Question

The area 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)

According to the question ,

Required area$=2 \int\limits_{0}^{16} x\, d y$
$=2 \int\limits_{0}^{16} \sqrt{y} \,d y $
$=2\left[\frac{y^{3 / 2}}{3 / 2}\right]^{16}_{0}$
$=\frac{4}{3}\left[16^{3 / 2}-0^{3 / 2}\right]$
$=\frac{4}{3} \times 64=\frac{256}{3}$ sq units.