Question

The area bounded by the parabola x = 4 − y² and y-axis, in square units, is
(a) $\frac{3}{32}$

(b) $\frac{32}{3}$

(c) $\frac{33}{2}$

(d) $\frac{16}{3}$

Answer 1

$\left(\mathrm{b}\right)\frac{32}{3}$



The points of intersection of the parabola and the y-axis are A(0, 2) and C(0, −2).
Therefore, the area of the required region ABCO,
$$\begin{gathered} A={\int }_{-2}^{2}x\mathit{d}y \\ ={\int }_{-2}^2\left(4-y^2\right)\mathit{d}y \\ ={\left[4y-\frac{y^3}{3}\right]}_{-2}^2 \\ =\left[4\left(2\right)-\frac{{\left(2\right)}^3}{3}\right]-\left[4\left(-2\right)-\frac{{\left(-2\right)}^3}{3}\right] \\ =\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right) \\ =8-\frac{8}{3}+8-\frac{8}{3} \\ =16-\frac{16}{3} \\ =\frac{32}{3}\mathrm{square}\mathrm{units} \end{gathered}$$