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ilikephysics2 Group Title

HELP Find the area of the region that lies inside both curves r = 8sin(2theta), r = 8sin(theta)

  • 2 years ago
  • 2 years ago

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  1. roadjester Group Title
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    let me guess, third year calculus?

    • 2 years ago
  2. ilikephysics2 Group Title
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    yeah why?

    • 2 years ago
  3. roadjester Group Title
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    It was torture for me :) let me see if I can help, I need some scratch paper though

    • 2 years ago
  4. ilikephysics2 Group Title
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    ok thanks man its so tough

    • 2 years ago
  5. roadjester Group Title
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    i can relate, gimmie a sec

    • 2 years ago
  6. ilikephysics2 Group Title
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    ok

    • 2 years ago
  7. roadjester Group Title
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    have you done double integrals?

    • 2 years ago
  8. ilikephysics2 Group Title
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    Not yet

    • 2 years ago
  9. roadjester Group Title
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    dang, okay gotta think of how to do this using single...

    • 2 years ago
  10. ilikephysics2 Group Title
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    we haven't gotten that far we are in chapter 9 with areas of regions and in curves

    • 2 years ago
  11. roadjester Group Title
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    give me as much info as you can then

    • 2 years ago
  12. ilikephysics2 Group Title
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    well the question is just right there lol, idk what else you want

    • 2 years ago
  13. ilikephysics2 Group Title
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    we have done polar coordinates, taylor series , maclaurin series

    • 2 years ago
  14. roadjester Group Title
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    okay good, you can't do this in rectangular, polar is your best, no only shot unless you want roots. it's also cleaner. I just need to remember how to do a single integral sadly. I haven't done Calc 3 in a long time

    • 2 years ago
  15. ilikephysics2 Group Title
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    Yeah I hear you, I'm looking at this problem and I don't even know where to start, if you can show me step by step what you do, that would be so helpful

    • 2 years ago
  16. roadjester Group Title
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    \[8\sin(2\theta)=8\sin(\theta)\] \[\sin(2\theta)=\sin(\theta)\] \[2\sin(\theta)\cos(\theta)=\sin(\theta)\] \[2\sin(\theta)\cos(\theta)-\sin(\theta)=0\] \[\sin(\theta)(2\cos(\theta)-1)=0\] \[\sin(\theta)=0\] and \[2\cos(\theta)-1=0\] Solve for theta and these are your limits of integration.

    • 2 years ago
  17. roadjester Group Title
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    Notice how I cancelled the 8 but didn't cancel the sin(theta). You never want to cancel out anything that isn't a constant since it might be a potential solution.

    • 2 years ago
  18. ilikephysics2 Group Title
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    ok I see

    • 2 years ago
  19. roadjester Group Title
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    Can you pick it up from there since you've done polar coordinates? I don't know if you've done integration using polar yet or not.

    • 2 years ago
  20. ilikephysics2 Group Title
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    so its 2 cos(theta) = 1, so cos(theta) = 1/2?

    • 2 years ago
  21. roadjester Group Title
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    But believe me, you'll need a strong grip on polar since double and triple integrals build on that. Cylindrical is like a second layer to polar and spherical...well just be strong in polar and double and triple will be easier for you

    • 2 years ago
  22. roadjester Group Title
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    Yes

    • 2 years ago
  23. ilikephysics2 Group Title
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    OK so i have sin(theta) = 0 and cos(theta) = 1/2

    • 2 years ago
  24. roadjester Group Title
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    yep, and notice how both functions are even( you're in polar not rectangular) so you need to take into account the negative answer

    • 2 years ago
  25. ilikephysics2 Group Title
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    Ok so now what?

    • 2 years ago
  26. roadjester Group Title
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    what are your values for theta?

    • 2 years ago
  27. ilikephysics2 Group Title
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    0 and 1/2?

    • 2 years ago
  28. roadjester Group Title
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    come on, your values for theta, not sin(theta) and cos(theta). Think trig or geometry. This is calculus so if you're willing to take calc, especially calc 3, you should know how to take the inverse of a trig function.

    • 2 years ago
  29. ilikephysics2 Group Title
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    I thought you meant those values though

    • 2 years ago
  30. roadjester Group Title
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    theta refers to the angle. What are the angles? Stop and don't think calculus right now. Go back to your geometry or trig. How did you find an angle using trig functions like sine and cosine?

