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Question
Using definite integrals, find the area of the circle x2 + y2 = a2.
Solution
Area of the circle x2 + y2 = a2 will be the 4 times the area enclosed between x = 0 and x = a in the first quadrant which is shaded.
\[A = 4 \int_0^a \left| y \right| d x\]
\[ = 4 \int_0^a \left( \sqrt{a^2 - x^2} \right) d x\]
\[ = 4 \left[ \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} \right]_0^a \]
\[ = 4\left[ 0 + \frac{1}{2} a^2 \sin^{- 1} 1 \right]\]
\[ = 4\left[ \frac{1}{2} a^2 \frac{\pi}{2} \right] ................\left( \because \sin^{- 1} 1 = \frac{\pi}{2} \right)\]
\[ = a^2 \pi\text{ sq units }\]
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