let the region r be the area enclosed the function f(x)=x² - 2, the horizontal line y = 0, and the y - axis…

Explanation:

Step1: Find intersection points with x - axis

Set $y = x^{2}-2=0$. Then $x^{2}=2$, so $x=\pm\sqrt{2}$. Since we are considering the region with the y - axis, for the integral with respect to $x$, the left - hand limit is $0$ and the right - hand limit is $\sqrt{2}$. The area between a curve $y = f(x)$ and the $x$ - axis from $x = a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)|dx$. Here, for $0\leq x\leq\sqrt{2}$, $y=x^{2}-2\leq0$, so $|x^{2}-2|=2 - x^{2}$. The integral with respect to $x$ is $\int_{0}^{\sqrt{2}}(0-(x^{2}-2))dx=\int_{0}^{\sqrt{2}}(2 - x^{2})dx$.

Step2: Solve for $x$ in terms of $y$

From $y=x^{2}-2$, we get $x=\sqrt{y + 2}$ (since we are in the first - quadrant part considering the y - axis). The lower limit for $y$ is $- 2$ and the upper limit is $0$. The area between a curve $x = g(y)$ and the $y$ - axis from $y = c$ to $y = d$ is given by $A=\int_{c}^{d}g(y)dy$. The integral with respect to $y$ is $\int_{-2}^{0}\sqrt{y + 2}dy$.

Answer:

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