The area bounded by the parabolas $\mathrm{y}=9 x^2, \mathrm{y}=\frac{x^2}{16}$ and the line $\mathrm{y}=1$ is

Given the parabolas $y = 9x^2$ and $y = \frac{x^2}{16}$ and the line $y = 1$, the area is determined by integrating with respect to $y$ due to the symmetry about the $y$-axis.

Solving for $x$ in terms of $y$ yields the right boundary $x = 4\sqrt{y}$ from $y = \frac{x^2}{16}$ and the left boundary $x = \frac{\sqrt{y}}{3}$ from $y = 9x^2$ in the first quadrant. The integral for half the area is $A_{half} = \int_0^1 \left( 4\sqrt{y} - \frac{\sqrt{y}}{3} \right) dy = \frac{11}{3} \int_0^1 \sqrt{y} dy$.

$\int_0^1 y^{1/2} dy = \left[ \frac{2}{3} y^{3/2} \right]_0^1 = \frac{2}{3}$
Thus, $A_{half} = \frac{11}{3} \cdot \frac{2}{3} = \frac{22}{9}$.

Due to symmetry, the total area is $2 \cdot \frac{22}{9} = \frac{44}{9}$ square units.

Answer: B