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Math Help - sketch regions

  1. #1
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    sketch regions

    Sketch the region bounded by the semicircle y=sqrt(2x-x^2)
    and the x-axis and then evaluate the given integral.........

    then the class notes say therefore this is "a circle centred at (1,0) with radius 1.

    how do you know this is a circle with radius one?
    how do you know where it is centred?
    starting a table of values is the only way I know to work out how to plot this
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by samdmansam View Post
    Sketch the region bounded by the semicircle y=sqrt(2x-x^2)
    and the x-axis and then evaluate the given integral.........

    then the class notes say therefore this is "a circle centred at (1,0) with radius 1.

    how do you know this is a circle with radius one?
    how do you know where it is centred?
    starting a table of values is the only way I know to work out how to plot this
    You could do this:

    y=\sqrt{2x-x^2}\implies y^2=2x-x^2

    Then we see that x^2-2x+y^2=0

    Completing the square gives us x^2-2x+1+y^2=1\implies (x-1)^2+y^2=1

    Now comparing this to the standard form of a circle: (x-x_0)^2+(y-y_0)^2=r^2, where (x_0,y_0) is the center of the circle and r is the radius of the circle, we now see that (x-1)^2+y^2=1 is centered at (1,0) and has a radius of one.

    Now note that we are not graphing the entire circle. Its just the upper half, since that is hinted by the positive square root. If it was -\sqrt{2x-x^2}, that would be dealing with the semicircle below the x axis.

    Now, if we were to integrate this guy, it would require some extra thinking:

    \int_0^2\sqrt{2x-x^2}\,dx

    This is the same thing as saying \int_0^2\sqrt{1-(x-1)^2}\,dx

    Now apply the trig substitution x-1=\sin\vartheta

    Can you take it from here?

    --Chris
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