Q104 / 180 · KCET 2017 · Mathematics
The area of the region bounded by the curve \( y=x^{2} \) and the line \( y=16 \) is
- A\( \frac{32}{3} \) sq. units
- Correct answerB\( \frac{256}{3} \) sq. units
- C\( \frac{64}{3} \) sq. units
- D\( \frac{128}{3} \) sq. units
Solution
Given curve, $y=x^{2} \rightarrow(1)$
and given line, $y=16 \rightarrow(2)$
Therefore, required area is given by 2 Area of region $\mathrm{OAC}$ $$ \begin{array}{l} =2 \int_{0}^{16} \sqrt{y} d y \\ =2 \times \frac{2}{3} \times\left[y^{\frac{3}{2}}\right]_{0}^{16}=2 \times \frac{2}{3} \times(16)^{\frac{3}{2}} \\ =2 \times \frac{2}{3} \times(4)^{3}=\frac{256}{3} \end{array} $$
Therefore, required area is given by 2 Area of region $\mathrm{OAC}$ $$ \begin{array}{l} =2 \int_{0}^{16} \sqrt{y} d y \\ =2 \times \frac{2}{3} \times\left[y^{\frac{3}{2}}\right]_{0}^{16}=2 \times \frac{2}{3} \times(16)^{\frac{3}{2}} \\ =2 \times \frac{2}{3} \times(4)^{3}=\frac{256}{3} \end{array} $$
Hence, the correct answer is option: 'B'.