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    hey guys. i was wondering is all the good mathematicians out there could help me on a hard question. (if it helsp this question is question 7 on an OCR past P3 paper dated 16 January 2002). i dont have a clue on thsi so could someone go through it step by sytep.

    by using substitution y=a sinx to prove that the

    $ (limits a and a/2) (a² - y²)^½ dx = (a²/2)[(n/3) - (3^½)/2]

    $ means integral
    ^½ means to the power of a half
    n means pi (i.e. 3.14)
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    you've written dx, don't yoou mean dy?

    y = a sinx => dy/dx = a cosx

    $ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

    = $ (limits pi/6 and pi/2) a² cos²x dx
    = $ (limits pi/6 and pi/2) a²/2 (1+cos2x) dx

    =a²/2[x + 1/2 sin2x] (limits pi/6 and pi/2)

    = a²/2(pi/3 - root3/2)
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    What does the dollar sign mean? :confused:
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    yeah sorry i meant dy not dx.

    by the way the dollar sign means intergral. u know... the squigly line thing !
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    (Original post by elpaw)
    you've written dx, don't yoou mean dy?

    y = a sinx => dy/dx = a cosx

    $ (limits a and a/2) (a² - y²)^½ dy = $ (limits pi/6 and pi/2) (a²-a²sin²x)^½ a cosx dx

    = $ (limits pi/6 and pi/2) a² cos²x dx
    = $ (limits pi/6 and pi/2) a/2 (1+cos2x) dx

    =[ax/2 + a/4 sin2x] (limits pi/6 and pi/2)

    = a²/2(pi/3 - root3/2)
    cheers elpaw that was great! u did it just like that! iwas expecting a few hours before someone got the answer for me. no wonder ur at oxford!. i bow down to ur greatness.
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    however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
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    (Original post by DLo)
    however maybe its cos i'm thick or something but how do u get the limits t oequall pi/6 and pi/2?
    The original limits are wrt y. The new limits are wrt x.

    using the substitution y=asinx,

    at y=a/2, a/2=asinx => sinx = ½ => x=pi/6
    at y=a, a=asinx => sinx = 1 => x=pi/2
 
 
 
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