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anonymous

  • 5 years ago

Can someone help me solve this iterated integral? Tanks.

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  1. anonymous
    • 5 years ago
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    Here's the equation.

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  2. anonymous
    • 5 years ago
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    the one marked red?

  3. anonymous
    • 5 years ago
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    Yes, xD

  4. anonymous
    • 5 years ago
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    use polar

  5. anonymous
    • 5 years ago
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    substitutions

  6. anonymous
    • 5 years ago
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    I have a test in 4days time on this stuff, double/triple integrals etc

  7. anonymous
    • 5 years ago
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    Oh, okay, thanks. Good luck to you! I'll get back here if I can't do it lol.

  8. anonymous
    • 5 years ago
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    let A be the angle

  9. anonymous
    • 5 years ago
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    x=rcos(A) y= rsin(A)

  10. anonymous
    • 5 years ago
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    the jacobian of the substitution is "f"

  11. anonymous
    • 5 years ago
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    edit "r"

  12. anonymous
    • 5 years ago
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    how would I get r? is it = 1? I figured out theta, which is from 0 to pi/4

  13. anonymous
    • 5 years ago
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  14. anonymous
    • 5 years ago
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    not the best picture lol

  15. anonymous
    • 5 years ago
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    lol, i'll upload mine

  16. anonymous
    • 5 years ago
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    the angle goes from 90degrees to 45degrees actually

  17. anonymous
    • 5 years ago
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    but the difference is still the same

  18. anonymous
    • 5 years ago
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    and our r goes from 0<=r<=1

  19. anonymous
    • 5 years ago
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    because we are inside the disc x^2 +y^2 <1 , so [rcos(A)]^2 +[ rsin(A)]^2 <1 , r^2<1 but r is also positive

  20. anonymous
    • 5 years ago
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    This is what I came up with, is it correct?

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  21. anonymous
    • 5 years ago
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    Elec, your drawing is correct, but the sector described is actually the portion you have marked 45 degrees. Good job

  22. anonymous
    • 5 years ago
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    I got it based on the 2 iterated integrals below in pic. ok, thx, i think i can do it now, 0<=r<=1

  23. anonymous
    • 5 years ago
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    wait, the region I drew was correct I am fairly sure

  24. anonymous
    • 5 years ago
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    I know it doesnt impact the final answer , but the integral has x as the lower limit, and the semicircle as the upper limit

  25. anonymous
    • 5 years ago
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    is it 2y+2x

  26. anonymous
    • 5 years ago
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    \[= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int\limits_{0}^{1} r^2 dr \]

  27. anonymous
    • 5 years ago
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    = \[\frac{\pi}{4} \times \frac{1}{3} = \frac{\pi}{12}\]

  28. anonymous
    • 5 years ago
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    What about the region I drew? Is that acceptable?

  29. anonymous
    • 5 years ago
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    Oh yeah, you're right, I was looking at it cross ways. Choose some x, and the y goes from x to the semicircle.

  30. anonymous
    • 5 years ago
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    Revan, elec's drawing is right. Also you have a sq rt 2/2 in there, that I think you got from your other problem.

  31. anonymous
    • 5 years ago
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    But isn't it, 0<=y<=x?

  32. anonymous
    • 5 years ago
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    no!

  33. anonymous
    • 5 years ago
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    the region you drew was y<=x and y<=sqrt(1-x^2)

  34. anonymous
    • 5 years ago
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    the intersection of those region , thats what your picture shows

  35. anonymous
    • 5 years ago
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    even though they would both give the same answer it is very likely you would lost a mark or two , because your drawing and the limits of the angle dont reflect the correct region

  36. anonymous
    • 5 years ago
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    Out of curiosity, revan, where did you get sq rt 2/2 from?

  37. anonymous
    • 5 years ago
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    http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.jpg from the two integrals at the bot of the pic. I have to sketch Region of the sum of the two and then replace the sum by an equivalent single iterated integral and evalute

  38. anonymous
    • 5 years ago
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    Part of confusion might be: that's not the question you posted here for this post.

  39. anonymous
    • 5 years ago
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    Lol, sorry >.< The question I posted is the single iterated integral. I just needed help solving it.

  40. anonymous
    • 5 years ago
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    yes, the way you did was very wrong lols

  41. anonymous
    • 5 years ago
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    :D thats why i posted lol

  42. anonymous
    • 5 years ago
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    \[\int\limits(ax+b)^n = \frac{(ax+b)^{n+1} } {a(n+1)} +C \]

  43. anonymous
    • 5 years ago
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    only works for linear functions

  44. anonymous
    • 5 years ago
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    raised to a power

  45. anonymous
    • 5 years ago
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    Now elec, is writing wrong thing

  46. anonymous
    • 5 years ago
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    You said to do poloar coordinates. So, I think I can solve it now. I'll post it when I'm done.

  47. anonymous
    • 5 years ago
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    Btw, y is r from 0 to 1? o.0

  48. anonymous
    • 5 years ago
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    nvm . ..

  49. anonymous
    • 5 years ago
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    Revan, if given this from scratch, you choose polar, but if given the integrals already compiled it is very easy to read.

  50. anonymous
    • 5 years ago
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    Yea, but I cant' solve it w/o using polar, but either system, the answer will be the same?

  51. anonymous
    • 5 years ago
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    What do you mean by 'solve'

  52. anonymous
    • 5 years ago
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    or find volume? that's what you mean?

  53. anonymous
    • 5 years ago
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    http://openstudy.com/updates/attachments/4dcf65a1dfd28b0b901b0dfd-revangale-1305438547251-photo1.jpg I'm supposed to combine the sum of the two iterated integrals at the bottom of this pic into a single iterated integral and evalute it.

  54. anonymous
    • 5 years ago
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    So, I drew the graph of the two iterated integrals together, to get the overall Region, and then use the overall Region to set up a single iterated integral. But, I couldn't solve it, unless it was in polar system. Is this the right approach?

  55. anonymous
    • 5 years ago
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    Well, just looking at the picture, it is a sector of a circle and it's a polar type thing. So sure. It is just a kind of trick question, your teacher wants to know if you understand what's going on

  56. anonymous
    • 5 years ago
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    Lol, okay, thx., yea I didn't know wat was going on.

  57. anonymous
    • 5 years ago
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    BTW ele engineer is right in using transformation but going from rectangular coordinates to polar is something that should be easy we know da---> r dr dtheta where x= r cos theta y = r sin theta and x^2 + y^2 = r ^2 and if you actually draw it out its easier to know the boundaries.

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