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Math Help - double intergration

  1. #1
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    double intergration

    hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?
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  2. #2
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    Quote Originally Posted by karljones View Post
    hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?
    (x^2+y^2-4y=0)
    Rewrite this as
    x^2 + y^2 - 4y + 4 = 4
    Or
    x^2 + (y-2)^2 = 4.......(1)
    Here r = 2.
    Now put x = r*cosθ and (y-2) = r*sinθ . Find the value of x or y at the angles π/3 and π/4.
    Then find the intg(x*dy) between the above limits to get the area.
    Last edited by sa-ri-ga-ma; March 15th 2010 at 04:07 AM.
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  3. #3
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    hey thanks for your reply sa ri ga ma. pls see attached. The area is in green, so i don't think r can be constant?
    Attached Thumbnails Attached Thumbnails double intergration-circle.jpg  
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  4. #4
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    Quote Originally Posted by karljones View Post
    hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?
    x^2 + y^2 - 4y = 0 \Rightarrow r^2 - 4r \sin \theta = 0 \Rightarrow r = 4 \sin \theta (r = 0 is trivial).
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  5. #5
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    So then i need to evalute the following:

    integral (theta = pi/3 and pi / 4) integral (r=0 and r= 4 sin theta) 1.r dr dtheta ???

    is this correct? many thanks.
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  6. #6
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    correct
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