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

• 8 sq units

• $20\pi$ sq units

• $16\pi$ sq units

• $256\pi$ sq units

Correct Option (a)

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

%

$\Rightarrow \sqrt {16 - {x^2}} = 0$
$\therefore x = \pm 4$
Required area
$= \int_{ - 4}^4 {\left( {\sqrt {16 - {x^2}} } \right)} dx$

$\left[ {\int {\sqrt {{a^2} - {x^2}} } dx = \frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{a}} \right)} \right]$

$= 2\int_0^4 {\left( {\sqrt {{4^2} - {x^2}} } \right)} dx$
$= 2\left[ {\frac{{x\sqrt {{4^2} - {x^2}} }}{2} + \frac{{{4^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_0^4$

$= 2\left( {0 + \frac{{8\pi }}{2} - 0 - 0} \right)$
$= 8\pi$ sq.units