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hmm

  • 2 years ago

Integrate

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  1. hmm
    • 2 years ago
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    \[\int\limits_{0}^{2\pi} \sqrt{2 \cos \theta + 2} d \theta\]

  2. hmm
    • 2 years ago
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    Find the area shared by the polar equations \[r = 2 \cos \theta\] \[r = 2 \sin \theta\] |dw:1327853300525:dw|

  3. amistre64
    • 2 years ago
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    polar integration eh ....

  4. amistre64
    • 2 years ago
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    A = 1/2 .... something or another, id have to refresh my memory

  5. hmm
    • 2 years ago
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    I asked the same question earlier but nobody could answer me and I felt pretty smart for that. :P

  6. amistre64
    • 2 years ago
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    http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx

  7. amistre64
    • 2 years ago
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    im guessing that since this is 2 parts that it might be a double integral tho, sound about right?

  8. amistre64
    • 2 years ago
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    |dw:1327854057363:dw| or maybe even this rendition?

  9. hmm
    • 2 years ago
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    The first question I did a double integral and got \[\theta \sqrt{2 \cos \theta + 2} -\theta \sin \theta - \cos \theta - \theta^2/2\] from 2 pi to zero. Since this was originally \[\int\limits_{0}^{2\pi} \sqrt{(1 + \cos^2 \theta) + \sin^2 \theta}\] for the arc length of the cardioid. It's not supposed to be negative, which I got.

  10. amistre64
    • 2 years ago
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    negative just means you got the bottom over the top i believe; just drop the sign and it should be good

  11. hmm
    • 2 years ago
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    The second questions about the polar area is the area between two circles whihc I drew on the top. I used the calculator to find the bound to be 0 and pi, but it might be wrong.

  12. shaan_iitk
    • 2 years ago
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    bound is 0 and pie/4

  13. hmm
    • 2 years ago
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    Nope, I got - 11.57 but the answer is 8.

  14. hmm
    • 2 years ago
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    Oh, I mgiht have gotten the upper and lower bound wrong. But I was trying to find the area without the shared part first then use that to be subtracted from the area of the entire circle but I am wondering if there's a better idea. I'll try the pi over 4 bound.

  15. hmm
    • 2 years ago
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    I still get a negative number for the polar area with the new upper and lower bound.

  16. amistre64
    • 2 years ago
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    if we move the circles around we might get an easier read? or at least I do

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  17. hmm
    • 2 years ago
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    The answer is pi/2 - 1

  18. amistre64
    • 2 years ago
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    this is just the quarter of a circle - the cresent part then

  19. shaan_iitk
    • 2 years ago
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    it intersects at tan(thetha) = 1 which means at pie/4 and 3*pie/4 ..

  20. hmm
    • 2 years ago
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    Ah, I don't understand what you just said. :O

  21. amistre64
    • 2 years ago
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    (pi/4)-(1-pi/4) = 2pi/4 - 1 pi/2 - 1

  22. hmm
    • 2 years ago
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    How did you get that?

  23. amistre64
    • 2 years ago
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    me?

  24. hmm
    • 2 years ago
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    Yep

  25. amistre64
    • 2 years ago
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    |dw:1327854888415:dw|you end up with a 1x1 square with an area of 1;

  26. amistre64
    • 2 years ago
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    |dw:1327854924962:dw|

  27. hmm
    • 2 years ago
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    How do I prove that in equations?

  28. amistre64
    • 2 years ago
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    id move the circles around to equavelent positions and do:\[\int_{0}^{pi/2}\frac{1}{2}(R^2-r^2)dr\]

  29. amistre64
    • 2 years ago
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    whre R = (x^2+y^2=1) and r = ((x-1)^2+(y-1)^2=1)

  30. amistre64
    • 2 years ago
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    but state it in sin cos stuff maybe lol

  31. hmm
    • 2 years ago
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    I tried that but I must have made a mistake in the integration because I got \[(\sin2 \theta )/2 -3 \theta\] at the end.

  32. hmm
    • 2 years ago
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    I integrated \[ (2\cos \theta) ^2 - (2 \sin \theta )^2\] /2

  33. amistre64
    • 2 years ago
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    keep the same equations perhaps and chnge integraion to -pi/2 to 0

  34. hmm
    • 2 years ago
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    Nope, still doesn't work.

  35. amistre64
    • 2 years ago
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    yeah, polars im still a bit rusty on, got the concept but terrible at polarizing it

  36. hmm
    • 2 years ago
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    Yeah, I don't polar stuff either.

  37. amistre64
    • 2 years ago
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    ill reread my book at the house and know this stuff better tomorrow :) i cant seem to make sense of what I find online ....

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