The area bounded by the parabola x = 4 − y2 and y-axis, in square units, is ____________ .

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KollegeApply · Accepted Answer

$\frac{32}{3}$ 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, A = \int_{- 2}^2 x d y = \int_{- 2}^2 \left( 4 - y^2 \right) d y = \left[ 4y - y^$\frac{3}{3}$ \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 - $\frac{16}{3}$ = $\frac{32}{3}$\text{ square units } [Solution diagram: 43939_solution_1.png]

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