Question 1050269
*[illustration fc9.JPG].
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The red circle is the circular garden.
The purple circle is the 3m radius from the edge of the circular garden fence.
The equations of the two circles are,
{{{x^2+y^2=(2.5)^2}}}
1.{{{x^2+y^2=6.25}}}
{{{(x-2.5)^2+y^2=(3)^2}}}
{{{(x-2.5)^2+y^2=9}}}
To find the intersection points substitute from eq. 1 into eq. 2,
{{{y^2=6.25-x^2}}}
So,
{{{(x-2.5)^2+6.25-x^2=9}}}
{{{x^2-5x+6.25+6.25-x^2=9}}}
{{{x^2-5x+12.5-x^2=9}}}
{{{-5x=-3.5}}}
{{{x=0.7}}}
So then integrate to get the sliver area (shown in green),
the limits of integration would be {{{x=0.7}}} and {{{x=2.5}}}.
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*[illustration fc10.JPG].
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{{{A=int((y[2]-y[1]),dx,x=0.7,2.5)}}}
{{{A=int((sqrt(9-(x-2.5)^2)-sqrt(6.25-x^2)),dx,x=0.7,2.5)}}}
{{{A=1.874}}} <-- Obtained numerically
Double that plus half of the area of the large circle will be the complete area that the goat can graze,
{{{A[tot]=2(1.874)+(pi(3)^2)/2}}}
{{{A[tot]=17.89}}}