Area Bounded by Parabolic Curves

To find the area of the region bounded by the curves, we first rewrite the equations of the given curves in terms of $y$:

1. First curve: $x = -3y^2$

2. Second curve: $x = 1 - 4y^2$

To determine the points of intersection of these two parabolas, we set the two expressions for $x$ equal to each other: $-3y^2 = 1 - 4y^2$

Rearranging the terms: $4y^2 - 3y^2 = 1$ $y^2 = 1 \implies y = \pm 1$

Thus, the curves intersect at $y = -1$ and $y = 1$.

For any $y$ in the interval $[-1, 1]$, notice that $1 - 4y^2 \ge -3y^2$ because: $(1 - 4y^2) - (-3y^2) = 1 - y^2 \ge 0$

Therefore, the curve $x = 1 - 4y^2$ lies to the right of $x = -3y^2$.

The area $A$ bounded by the two curves can be found by integrating with respect to $y$: $A = \int_{-1}^{1} \left[ (1 - 4y^2) - (-3y^2) \right] dy$ $A = \int_{-1}^{1} (1 - y^2) \, dy$

Since $f(y) = 1 - y^2$ is an even function, we can simplify the integral as: $A = 2 \int_{0}^{1} (1 - y^2) \, dy$

Evaluating the integral: $A = 2 \left[ y - \frac{y^3}{3} \right]_{0}^{1}$ $A = 2 \left( 1 - \frac{1}{3} \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}$

Thus, the correct option is C.