    • 2 years ago
  31. ilikephysics2 Group Title
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    The circle, so wouldn't it be 30 degrees

    • 2 years ago
  32. roadjester Group Title
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    |dw:1351461552476:dw|

    • 2 years ago
  33. ilikephysics2 Group Title
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    I barely remember the unit circle

    • 2 years ago
  34. roadjester Group Title
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    That is a right triangle. Forget the unit circle, focus on the triangle. Cosine of which angle will give you 1/2? What two sides do you use for cosine? |dw:1351461655567:dw|

    • 2 years ago
  35. ilikephysics2 Group Title
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    I thought it was 30 degrees?

    • 2 years ago
  36. roadjester Group Title
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    Oh and use that graph for sine. It's crappy but it was the best I had. Ok, what is cosine? Think back to your geometry.

    • 2 years ago
  37. ilikephysics2 Group Title
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    sqrt(3)/2?

    • 2 years ago
  38. ilikephysics2 Group Title
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    That is the best I could remember

    • 2 years ago
  39. roadjester Group Title
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    no, no, the DEFINITION of cosine

    • 2 years ago
  40. ilikephysics2 Group Title
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    like which degree or just the definition?

    • 2 years ago
  41. roadjester Group Title
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    forget all numbers

    • 2 years ago
  42. ilikephysics2 Group Title
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    A/H

    • 2 years ago
  43. roadjester Group Title
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    forget angles, forget numbers, just the textbook definition

    • 2 years ago
  44. roadjester Group Title
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    GOOD.

    • 2 years ago
  45. roadjester Group Title
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    Okay, so what is A/H?

    • 2 years ago
  46. roadjester Group Title
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    what does it stand for?

    • 2 years ago
  47. ilikephysics2 Group Title
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    sqrt(3)/2

    • 2 years ago
  48. ilikephysics2 Group Title
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    adjacent over hypotenuse

    • 2 years ago
  49. roadjester Group Title
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    good, now when we had \[2\cos(\theta)-1=0\] how did you solve it?

    • 2 years ago
  50. ilikephysics2 Group Title
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    I got cos(theta) = 1/2 by adding then dividing

    • 2 years ago
  51. roadjester Group Title
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    good, now look at that triangle I drew. Do you see a 1 for the adjacent, and 2 for the hypotenuse?

    • 2 years ago
  52. ilikephysics2 Group Title
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    Oh yeah so 1/2

    • 2 years ago
  53. roadjester Group Title
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    right, when you have a fraction, that's what that fraction means

    • 2 years ago
  54. ilikephysics2 Group Title
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    yes

    • 2 years ago
  55. roadjester Group Title
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    so what angle, 30, 60, or 90, will give you cos(theta)=1/2?

    • 2 years ago
  56. ilikephysics2 Group Title
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    Nope 60 degrees since its a/h which is 1/2

    • 2 years ago
  57. roadjester Group Title
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    |dw:1351462522741:dw|

    • 2 years ago
  58. ilikephysics2 Group Title
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    60 degrees will give us 1/2

    • 2 years ago
  59. roadjester Group Title
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    got it! so theta is 60 degrees or pi/3

    • 2 years ago
  60. ilikephysics2 Group Title
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    ok yes cause they are equivalent

    • 2 years ago
  61. roadjester Group Title
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    ok, so can you do the same for\[\sin(\theta)=0\]

    • 2 years ago
  62. roadjester Group Title
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    Hint: you can't do this using a triangle

    • 2 years ago
  63. ilikephysics2 Group Title
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    its just 0

    • 2 years ago
  64. roadjester Group Title
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    bingo

    • 2 years ago
  65. roadjester Group Title
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    those are your limits of integration

    • 2 years ago
  66. ilikephysics2 Group Title
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    so 0 to pi/3?

    • 2 years ago
  67. roadjester Group Title
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    yes, but do you remember the area(not under a curve) between two curves?

    • 2 years ago
  68. roadjester Group Title
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    how would you integrate if I said y=x^2 and y=3-4x^2?

    • 2 years ago
  69. ilikephysics2 Group Title
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    like Big R ^2 - Little r ^2?

    • 2 years ago
  70. roadjester Group Title
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    YES

    • 2 years ago
  71. ilikephysics2 Group Title
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    so we have 0 to pi/3 of our 2 functions?

