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Last integration question in the book

  1. May 10, 2007 #1
    its the last question in the last book so its lengthy, and not specific to one thing,

    anyway a curve (that looks like an infinity symbol) is given by
    x = 3cost and y = 9sin2t, 0≤t<2pi

    a) find the cartesian in form y^2 = f(x)

    i done that no problem and got y^2 = 4x^2(9-x^2)

    b) show the shaded area enclosed by the curve and the x axis is given by
    [tex]\int_{0}^{\frac{\pi}{2}}Asin2t sint dt[/tex]
    stating the value of A

    the infinity sign shaped curve is cut in half horizontally by the x axis, (and vertically by the y and by itself) the shaded sector is the top right one.

    as the area = ∫ydx
    y = 9sin2t

    dx/dt = -3sint
    so dx = -3sintdt

    therefore area = ∫(9sin2t)(-3sintdt)
    = ∫ -27sin2tsintdt

    so A = -27
    but the answer gives +27...

    if i took +27 as my answer i can get the rest of the question right, but i get minus which doesnt make sense

    if you need the diagram clarified i can draw it if you wish

    thanks in advance
  2. jcsd
  3. May 10, 2007 #2


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    Staff Emeritus
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    Gold Member

    To find area, you should always take the absolute value of ydx (so if it goes under the curve, for example, it doesn't mess you up).

    The reason you're gettting a negative sign here is because at t=0, x=3. At t=Pi/2, x=0. So basically, you're integrating in the negative direction. Going in the opposite direction would give you the positive answer you would expect (since it is all above the x-axis anyway)
  4. May 10, 2007 #3
    ahhhh i get it so what ive found was
    [tex]\int_{\frac{\pi}{2}}^{0}-27sin2t sint dt[/tex]

    which is equal to
    [tex]\int_{0}^{\frac{\pi}{2}}-(-27)sin2t sint dt[/tex]

    which ofcourse is +27

    thankyou so much that was really bugging me
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