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anonymous

  • 5 years ago

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx. a) Find the area of R b)Find the volume of the solid generated when R is revolved about the x-axis c)Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares

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  1. amistre64
    • 5 years ago
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    thats a little section beside the sin cos stuff eh

  2. amistre64
    • 5 years ago
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    since sin and cos = each other at pi/4; take your integrals from 0 to pi/4

  3. amistre64
    • 5 years ago
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    [S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

  4. amistre64
    • 5 years ago
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    to revolve it around the x axis; we do a sum of areas [S] 2pi [f(x)]^2 dx

  5. amistre64
    • 5 years ago
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    take the cos first and subtract out the sin next; like cutting a hole out of a donut

  6. amistre64
    • 5 years ago
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    ack!!... pi [f(x)]^2 lol

  7. amistre64
    • 5 years ago
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    pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]

  8. amistre64
    • 5 years ago
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    makes sense?

  9. anonymous
    • 5 years ago
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    i think the integral will involve integration from x =0 to 90 degree since it's given that the problem is about to solve at the first quadrant. for question a) integration \[\int\limits_{0}^{\pi/2}\int\limits_{\sin x}^{\cos x} dy dx\]

  10. anonymous
    • 5 years ago
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    seems to me because the area enclosed by y-axis the interval should be from pi/4 yo pi/2...

  11. amistre64
    • 5 years ago
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  12. anonymous
    • 5 years ago
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    nice drawing ! so, shaded area - [pi/4,pi/2]. am I wrong?

  13. anonymous
    • 5 years ago
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    amistre, is that a (cos(x))² or is it cos (x²)

  14. amistre64
    • 5 years ago
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    your wrong :) 0 to pi/4

  15. amistre64
    • 5 years ago
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    cos^2(x)

  16. anonymous
    • 5 years ago
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    ok

  17. anonymous
    • 5 years ago
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    i'm sorry, it's from o to pi/4 :)

  18. amistre64
    • 5 years ago
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    the double integral? dunno bout it ;)

  19. anonymous
    • 5 years ago
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    i thought the question want to integrate between cos x , sin x and x-axis. silly me :P you can find area under curve using double integration.

  20. amistre64
    • 5 years ago
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    cos(2t) = 2cos^2 - 1 cos^2 = (1+cos(2t))/2 might help

  21. amistre64
    • 5 years ago
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    just havent had any experience with doubles yet ;)

  22. anonymous
    • 5 years ago
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    for part A did you come up with sqrt(2) - 1?

  23. amistre64
    • 5 years ago
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    [S] cos^2(x) dx [S] 1/2 dt + [S] 2cos(2x)/4 dt ; [0,pi/4] is the first one right?

  24. amistre64
    • 5 years ago
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    whaa !?!

  25. amistre64
    • 5 years ago
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    [S] cos(t) dt -> sin(t). sin(pi/4) - sin(0) = 1/sqrt(2) - 0

  26. anonymous
    • 5 years ago
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    or .414

  27. amistre64
    • 5 years ago
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    [S] sin(t) dt -> -cos(t). -cos(pi/4) + cos(0) = -1/sqrt(2) +1

  28. anonymous
    • 5 years ago
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    i got sqrt (2) - 1 for question a

  29. amistre64
    • 5 years ago
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    1/sqrt(2) - (-1/sqrt(2) +1) 1/sqrt(2) + 1/sqrt(2) -1 (2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1 right?

  30. anonymous
    • 5 years ago
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    1/sqrt(2) + 1/sqrt(2) -1 =2/sqrt (2) - 1 =sqrt (2) - 1

  31. amistre64
    • 5 years ago
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    i see it ;)

  32. amistre64
    • 5 years ago
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    melor got it right :)

  33. anonymous
    • 5 years ago
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    okay I understand part a :)

  34. amistre64
    • 5 years ago
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    i invented a sqrt(2) in the numerator lol

  35. amistre64
    • 5 years ago
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    part b is integrateing the areas of the "donut", ie torus

  36. amistre64
    • 5 years ago
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    pi [S] cos^2 - sin^2 from 0 to pi/4 right?

  37. anonymous
    • 5 years ago
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    what's the difference between that, and having a 2 pi in front? I always get mixed up when to put pi or 2pi

  38. amistre64
    • 5 years ago
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    which is just cos(2t) thich integrates to (pi/2).sin(2t) from 0 to pi/4

  39. amistre64
    • 5 years ago
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    2pi is a surface area

  40. amistre64
    • 5 years ago
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    you want to add areas to get volume.... area of a circle = pi [f(x)]^2

  41. anonymous
    • 5 years ago
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    okay!

  42. amistre64
    • 5 years ago
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    2pi [f(x)] amounts to surface area :)

  43. amistre64
    • 5 years ago
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    adding together all the circumfrences of the circles :)

  44. anonymous
    • 5 years ago
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    so we're working with pi[S] cos²x - sin²x dx [0,pi/4] ?

  45. amistre64
    • 5 years ago
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    yes

  46. anonymous
    • 5 years ago
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    gosh, I feel silly. How do you integrate that? :/

  47. amistre64
    • 5 years ago
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    cos^2 - sin^2 = cos(2x) right? 2 cos(2x) ups to sin(2x) so we need to multiply by a comvenient form of 1; say 2/2

  48. amistre64
    • 5 years ago
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    keep the top 2 to use in the integral and slide out the bottom to under the pi

  49. amistre64
    • 5 years ago
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    we get in the end; (pi/2) sin(2x) right?

  50. amistre64
    • 5 years ago
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    go you recall your trig identitites?

  51. anonymous
    • 5 years ago
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    ehh, cos²x-sin²x= cos(2x)?

  52. amistre64
    • 5 years ago
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    the volume should be pi ------ 2sqrt(2)

  53. anonymous
    • 5 years ago
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    cos(x+y)=cos x cos y - sin x sin y substitue x into y you get, cos (2x) = cos^2 - sin^2 :)

  54. amistre64
    • 5 years ago
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    cos(a+b) = cos(a)cos(b) - sin(a)sin(b) cos(a+a); or cos(2a) = cos(a)cos(a)-sin(a)sin(a cos(2a)=cos^2(a)-sin^2(a)

  55. amistre64
    • 5 years ago
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    D(sin(2x)) = 2 cos(2x)

  56. amistre64
    • 5 years ago
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    2 cos(2x) -------- = cos(2x) right? 2

  57. anonymous
    • 5 years ago
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    I plugged pi[S] cos²x - sin²x dx [0,pi/4] into wolframalpha and it gave me pi/2?

  58. amistre64
    • 5 years ago
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    pi/2 [S] 2cos(2x) dx -> pi/2 sin(2x)

  59. amistre64
    • 5 years ago
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    at 0 = its = 0

  60. amistre64
    • 5 years ago
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    so the volume of the solid is pi/2 * sin(2(45)) = pi/2

  61. amistre64
    • 5 years ago
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    sin(2(pi/4)) = sin(pi/2) = 1

  62. amistre64
    • 5 years ago
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    so yes; the volume of the solid of rotation is pi/2 :)

  63. anonymous
    • 5 years ago
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    sweet!

  64. amistre64
    • 5 years ago
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    i gots no idea what c is saying

  65. anonymous
    • 5 years ago
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    c is asking to take the graph that we had, using the area R as the base of a solid, placing square cross sections on top of the graph to create a solid... haha I have no clue how to explain it. and I don't know how to solve it :(

  66. anonymous
    • 5 years ago
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    I guess for a square, every side is equal, so the length of each individual square placed is going to fit depending on the base R

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