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Thread: area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant

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    area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant

    Evaluate ydA where R is region between (y^2) + (x-2)^2 = 4 and y = x in first quadrant ...My ans is 32/3 , but given ans is 4/3.. What's wrong with my working?area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant-img_20160915_114715-1-.jpg
    Last edited by xl5899; Sep 14th 2016 at 08:14 PM.
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    Re: area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant

    Hey xl5899.

    I think you have the wrong part of the area - are you integrating above the line y = x or below it [in the first quadrant]?
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    Re: area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant

    A chiro pointed out there are two different regions bounded by that circle and line- the part of the disk above the line and the part below the line. But in neither is the integral of a single function from 0 to 4. If you are finding the area of the part of the disk above the line, your integral stops where the line intersects the circle. If you are finding the area of the part of the disk below the line, you will have one integrand before the x value where the line intersects the circle and another after.
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    Re: area bounded by (y^2) + (x-2)^2 = 4 and y = x in first quadrant

    Quote Originally Posted by chiro View Post
    Hey xl5899.

    I think you have the wrong part of the area - are you integrating above the line y = x or below it [in the first quadrant]?
    it's below it [in the first quadrant] , is it wrong ?
    sorry for the wrongly sketched diagram...
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