The area of the region bounded by y=-sqrt{16- | vedclass

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(A) The equation $y=-\sqrt{16-x^{2}}$ represents the lower semi-circle of the circle $x^{2}+y^{2}=16$,which has a radius $r=4$ and center at the origi

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$8 \pi$ sq units

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(A) The equation $y=-\sqrt{16-x^{2}}$ represents the lower semi-circle of the circle $x^{2}+y^{2}=16$,which has a radius $r=4$ and center at the origin $(0,0)$.Since the area is bounded by this curve and the $X$-axis,we are looking for the area of the semi-circle below the $X$-axis.The area of a full circle is $\pi r^{2} = \pi(4)^{2} = 16\pi$.Therefore,the area of the semi-circle is $\frac{1}{2} 	imes 16\pi = 8\pi$ sq units.Alternatively,using integration:$	ext{Area} = \left| \int_{-4}^{4} (-\sqrt{16-x^{2}}) dx ight|$$= \left| \left[ \frac{x}{2} \sqrt{16-x^{2}} + \frac{16}{2} \sin^{-1} \frac{x}{4} ight]_{-4}^{4} ight|$$= \left| [0 + 8 \sin^{-1}(1)] - [0 + 8 \sin^{-1}(-1)] ight|$$= \left| 8(\frac{\pi}{2}) - 8(-\frac{\pi}{2}) ight| = |4\pi + 4\pi| = 8\pi$ sq units.

The area of the region bounded by $y=-\sqrt{16-x^{2}}$ and the $X$-axis is

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