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The area (in sq. units) of the region bounded by the curve $y = \sqrt{16-x^2}$ and x-axis is |
$8π$ $16π$ $24π$ $32π$ |
$8π$ |
The correct answer is Option (1) → $8π$ Region is the upper semicircle of radius $4$ (since $y=\sqrt{16-x^{2}}$, $-4\le x\le4$). Area $=\displaystyle \int_{-4}^{4}\sqrt{16-x^{2}}\,dx=\frac{1}{2}\pi(4)^{2} = 8\pi$ |
