1. **State the problem:** Find the area of the region bounded by the curve $x = 2 - y^2$ and the y-axis. 2. **Understand the boundaries:** The curve is $x = 2 - y^2$, a parabola opening leftwards with vertex at $(2,0)$. The y-axis is $x=0$. 3. **Find intersection points:** Set $x=0$ to find $y$ where the curve meets the y-axis: $$0 = 2 - y^2 \implies y^2 = 2 \implies y = \pm \sqrt{2}$$ 4. **Set up the integral:** The area between the curve and y-axis is the integral of $x$ from $y = -\sqrt{2}$ to $y = \sqrt{2}$: $$\text{Area} = \int_{-\sqrt{2}}^{\sqrt{2}} (2 - y^2) \, dy$$ 5. **Calculate the integral:** $$\int (2 - y^2) \, dy = 2y - \frac{y^3}{3} + C$$ Evaluate from $-\sqrt{2}$ to $\sqrt{2}$: $$\left[2y - \frac{y^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(2\sqrt{2} - \frac{(\sqrt{2})^3}{3}\right) - \left(-2\sqrt{2} + \frac{(-\sqrt{2})^3}{3}\right)$$ 6. **Simplify:** $$(\sqrt{2})^3 = \sqrt{2} \times \sqrt{2} \times \sqrt{2} = 2\sqrt{2}$$ So, $$= \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) - \left(-2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) = \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) + \left(2\sqrt{2} + \frac{2\sqrt{2}}{3}\right)$$ $$= 2\sqrt{2} + 2\sqrt{2} - \frac{2\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = 4\sqrt{2}$$ 7. **Final answer:** The area is $4\sqrt{2}$ square units. **Note:** The options are fractions with denominators 3, so rewrite $4\sqrt{2}$ as $\frac{12\sqrt{2}}{3}$, which is not listed. Re-examining step 6 carefully: $$\left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) - \left(-2\sqrt{2} + \frac{-2\sqrt{2}}{3}\right) = \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) + \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) = 4\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{12\sqrt{2} - 4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}$$ So the correct area is $\frac{8\sqrt{2}}{3}$ square units. **Answer: (c) $\frac{8\sqrt{2}}{3}$**