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# Find the perimeter of the circle with radius $a$.

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## 1 Answer

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Toolbox:
• Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx$ or $\int \limits _a^b y dx$
• If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
Equation of the circle is $x^2+y^2=x^2$ .
By symmetry, the arc length (circumference ) is 4x length of are in the arc length in the first quadrent
$L=4 \int \limits _0^a \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2 \normalsize dx}$
Step 2:
Now $x^2+y^2=a^2$
$\therefore \;2x+2ydy=0$
=>$\large\frac{dy}{dx}=\frac{-x}{y}$
Step 3:
$1+\bigg(\large\frac{dy}{dx}\bigg)$$=1+ \large\frac{x^2}{y^2} \qquad=\large\frac{x^2+y^2}{y^2} \qquad=\large\frac{a^2}{a^2-x^2} Step 4: L= 4 \int \limits_0^a \large\frac{a}{\sqrt {a^2-x^2}}\;$$dx$
$\quad= 4a \sin^{-1} \large\frac{x}{a}\bigg]_0^a$
$\quad=4a \times \large\frac{\pi}{2}$
$\quad=2 \pi a \;units$
answered Aug 17, 2013 by

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