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Math Help - integration to find area please

  1. #1
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    integration to find area please

    hi

    I want to find the area inside the circle r=2, with x>0 and y>1

    I evaluate the

    integral (from x=0 to x=2) of the ingral (y=1 to y=+sqrd.root.of(4-x^2)of dydx

    and come up with pi-2 = 1.14 but book answer is 1.23

    pls help!!
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  2. #2
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    Quote Originally Posted by pepsi View Post
    hi

    I want to find the area inside the circle r=2, with x>0 and y>1

    I evaluate the

    integral (from x=0 to x=2) of the ingral (y=1 to y=+sqrd.root.of(4-x^2)of dydx

    and come up with pi-2 = 1.14 but book answer is 1.23

    pls help!!
    The integral \int_0^2\sqrt{4-x^2}dx is the area of \frac{1}{4} of the circle. So the total area should be:

    4\int_0^2\sqrt{4-x^2}dx

    The indefinite integral is I(x)=\frac{1}{2}\sqrt{4-x^2}x+2sin^{-1}\left(\frac{x}{2}\right)

    http://integrals.wolfram.com/index.j...D&random=false

    So did you calculate 4[I(2)-I(0)=4\pi

    Your book must be wrong. The area of a circle with radius 2 is 4\pi. You can show this using integration as I did above.
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    The integral \int_0^2\sqrt{4-x^2}dx is the area of \frac{1}{4} of the circle. So the total area should be:

    4\int_0^2\sqrt{4-x^2}dx

    The indefinite integral is I(x)=\frac{1}{2}\sqrt{4-x^2}x+2sin^{-1}\left(\frac{x}{2}\right)

    Wolfram Mathematica Online Integrator

    So did you calculate 4[I(2)-I(0)=4\pi

    Your book must be wrong. The area of a circle with radius 2 is 4\pi. You can show this using integration as I did above.

    thanks for your reply,

    however the question is not to find the area of the circle but that which is defined by x>0 and y>1 and the cirlce, that is, a subsection of the first orthant...
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