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  • The area of the region bounded by the curve y=\sqrt{16-x^2} and x-axis is. A. 8π sq units

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

A. 8π sq units

B. 20π sq units
C. 16π sq units
D. 256π sq units

 

Answers (1)

A)

\text { The curve } y=\sqrt{16-x^{2}} \text { and } x \text { -axis; } y=0

\begin{array}{l} \Rightarrow \sqrt{16-\mathrm{x}^{2}}=0 \\ \therefore \mathrm{x}=\pm 4 \end{array}

\\\text{Required area}\\ =\int_{-4}^{4}\left(\sqrt{16-\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ \\$\left[\int \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{\mathrm{x} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}{2}+\frac{\mathrm{a}^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]$ \\$=2 \int_{0}^{4}\left(\sqrt{4^{2}-x^{2}}\right) d x$ \\$=2\left[\frac{\mathrm{x} \sqrt{4^{2}-\mathrm{x}^{2}}}{2}+\frac{4^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4}\right)\right]_{0}^{4}$ \\$=2\left(0+\frac{8 \pi}{2}-0-0\right)$ \\$=8 \pi$ sq.units

 

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