Integral 2,-2 sqrt (4-x^2)dx  ,circle?

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  • Ian H
    Lv 7
    1 month ago

    y = 2 - 2√[4 – x^2] is the lower half of an ellipse 

    https://www.wolframalpha.com/input/?i=y+%3D+2+-+2%... 

    2√[4 – x^2] = 2 - y 

    4(4 – x^2) = (2 – y)^2 

    16 – 4x^2 = (y – 2)^2 

    4x^2 + (y – 2)^2 = 16 

    x^2/2^2 + (y – 2)^2/4^2 = 1 

    The related full ellipse is centred at (0, -2) 

    semi-axes 2 along x and 4 along y 

    https://www.wolframalpha.com/input/?i=x%5E2%2F2%5E... 

    For the integral, use x = 2sin(u) = 2s, dx = 2cos(u)du and  

    2[cos(u)]^2 du = [1 + cos(2u)] du and 

    sin(2u) = 2sin u * cos u and  

    2sin u = x and 2 cos u = √(4 – x^2) 

    Applying limits x from -2 to 2, I = 8 – 4𝛑 

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  • 1 month ago

    As david said it is a semicircle, so you could find what the area would be for a circle of radius 2 then cut it in half.  To actually carry out the calculation, are you familiar with trigonometric substitution?  if not I could explain it.   

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  • david
    Lv 7
    1 month ago

    not a circle ... only the top half === a semicircle.

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