[SOLVED] Area inside a circle.

**olyviab1** · April 7th 2010, 07:59 PM

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?

**Prove It** · April 7th 2010, 09:41 PM

Originally Posted by olyviab1

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.

**olyviab1** · April 7th 2010, 09:54 PM

Originally Posted by Prove It

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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