Question

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

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

Answer 1

Answer: (A)

$x^2+y^2=16$
$\frac{1}{2}\pi (4{)}^2=8\pi$