Q113 / 180 · KCET 2007 · Mathematics
The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is
- A$16 \mathrm{sq}$ units
- B32 sq units
- Correct answerC$\frac{32}{3}$ sq units
- D$\frac{16}{3}$ sq units
Solution
Given curve $x=4-y^{2}$ and $Y$-axis.
The formed area is $A B C O A$ $$ \begin{aligned} &=2 \times \text { area of } A B O A \\ &=2 \times \int_{0}^{4} \sqrt{4-x} d x \\ &=2\left[-\frac{(4-x)^{3 / 2}}{3 / 2}\right]_{0}^{4} \\ &=2\left[-\frac{2}{3} \times 0+\frac{2}{3}(4)^{3 / 2}\right] \\ &=\frac{4}{3} \times 8=\frac{32}{3} \text { sq units } \end{aligned} $$
The formed area is $A B C O A$ $$ \begin{aligned} &=2 \times \text { area of } A B O A \\ &=2 \times \int_{0}^{4} \sqrt{4-x} d x \\ &=2\left[-\frac{(4-x)^{3 / 2}}{3 / 2}\right]_{0}^{4} \\ &=2\left[-\frac{2}{3} \times 0+\frac{2}{3}(4)^{3 / 2}\right] \\ &=\frac{4}{3} \times 8=\frac{32}{3} \text { sq units } \end{aligned} $$
Hence, the correct answer is option: 'C'.