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Thread: [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; Apr 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 $\displaystyle (x, y) = (1, 0)$.

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

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


    By subtracting the first equation from the second we find

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

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

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

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

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

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



    Putting the equations as $\displaystyle y$ in terms of $\displaystyle x$ gives:

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

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


    So we can finally say:

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

    You will need to use trigonometric substitution to solve these.

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

    and $\displaystyle x = \sin{\theta}$ for the second.
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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 $\displaystyle (x, y) = (1, 0)$.

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

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


    By subtracting the first equation from the second we find

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

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

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

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

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

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



    Putting the equations as $\displaystyle y$ in terms of $\displaystyle x$ gives:

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

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


    So we can finally say:

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

    You will need to use trigonometric substitution to solve these.

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

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


    Thank you so much your explanation is great.
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