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    I'm trying to find out the equation of a specific bit of a circle...

    I have the circle y^2=2x-x^2 (1) which is a circle with radius 1 centred around (1,0).

    I need an expression, x=f(y), for the the top right quarter of the circle. For the life in me I can't find it.

    The original equation i was given was y=\sqrt{2x-x^2} (2), which I decided was equivalent to (1) if x,y\ge 0.

    I've tried simply rearranging and thought about inverses and such but I think I'm going round in circles.

    Any hints?
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    http://www.thestudentroom.co.uk/show...6&postcount=34

    ^ there I've found limits for a similar circle

    edit: I realise I didn't read, you want to integrate wrt x first for some reason:p:
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    Ummm... you could just specify the values of x and y so that you can only get that part of the circle?
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    (Original post by nota bene)
    http://www.thestudentroom.co.uk/show...6&postcount=34

    ^ there I've found limits for a similar circle
    I like how you knew I was finding limits :p:
    I'm not sure I can usr your example, though.
    For my problem the logical way would be to fix values of x and find limits of y - it's the easy way. Differentiate w.r.t. y then w.r.t. x.

    However for the question i'm trying to solve I'm given the double integral and have to reverse it, ie fix y and vary x.

    It's a good question really - still don't know where to go from here.
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    (Original post by grace_)
    Ummm... you could just specify the values of x and y so that you can only get that part of the circle?
    I need an expresion of the form x=f(y).... which isn't easy.
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    I think i've found it.... x=\sqrt{y^2-2y}.

    It seems right.
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    (Original post by kexy)
    I like how you knew I was finding limits :p:
    Not too hard to guess when I remembered the thread from yesterday:p:

    It seems right
    I think so too, but don't ask me to prove it; I'm having a dumb day!
 
 
 
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Updated: March 5, 2008

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