1. Write both curves in terms of $x$

The given curves are: $x+2y^2=0 \quad \Rightarrow \quad x=-2y^2$ $x+3y^2=1 \quad \Rightarrow \quad x=1-3y^2$

So the region is bounded between the parabolas:

• left/right curve: $x=-2y^2$
• left/right curve: $x=1-3y^2$

2. Find the points of intersection

At intersection points, $-2y^2=1-3y^2$ $y^2=1$ $y=\pm 1$

Corresponding $x$ values: $x=-2(1)=-2$ So the curves intersect at $(-2,1)$ and $(-2,-1)$.

3. Determine which curve is on the right

For a fixed $y$ between $-1$ and $1$, $x_1=-2y^2, \qquad x_2=1-3y^2$ Now, $x_2-x_1=(1-3y^2)-(-2y^2)=1-y^2$ Since $-1\le y\le 1$, we have $1-y^2\ge 0$. Thus, $1-3y^2 \ge -2y^2$ so the right curve is $x=1-3y^2$ and the left curve is $x=-2y^2$.

4. Set up the area integral

Area between curves when integrating with respect to $y$ is: $A=\int_{-1}^{1}\big(\text{right} - \text{left}\big)\,dy$ Hence, $A=\int_{-1}^{1}\left[(1-3y^2)-(-2y^2)\right]dy$ $A=\int_{-1}^{1}(1-y^2)\,dy$

5. Evaluate the integral

$A=\int_{-1}^{1}(1-y^2)\,dy$ $=\left[y-\frac{y^3}{3}\right]_{-1}^{1}$

At $y=1$: $1-\frac{1}{3}=\frac{2}{3}$ At $y=-1$: $-1-\left(-\frac{1}{3}\right)=-\frac{2}{3}$

Therefore, $A=\frac{2}{3}-\left(-\frac{2}{3}\right)=\frac{4}{3}$

6. Match with the options

$\boxed{\frac{4}{3}}$ So the correct option is D.