Question

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

Answer 1

To find: area of the region bounded by the curve y = $\sqrt{16-x^2}$ and x-axis



The area of the region CABC = 2(Area of the region OABO)
Thus,
$$\begin{gathered} \mathrm{Required}\mathrm{area}=2{\int }_0^4\left(y\right)dx \\ =2{\int }_0^4\left(\sqrt{16-x^2}\right)dx \\ =2{\left(\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}\right)}_0^4 \\ =2\left[\left(\frac{4}{2}\sqrt{16-4^2}+8{\sin }^{-1}\frac{4}{4}\right)-\left(\frac{0}{2}\sqrt{16-0^2}+8{\sin }^{-1}\frac{0}{4}\right)\right] \\ =2\left[\left(2\sqrt{16-16}+8{\sin }^{-1}1\right)-\left(0+0\right)\right] \\ =2\left[\left(0+8\times \frac{\mathrm{\pi }}{2}\right)\right] \\ =2\left[\left(4\mathrm{\pi }\right)\right] \\ =8\mathrm{\pi } \\ \mathrm{Thus},\mathrm{Area}=8\mathrm{\pi }\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, the correct option is (a).