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anonymous

  • 4 years ago

Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?

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  1. nikvist
    • 4 years ago
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    \[r_1=A\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{A}{2}\right)^2=\left(\frac{A}{2}\right)^2\]\[r_2=B\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{B}{2}\right)^2=\left(\frac{B}{2}\right)^2\]\[S=\frac{\pi}{4}\cdot\left(\min{\left\{A,B\right\}}\right)^2\]

  2. perl
    • 4 years ago
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    this does not work

  3. perl
    • 4 years ago
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    try A = 2 , B = 1

  4. perl
    • 4 years ago
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    should be 3pi/2 , not sure how you got your answer

  5. perl
    • 4 years ago
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    ok looks like you did a polar conversion r = A sin theta r^2 = r* A sin theta x^2 + y^2 = A r* sin theta x^2 + y^2 = Ay x^2 + y^2 - Ay = 0 complete square x^2 + (y -A/2)^2 = (A/2)^2 similiarly x^2 + (y-B/2)^2 = (B/2)^2 directly integrating polar is much easier, btw so now you have two cartesian equations , and you did ... ?

  6. perl
    • 4 years ago
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    im not sure how to to integrate it using cartesian equations, hmmm

  7. perl
    • 4 years ago
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    ok i guess it can be done, but it is tricky

  8. perl
    • 4 years ago
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    wlog A > B > 0 so

  9. nikvist
    • 4 years ago
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    These are circles. Why do you integrate?

  10. perl
    • 4 years ago
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    |dw:1328800285296:dw|

  11. perl
    • 4 years ago
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    you want the area inside bounded by the two circles, you changed to cartesian so i integrated cartesian. the polar integration is much easier

  12. perl
    • 4 years ago
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    |dw:1328800478524:dw|

  13. perl
    • 4 years ago
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    i am not sure why you changed to cartesian, but either way works

  14. nikvist
    • 4 years ago
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    Perl,can you draw these circles?

  15. perl
    • 4 years ago
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    |dw:1328800689137:dw|

  16. nikvist
    • 4 years ago
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    Which does region lie inside both circles?

  17. perl
    • 4 years ago
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    lol

  18. perl
    • 4 years ago
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    I guess i solved a related question

  19. perl
    • 4 years ago
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    yours is still wrong though

  20. perl
    • 4 years ago
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    |dw:1328801098170:dw|

  21. perl
    • 4 years ago
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    |dw:1328801200075:dw|

  22. nikvist
    • 4 years ago
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    limits of integration are 0 and pi, not 0 and 2pi.

  23. perl
    • 4 years ago
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    ok

  24. perl
    • 4 years ago
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    ok i agree now

  25. perl
    • 4 years ago
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    so i calculated area twice?

  26. nikvist
    • 4 years ago
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    Between pi and 2pi circles do not exist.

  27. perl
    • 4 years ago
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    ok so you did integrate using polar, good , so you were just using the cartesian as an aid to graph it

  28. perl
    • 4 years ago
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    i set A sin theta = B sin theta

  29. perl
    • 4 years ago
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    solutions are pi*k

  30. perl
    • 4 years ago
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    why dont circles exist between pi and 2pi?

  31. nikvist
    • 4 years ago
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    because r cannot be negative

  32. perl
    • 4 years ago
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    yes it can in polar coordinates

  33. perl
    • 4 years ago
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    errr (r, theta ) = (-r, theta + pi)

  34. perl
    • 4 years ago
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    it just keeps going round and round in circles :)

  35. perl
    • 4 years ago
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    but clearly i went twice around

  36. perl
    • 4 years ago
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    that is why I got pi instead of pi/2

  37. nikvist
    • 4 years ago
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    \[r=\sqrt{x^2+y^2}\ge0\]

  38. perl
    • 4 years ago
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    in polar we allow negative r

  39. perl
    • 4 years ago
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    well if you have a TI 84 or graphing calculator, you can graph r = sin theta , and watch what happens when theta > pi . i think

  40. perl
    • 4 years ago
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    anyways, we define (-r, theta +pi) = (r,theta) , so its not a problem.

  41. perl
    • 4 years ago
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    for instance, suppose theta = 3pi/2, and given r(theta) = sin theta . we have ( r, theta ) = ( sin 3pi/2, 3pi/2) = (-1, 3pi/2) = (1 , 3pi/2 + - pi) = ( 1, pi/2) , so we can make any negative radius positive, using that trick

  42. nikvist
    • 4 years ago
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    For this problem, your solution is very complicated and unnecessary.

  43. perl
    • 4 years ago
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    which one? i was solving a different problem

  44. perl
    • 4 years ago
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    finding the region bounded by the two semicircles, i think its called the 'lune'

  45. nikvist
    • 4 years ago
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    these are not semicircles

  46. perl
    • 4 years ago
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    right, i solved a different problem , the area bounded by the arcs . the lune

  47. perl
    • 4 years ago
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    might not be the lune either not sure what you would call it, the area bounded by the inner and outer circles?

  48. perl
    • 4 years ago
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    and i got | a^2 - b^2) |* pi/4 as the area of the region bounded by the inner and outer circles

  49. nikvist
    • 4 years ago
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    Find the area of the region that lies inside both curve

  50. perl
    • 4 years ago
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    yes i misread it

  51. perl
    • 4 years ago
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    i have the answer now, thanks

  52. perl
    • 4 years ago
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    but i answered a different question , which is interesting

  53. perl
    • 4 years ago
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    i see , you didnt do any integration. you just used the area formula, pi * radius ^2 . nice

  54. perl
    • 4 years ago
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    i Have a difficult question, if you can help me...

  55. perl
    • 4 years ago
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    its elementary though

  56. perl
    • 4 years ago
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    Can i post it here?

  57. perl
    • 4 years ago
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    solve for all real and non real the system of equations. xy - z/3 = xyz +1 yz - x/3 = xyz -1 xz - y/3 = xyz - 9

  58. perl
    • 4 years ago
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    if you have a solution, please email riemanngalois@gmail.com I already have the integer solution (3,0,-3) and i know that you can plug it into wolfram. I want to know how did wolfram solve this. thanks

  59. perl
    • 4 years ago
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    as in , i would like to know the other five solutions, i believe 4 real solutions and 2 non real

  60. nikvist
    • 4 years ago
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    brute force solution

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  61. perl
    • 4 years ago
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    wow, i wish i could do that. did you use a computer algebra system for help, like maple or matlab?

  62. perl
    • 4 years ago
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    yes so we have to test the solutions, to see if any are extraneous. looks good so you did equation 1 - equation 2 , then you solved equation 3 . the logic of this gets me a bit confused. you got the quadratic of z, treating y as a constant. then z = f(y). so then x = g(y) back substituting. so shouldnt it be ( g(y), y, f(y) ) , infinite solutions>?

  63. perl
    • 4 years ago
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    ok so you did (1) - (2), (3) -> you got z = f(y) , x = g(y) . you can then plug these into any of the three equations? does it matter?

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