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ilikephysics2

  • 2 years ago

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

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  1. roadjester
    • 2 years ago
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    let me guess, third year calculus?

  2. ilikephysics2
    • 2 years ago
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    yeah why?

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

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

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

  6. ilikephysics2
    • 2 years ago
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    ok

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

  8. ilikephysics2
    • 2 years ago
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    Not yet

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

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

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

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

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

  14. roadjester
    • 2 years ago
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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

  15. ilikephysics2
    • 2 years ago
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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

  16. roadjester
    • 2 years ago
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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.

  17. roadjester
    • 2 years ago
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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.

  18. ilikephysics2
    • 2 years ago
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    ok I see

  19. roadjester
    • 2 years ago
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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.

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

  21. roadjester
    • 2 years ago
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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

  22. roadjester
    • 2 years ago
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    Yes

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

  24. roadjester
    • 2 years ago
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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

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

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

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

  28. roadjester
    • 2 years ago
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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.

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

  30. roadjester
    • 2 years ago
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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?

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

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

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

  34. roadjester
    • 2 years ago
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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|

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

  36. roadjester
    • 2 years ago
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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.

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

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

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

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

  41. roadjester
    • 2 years ago
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    forget all numbers

  42. ilikephysics2
    • 2 years ago
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    A/H

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

  44. roadjester
    • 2 years ago
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    GOOD.

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

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

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

  48. ilikephysics2
    • 2 years ago
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    adjacent over hypotenuse

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

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

  51. roadjester
    • 2 years ago
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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?

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

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

  54. ilikephysics2
    • 2 years ago
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    yes

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

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

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

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

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

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

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

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

  63. ilikephysics2
    • 2 years ago
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    its just 0

  64. roadjester
    • 2 years ago
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    bingo

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

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

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

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

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

  70. roadjester
    • 2 years ago
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    YES

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

  72. roadjester
    • 2 years ago
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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

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

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

  75. ilikephysics2
    • 2 years ago
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    No I graphed it

  76. roadjester
    • 2 years ago
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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

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

  78. roadjester
    • 2 years ago
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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

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

  80. roadjester
    • 2 years ago
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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

  81. roadjester
    • 2 years ago
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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

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

  83. roadjester
    • 2 years ago
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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

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

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

  86. roadjester
    • 2 years ago
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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?

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

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

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

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

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

  92. ilikephysics2
    • 2 years ago
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    Yeah I have that

  93. roadjester
    • 2 years ago
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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.

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

  95. roadjester
    • 2 years ago
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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

  96. ilikephysics2
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

  107. roadjester
    • 2 years ago
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    OUGH

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