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The total area enclosed by the circle $x^2 + y^2 = a^2$ is
A$2\pi a$
B$4\pi a^2$
C$\frac{\pi a^2}{2}$
D$\pi a^2$
Answer & Solution
Correct answer: D. $\pi a^2$
1. By symmetry the area is $4$ times the first-quadrant region: $A = 4\int_0^a \sqrt{a^2 - x^2}\,dx$.
2. Using $\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$, evaluate from $0$ to $a$.
3. At $x = a$ the bracket is $\frac{a^2}{2}\sin^{-1}1 = \frac{a^2}{2}\cdot\frac{\pi}{2}$.
4. So $A = 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2$.
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.2_
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