The problem requires finding the value of the double integral $\int_{0}^{2}\int_{y^2}^{y+2} \mathrm{dx}\mathrm{dy}$ by potentially changing the order of integration. We will determine the region of integration and express the integral in the order $dy \, dx$.

Integral Region Analysis

The given integral is defined by the bounds:

• $0 \le y \le 2$
• $y^2 \le x \le y+2$

This defines a region R in the xy-plane bounded by the curves $y=0$, $y=2$, $x=y^2$, and $x=y+2$. The region is specifically bounded below by the x-axis ($y=0$), on the right by the line $x=y+2$, and on the left by the parabola $x=y^2$. The upper limit for y is 2.

Identifying Boundaries

Let's identify the key boundary curves and intersection points relevant to the region:

• Curve 1: $x = y^2$ (parabola opening right). For $y \ge 0$, this can be written as $y = \sqrt{x}$.
• Curve 2: $x = y+2$. This can be written as $y = x-2$.
• Lower bound for y: $y=0$.
• Upper bound for y: $y=2$.

Intersection of $x=y^2$ and $x=y+2$: $y^2 = y+2 \implies y^2-y-2=0 \implies (y-2)(y+1)=0$. Since $y \ge 0$, the relevant intersection is at $y=2$, which gives $x=4$. The point is $(4, 2)$.

Intersection of $y=0$ and $x=y+2$: $x = 0+2 = 2$. The point is $(2, 0)$.

Intersection of $y=0$ and $x=y^2$: $x = 0^2 = 0$. The point is $(0, 0)$.

The region is bounded by the curve $x=y^2$ from $(0,0)$ to $(4,2)$, the line $x=y+2$ from $(2,0)$ to $(4,2)$, and the line $y=0$ from $(0,0)$ to $(2,0)$.

Changing Integration Order

To change the order to $dy \, dx$, we need to express the bounds for $y$ in terms of $x$. The range of $x$ in the region is from $0$ to $4$. We must split the region into two parts based on the nature of the lower boundary for $y$:

1. For $0 \le x \le 2$:
   The lower bound for $y$ is $y=0$.
   The upper bound for $y$ is given by the parabola $x=y^2$, so $y=\sqrt{x}$.
   The integral for this part is: $\int_{0}^{2}\int_{0}^{\sqrt{x}} \mathrm{dy}\mathrm{dx}$.
2. For $2 \le x \le 4$:
   The lower bound for $y$ is given by the line $x=y+2$, so $y=x-2$.
   The upper bound for $y$ is given by the parabola $x=y^2$, so $y=\sqrt{x}$.
   The integral for this part is: $\int_{2}^{4}\int_{x-2}^{\sqrt{x}} \mathrm{dy}\mathrm{dx}$.

Final Integral Expression

Combining the two parts gives the equivalent double integral with the order of integration changed to $dy \, dx$:

This matches Option 3.