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Math Help - [SOLVED] Area inside a circle.

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    Question [SOLVED] Area inside a circle.

    a circular corral of unit radius(radius = 1) is enclosed by a fence. A goat inside the corral is tied to the fence with a rope of length a = 1.5 . What is the area of the region (inside the corral) that the goat can graze?
    Last edited by mr fantastic; April 9th 2010 at 02:17 AM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by olyviab1 View Post
    a circular corral of unit radius(radius = 1) is enclosed by a fence. A goat inside the corral is tied to the fence with a rope of length a = 1.5 . What is the area of the region (inside the corral) that the goat can graze?
    It always helps to draw a picture.


    You can have the second circle be centred anywhere on the first, I've just chosen the point (x, y) = (1, 0).

    It's important to note that the equations of the circles are

    x^2 + y^2 = 1 and (x - 1)^2 + y^2 = \frac{9}{4}.


    By subtracting the first equation from the second we find

    (x - 1)^2 + y^2 - (x^2 + y^2) = \frac{9}{4} - 1

    x^2 - 2x + 1 + y^2 - x^2 - y^2 = \frac{5}{4}

    -2x + 1 = \frac{5}{4}

    -2x = \frac{1}{4}

    x = -\frac{1}{8}.

    So the equations intersect at x = -\frac{1}{8}.



    Putting the equations as y in terms of x gives:

    y = \pm \sqrt{1 - x^2} and y = \pm \sqrt{\frac{9}{4} - (x - 1)^2}.

    Since these are symmetrical about the x axis, we can work out the areas enclosed by the above semicircles, and then double it.


    So we can finally say:

    A = 2\left(\int_{-\frac{1}{2}}^{-\frac{1}{8}}{\sqrt{\frac{9}{4} - (x - 1)^2}\,dx} + \int_{-\frac{1}{2}}^{1}{\sqrt{1 - x^2}\,dx}\right)

    You will need to use trigonometric substitution to solve these.

    Try x - 1 = \frac{3}{2}\sin{\theta} for the first

    and x = \sin{\theta} for the second.
    Attached Thumbnails Attached Thumbnails [SOLVED] Area inside a circle.-goat-problem.jpg  
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  3. #3
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    Quote Originally Posted by Prove It View Post
    It always helps to draw a picture.


    You can have the second circle be centred anywhere on the first, I've just chosen the point (x, y) = (1, 0).

    It's important to note that the equations of the circles are

    x^2 + y^2 = 1 and (x - 1)^2 + y^2 = \frac{9}{4}.


    By subtracting the first equation from the second we find

    (x - 1)^2 + y^2 - (x^2 + y^2) = \frac{9}{4} - 1

    x^2 - 2x + 1 + y^2 - x^2 - y^2 = \frac{5}{4}

    -2x + 1 = \frac{5}{4}

    -2x = \frac{1}{4}

    x = -\frac{1}{8}.

    So the equations intersect at x = -\frac{1}{8}.



    Putting the equations as y in terms of x gives:

    y = \pm \sqrt{1 - x^2} and y = \pm \sqrt{\frac{9}{4} - (x - 1)^2}.

    Since these are symmetrical about the x axis, we can work out the areas enclosed by the above semicircles, and then double it.


    So we can finally say:

    A = 2\left(\int_{-\frac{1}{2}}^{-\frac{1}{8}}{\sqrt{\frac{9}{4} - (x - 1)^2}\,dx} + \int_{-\frac{1}{2}}^{1}{\sqrt{1 - x^2}\,dx}\right)

    You will need to use trigonometric substitution to solve these.

    Try x - 1 = \frac{3}{2}\sin{\theta} for the first

    and x = \sin{\theta} for the second.


    Thank you so much your explanation is great.
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