Question
Using definite intergeals, find the area of the circle $x^2+ y^2 = a^2$.
✓
Answer
Area of the circle $x^2+ y^2 = a^2$ will be thr $4$ times the area enclosed between $x = 0$ and $x = a$ in the first quadrant which is shaded.
$\text{A}=4\int\limits_{0}^{a}|\text{y}|\text{dx} $
$=4\int\limits_{0}^{a}\Big(\sqrt{\text{a}^{2}-\text{x}^{2}}\Big)\text{dx}$
$=4\Big[\frac{1}{2}\text{x}\sqrt{\text{a}^{2}-\text{x}^{2}}+\frac{1}{2}\text{a}^{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Big]^{\text{a}}_{0}$
$=4\big[0+\frac{1}{2}\text{a}^{2}\sin^{-1}1\big] $
$=4\Big[\frac{1}{2}\text{a}^{2}\frac{\pi}{2}\Big] $
$=\text{a}^{2}\pi\ \text{sq.}\ \text{units}$
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