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joannaj93

Using a polar double integral, find the volume in the first octant below the surface z=xy, inside cylinder r=4cos(theta), and between the cylinders x^2 + y^2=4 and x^2 + y^2=8.

  • one year ago
  • one year ago

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  1. TuringTest
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    |dw:1352751755636:dw|

    • one year ago
  2. TuringTest
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    in polar the three graphs are\[r=2\]\[r=2\sqrt2\]\[r=4\cos\theta\]can you find the intersection points in terms of theta?|dw:1352752221119:dw|

    • one year ago
  3. Algebraic!
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    1st octant...

    • one year ago
  4. joannaj93
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    Make the equations equal to one another...

    • one year ago
  5. TuringTest
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    yes, and what do you get for theta?

    • one year ago
  6. joannaj93
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    arccos 2^(1/2)/2

    • one year ago
  7. TuringTest
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    well that's one, what angle does that correspond to? and what is the other intersection?

    • one year ago
  8. joannaj93
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    45 degrees

    • one year ago
  9. joannaj93
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    and 60 degrees

    • one year ago
  10. TuringTest
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    yes, 45 and 60, but you better use radians in calc|dw:1352754013710:dw|

    • one year ago
  11. TuringTest
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    |dw:1352754159926:dw|

    • one year ago
  12. TuringTest
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    so which fuunction is the outer radius and which one the inner radius for\[0\le\theta\le\frac\pi4\]?

    • one year ago
  13. joannaj93
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    r=2 is inner and r=2sqrt2 is outer...

    • one year ago
  14. TuringTest
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    yes, good :) and for the interval\[\frac\pi4\le\theta\le\frac\pi3\]what is it?

    • one year ago
  15. joannaj93
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    r=4cos (theta) is outer and r=2sqrt2 is inner...

    • one year ago
  16. TuringTest
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    I think you have the inner wrong|dw:1352754952806:dw|

    • one year ago
  17. joannaj93
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    oh, inner is r=2

    • one year ago
  18. TuringTest
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    right and what are your integrands?

    • one year ago
  19. joannaj93
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    I'm not too sure about that

    • one year ago
  20. TuringTest
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    you are integrating xy sub in x=rcost, y=rsint

    • one year ago
  21. TuringTest
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    above t=theta in case you didn't guess also you need to use dA for polar coordinate

    • one year ago
  22. joannaj93
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    so it'll be the double integral of r^2 cost sint...

    • one year ago
  23. TuringTest
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    that is the xy part, and what is dA in polar?

    • one year ago
  24. joannaj93
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    dr dt

    • one year ago
  25. joannaj93
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    sorry, r dr dt

    • one year ago
  26. TuringTest
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    yes, forget that extra r and you will be sorry :P so can you write your integrals yet?

    • one year ago
  27. joannaj93
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    I dont think so

    • one year ago
  28. TuringTest
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    do you see that it will be two integrals?

    • one year ago
  29. joannaj93
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    im getting three rs and thethas

    • one year ago
  30. TuringTest
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    there are two regions to consider|dw:1352756391639:dw|what are the bounds for r and theta on the first integral? what are they on the second? those will be the bounds of your integral.

    • one year ago
  31. joannaj93
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    r=2 to 2√2 and t= 0 to pi/4 and r=2√2 to 4cosθ and t= 0 to pi/3

    • one year ago
  32. TuringTest
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    on the second region the lower bound on r is not 2√2 as we have already discussed, and the lower bound on theta is not 0, but the upper bound of the previous integral

    • one year ago
  33. joannaj93
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    Sorry about that. give me a sec

    • one year ago
  34. joannaj93
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    for the second integral: r=2 to 4cost and t= pi/4 to pi/3

    • one year ago
  35. TuringTest
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    good :) so how 'bout setting up those integrals? (two double intergals)

    • one year ago
  36. joannaj93
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    \[\int\limits_{0}^{\pi/4}\int\limits\limits_{2}^{2\sqrt{2}}r ^{2}\cos \Theta \sin \Theta r dr d \Theta \]

    • one year ago
  37. TuringTest
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    that's the first one, yes

    • one year ago
  38. joannaj93
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    \[\int\limits_{\pi/4}^{\pi/3} \int\limits_{2}^{4\cos \Theta} r2\cosΘ\sinΘrdrdΘ\]

    • one year ago
  39. TuringTest
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    yes, good :) then add them together to find the total voume happy integrating

    • one year ago
  40. joannaj93
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    Oh, God this does not look like fun, but Thanks for the help :)

    • one year ago
  41. TuringTest
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    It's really not that hard if you can just remember to integrate with respect to theta and r separately, but I'm still too lazy to do it. Welcome!

    • one year ago
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