    • 2 years ago
  72. roadjester Group Title
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    now you have your limits of integration (in radians). You need to figure out which is the top function and which is the bottom

    • 2 years ago
  73. ilikephysics2 Group Title
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    top is 8 sin(theta) and bottom is 8sin(2theta)

    • 2 years ago
  74. roadjester Group Title
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    Did you guess that?

    • 2 years ago
  75. ilikephysics2 Group Title
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    No I graphed it

    • 2 years ago
  76. roadjester Group Title
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    Ok, so then you know we're missing the left part of the graph then correct? since we are only going from 0 to pi/3

    • 2 years ago
  77. ilikephysics2 Group Title
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    so then we have to go from -pi/3 to 0?

    • 2 years ago
  78. roadjester Group Title
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    Nope, don't go into negative. Use the trace on your calculator and you'll see that the graph keeps going from 0 to 2pi. A lot of graphs are like that but not all

    • 2 years ago
  79. ilikephysics2 Group Title
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    so whats our final integral going to look like then?

    • 2 years ago
  80. roadjester Group Title
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    Remember, the independent variable here isn't x like in rectangular coordinates, the angle theta (in radians) is the independent variable. Both x and y are dependent

    • 2 years ago
  81. roadjester Group Title
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    Good question. There are actually two limits I didn't want to get into yet, but since you brought it up, there are TWO solutions to both sin(theta)=0 and cos(theta)=1/2 (within the restriction that theta is from 0 to 2pi.) Without that restriction, theta would have an infinite number of solutions

    • 2 years ago
  82. ilikephysics2 Group Title
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    You realize this is not a double intgreal question right?

    • 2 years ago
  83. roadjester Group Title
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    yup, you're going to add the two integrals, just like if you had a break in the middle of a graph

    • 2 years ago
  84. ilikephysics2 Group Title
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    So integral from 0 to 2pi (8sin(theta)-8sin(2theta)) + 0 to pi/3 ((8sin(theta)-8sin(2theta))

    • 2 years ago
  85. roadjester Group Title
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    |dw:1351463619332:dw|

    • 2 years ago
  86. roadjester Group Title
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    notice how if you did top minus bottom( which is the x axis) you need to change your limits of integration?

    • 2 years ago
  87. roadjester Group Title
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    It's the same idea here, but using polar

    • 2 years ago
  88. ilikephysics2 Group Title
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    it is top-bottom though

    • 2 years ago
  89. ilikephysics2 Group Title
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    Can you draw out what the ending integral will look like

    • 2 years ago
  90. roadjester Group Title
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    true, you're right there, just that the limits(BOTH LIMITS) will change

    • 2 years ago
  91. roadjester Group Title
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    |dw:1351463857221:dw|

    • 2 years ago
  92. ilikephysics2 Group Title
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    Yeah I have that

    • 2 years ago
  93. roadjester Group Title
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    I focused on three points, but one of those points is actually a different limit of integration. For the graph of sine, what TWO values will give you 0? That's a big hint on your limits.

    • 2 years ago
  94. ilikephysics2 Group Title
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    Ok cool thanks I think I got it from here

    • 2 years ago
  95. roadjester Group Title
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    I hope I didn't hold back too much, but I wanted you to think and not just give you the answers. I hope I did okay

    • 2 years ago
  96. ilikephysics2 Group Title
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    it was good thanks, I still have to solve for the answer though, what was your result? Mine was 12.75516082

    • 2 years ago
  97. roadjester Group Title
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    uh...do you by any chance have the answer? Just a yes or no.

    • 2 years ago
  98. roadjester Group Title
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    Back of the book I mean

    • 2 years ago
  99. ilikephysics2 Group Title
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    yes thats my answer, oh no its webwork

    • 2 years ago
  100. ilikephysics2 Group Title
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    webwork didnt take my answer

    • 2 years ago
  101. roadjester Group Title
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    Uh, I got an 8...

    • 2 years ago
  102. roadjester Group Title
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    and no idea what webwork is

    • 2 years ago
  103. ilikephysics2 Group Title
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    what did you get 8 what?

    • 2 years ago
  104. roadjester Group Title
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    8 as the final answer

    • 2 years ago
  105. roadjester Group Title
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    Is that the answer in your text?

    • 2 years ago
  106. ilikephysics2 Group Title
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    ok but our answers are both wrong

    • 2 years ago
  107. roadjester Group Title
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    OUGH

    • 2 years ago
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