The area of the region (in sq. units), in the first quadrant bounded by the parabola $$y = 9x^2$$ and the lines $$x = 0$$, $$y = 1$$ and $$y = 4$$, is :

To find the area in the first quadrant bounded by the parabola $$ y = 9x^2 $$, the y-axis ($$ x = 0 $$), and the horizontal lines $$ y = 1 $$ and $$ y = 4 $$, we need to determine the region enclosed by these curves and lines. Since the boundaries include horizontal lines, it is convenient to integrate with respect to $$ y $$. This approach simplifies the calculation because the region can be described using $$ y $$ as the independent variable.

First, rewrite the equation of the parabola in terms of $$ x $$ as a function of $$ y $$. Given $$ y = 9x^2 $$, solve for $$ x $$:

$$ y = 9x^2 $$

$$ x^2 = \frac{y}{9} $$

$$ x = \sqrt{\frac{y}{9}} \quad \text{(since we are in the first quadrant, } x \geq 0\text{)} $$

$$ x = \frac{\sqrt{y}}{3} $$

The region is bounded on the left by the y-axis ($$ x = 0 $$), on the right by the parabola $$ x = \frac{\sqrt{y}}{3} $$, below by $$ y = 1 $$, and above by $$ y = 4 $$. Therefore, for each fixed $$ y $$ between 1 and 4, $$ x $$ ranges from 0 to $$ \frac{\sqrt{y}}{3} $$.

The area $$ A $$ is given by the integral:

$$ A = \int_{y=1}^{y=4} \left( \text{right} - \text{left} \right) dy = \int_{1}^{4} \left( \frac{\sqrt{y}}{3} - 0 \right) dy = \int_{1}^{4} \frac{\sqrt{y}}{3} dy $$

Factor out the constant $$ \frac{1}{3} $$:

$$ A = \frac{1}{3} \int_{1}^{4} \sqrt{y} dy $$

Recall that $$ \sqrt{y} = y^{1/2} $$. The antiderivative of $$ y^{1/2} $$ is $$ \frac{y^{3/2}}{3/2} = \frac{2}{3} y^{3/2} $$. Therefore:

$$ \int \sqrt{y} dy = \frac{2}{3} y^{3/2} $$

Now evaluate the definite integral:

$$ \int_{1}^{4} \sqrt{y} dy = \left[ \frac{2}{3} y^{3/2} \right]_{1}^{4} $$

Substitute the limits:

At $$ y = 4 $$:

$$ \frac{2}{3} (4)^{3/2} = \frac{2}{3} (4^{1/2})^3 = \frac{2}{3} (2)^3 = \frac{2}{3} \times 8 = \frac{16}{3} $$

At $$ y = 1 $$:

$$ \frac{2}{3} (1)^{3/2} = \frac{2}{3} (1) = \frac{2}{3} $$

Subtract the lower limit value from the upper limit value:

$$ \left[ \frac{2}{3} y^{3/2} \right]_{1}^{4} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3} $$

Now substitute back into the expression for $$ A $$:

$$ A = \frac{1}{3} \times \frac{14}{3} = \frac{14}{9} $$

Thus, the area is $$ \frac{14}{9} $$ square units.

Comparing with the options:

A. $$ \frac{7}{9} $$

B. $$ \frac{14}{3} $$

C. $$ \frac{7}{3} $$

D. $$ \frac{14}{9} $$

Hence, the correct answer is Option D.