he
    said for d[tex]\theta[/tex] is from 0 to pi/4 for the first one.......etc i don't understand how you get this value according to that graph
     
  15. May 11, 2010 #14

    Mark44

    Staff: Mentor

    The graph in your attachment is not very useful, as the circle part doesn't look much like part of a circle, and almost nothing is identified with its equation. To answer your question, what are the coordinates of the point where the vertical line intersects the circle? For the same point, what are the polar coordinates?
     
  16. May 11, 2010 #15
    1? cos^2+sin^2????
     
  17. May 11, 2010 #16

    Mark44

    Staff: Mentor

    Is that supposed to be a point? I'm looking for two numbers.
     
  18. May 11, 2010 #17
    i mean (1,1) that is where they intersect right?
     
  19. May 11, 2010 #18

    Mark44

    Staff: Mentor

    Right. Now, what are the polar coordinates of that point?
     
  20. May 11, 2010 #19
    em (cosx)^2+(sinx)^2?
     
  21. May 11, 2010 #20

    Mark44

    Staff: Mentor

    No. Polar coordinates are pairs of numbers (r, theta), with the first being the distance from the pole (origin) and the second the angle theta.

    In any case, cos2(x) +sin2(x) is identically equal to 1.
     
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