Question

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

• A. $8\pi$ sq units
• B. $20\pi$ sq units
• C. $16\pi$ sq units
• D. $256\pi$ sq units

Answer 1

Answer: (A)

We have, $y=\sqrt{16-x^2}$
On squaring both sides, we get
$y^2=16-x^2$
$x^2+y^2=16$
Let us sketch the figure of the curve which represents a circle. $(y\ge 0)$

$\therefore$ Area of shaded region
$=$ Area $(BOACB)$
$=\underset{-4}{\overset{4}{\int }}\sqrt{16-x^2}dx$
$=\underset{-4}{\overset{4}{\int }}\sqrt{(4{)}^2-x^2}dx$
$={\left[\frac{x}{2}\sqrt{16-x^2}+8{\sin }^{-1}\left(\frac{x}{1}\right)\right]}_{-4}^4$
$=[\left\{\frac{4}{2}\sqrt{16-16}+8{\sin }^{-1}\left(\frac{4}{4}\right)\right\}$
$-\left\{\frac{-4}{2}\sqrt{16-(-4{)}^2}+8{\sin }^{-1}\left(\frac{-4}{4}\right)\right\}]$
$=\left[2\times 0+8\times \frac{\pi }{2}+8\times \frac{\pi }{2}\right]$
$=8\pi$ sq units