let the region r be the area enclosed the function f(x)=x² - 2, the horizontal line y = 2, and the y - axis…

Explanation:

Step1: Find intersection points with respect to x

Set $f(x)=x^{2}-2 = 2$, then $x^{2}=4$, so $x = 2$ or $x=-2$. Since we are considering the region with the $y -$axis, we focus on $x\in[0,2]$. The upper - curve is $y = 2$ and the lower - curve is $y=x^{2}-2$. The area with respect to $x$ is given by the integral of the difference of the upper and lower functions. $$A_x=\int_{0}^{2}(2-(x^{2}-2))dx=\int_{0}^{2}(4 - x^{2})dx$$

Step2: Express $x$ in terms of $y$

From $y=x^{2}-2$, we get $x=\sqrt{y + 2}$ (since $x\geq0$ for the region we care about). The left - hand side is $x = 0$ and the right - hand side is $x=\sqrt{y + 2}$, and $y$ ranges from $- 2$ to $2$. The area with respect to $y$ is given by the integral of the right - hand function minus the left - hand function with respect to $y$. $$A_y=\int_{-2}^{2}\sqrt{y + 2}dy$$

Answer:

Integral in terms of $x$: $\int_{0}^{2}(4 - x^{2})dx$ Integral in terms of $y$: $\int_{-2}^{2}\sqrt{y + 2}dy$