Question

The area of the region bounded by the curve y = x² and the line y = 16 is
$$\begin{gathered} \left(\mathrm{a}\right)\frac{32}{3} \\ \left(\mathrm{b}\right)\frac{256}{3} \\ \left(\mathrm{c}\right)\frac{64}{3} \\ \left(\mathrm{d}\right)\frac{128}{3} \end{gathered}$$

Answer 1

To find: area of the region bounded by the curve y = x² and the line y = 16



The required area of the region COBC = 2(Area of the region bounded by AOBA)

Thus,
$$\begin{gathered} \mathrm{Area}=2{\int }_0^{16}xdy \\ =2{\int }_0^{16}\sqrt{y}dy \\ =2{\left(\frac{y^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right)}_0^{16} \\ =\frac{4}{3}{\left(y^{\frac{3}{2}}\right)}_0^{16} \\ =\frac{4}{3}\left({16}^{\frac{3}{2}}-0^{\frac{3}{2}}\right) \\ =\frac{4}{3}\left(4^3\right) \\ =\frac{4}{3}\left(64\right) \\ =\frac{256}{3} \\ \mathrm{Thus},\mathrm{Area}=\frac{256}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the correct option is (b).