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Find the area of the circle whose radius is $a$ .

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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:
Area of circle $x^2+y^2=a^2$
$\quad=4 \times$ area in first quadrant bounded by $x^2+y^2=a^2,x=0,x=a$
Step 2:
$A= 4 \int \limits _0^a ydx=4 \int \limits_0^a \sqrt {a^2-x^2}dx$
$\quad= 4 \bigg[\large\frac{a^2}{2} $$\sin ^{-1} \large\frac{x}{a} -\frac{x}{2} $$\sqrt {a^2-x^2}\bigg]_0^a$
$\quad= 4 \bigg[\large\frac{a^2}{2} $$\sin ^{-1} 1\bigg]$
$\quad=4. \large\frac{a^2}{2}.\large\frac{\pi}{2}$
$\quad= \pi a^2 sq.units$
answered Aug 16, 2013 by meena.p
 

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