1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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

Open in App
Solution

## 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, $\mathrm{Required}\mathrm{area}=2{\int }_{0}^{4}\left(y\right)dx\phantom{\rule{0ex}{0ex}}=2{\int }_{0}^{4}\left(\sqrt{16-{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}=2{\left(\frac{x}{2}\sqrt{16-{x}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\frac{x}{4}\right)}_{0}^{4}\phantom{\rule{0ex}{0ex}}=2\left[\left(\frac{4}{2}\sqrt{16-{4}^{2}}+8{\mathrm{sin}}^{-1}\frac{4}{4}\right)-\left(\frac{0}{2}\sqrt{16-{0}^{2}}+8{\mathrm{sin}}^{-1}\frac{0}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=2\left[\left(2\sqrt{16-16}+8{\mathrm{sin}}^{-1}1\right)-\left(0+0\right)\right]\phantom{\rule{0ex}{0ex}}=2\left[\left(0+8×\frac{\mathrm{\pi }}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=2\left[\left(4\mathrm{\pi }\right)\right]\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{Area}=8\mathrm{\pi }\mathrm{sq}.\mathrm{units}$ Hence, the correct option is (a).

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Area under the Curve
MATHEMATICS
Watch in App
Join BYJU'S Learning Program