The area of the region (in sq. units) bounded | vedclass

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(C) The given equation is $y = 2\sqrt{1 - x^2}$.Squaring both sides, we get $y^2 = 4(1 - x^2)$, which implies $x^2 + \frac{y^2}{4} = 1$.This is the eq

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$\pi$

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(C) The given equation is $y = 2\sqrt{1 - x^2}$.Squaring both sides, we get $y^2 = 4(1 - x^2)$, which implies $x^2 + \frac{y^2}{4} = 1$.This is the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a = 1$ and $b = 2$.The curve $y = 2\sqrt{1 - x^2}$ represents the upper half of the ellipse.The area of the full ellipse is $A = \pi ab = \pi(1)(2) = 2\pi$.The area bounded by the curve and the $X$-axis is the area of the upper half of the ellipse.Area $= \frac{1}{2} 	imes 2\pi = \pi 	ext{ sq. units}$.

The area of the region (in sq. units) bounded by the curve $y = 2\sqrt{1 - x^2}$ and the $X$-axis is . . . . . . .

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