Question

The area (in sq units) of the region bounded by the curve $y=x^2$ and the line $y=16$ is

• A. $\frac{32}{3}$ sq units
• B. $\frac{256}{3}$ sq units
• C. $\frac{64}{3}$ sq units
• D. $\frac{128}{3}$ sq units

Answer 1

Answer: (B)

Area of region $=2\underset{10}{\overset{16}{\int }}\sqrt{y}dy$
(putting $y=16$ in $y=x^2$, we get $x=\pm 4$ )

$=2{\left[\frac{y^{3/2}}{3/2}\right]}_0^{16}=\frac{2}{3}\times 2\left[16^{3/2}-0\right]$
$=\frac{4\times 64}{3}=\frac{256}{3}$ sq units