Find the area of the region defined by the conditions: $ \left\{ (x, y): 0 \leq y \leq \sqrt{9x}, y^2 \geq 3 - 6x \right\} \text{(in square units)} $
Show Hint
When finding the area between curves, first solve for the points of intersection and then integrate the difference between the curves over the appropriate interval.
The region is bounded by the conditions:
\[
0 \leq y \leq \sqrt{9x}, \quad y^2 \geq 3 - 6x
\]
First, let's solve for the points of intersection by solving the equations \( y = \sqrt{9x} \) and \( y^2 = 3 - 6x \).
From \( y = \sqrt{9x} \), we get:
\[
y^2 = 9x
\]
Substitute \( y^2 = 9x \) into the equation \( y^2 = 3 - 6x \):
\[
9x = 3 - 6x
\]
Solving for \( x \), we get:
\[
9x + 6x = 3 \quad \Rightarrow \quad 15x = 3 \quad \Rightarrow \quad x = \frac{1}{5}
\]
Substitute \( x = \frac{1}{5} \) into \( y^2 = 9x \):
\[
y^2 = 9 \times \frac{1}{5} = \frac{9}{5} \quad \Rightarrow \quad y = \sqrt{\frac{9}{5}}
\]
Thus, the points of intersection are at \( x = \frac{1}{5} \) and \( y = \sqrt{\frac{9}{5}} \).
Next, the area of the region can be found by integrating the distance between the curves over the interval from \( x = 0 \) to \( x = \frac{1}{5} \):
\[
A = \int_0^{1/5} \left( \sqrt{9x} - \sqrt{3 - 6x} \right) \, dx
\]
After performing the integration, we find that the area is given by:
\[
A = \frac{1}{3} \left( \frac{9}{5} \right)^{1/2}
\]
Thus, the correct answer is \( \frac{1}{3} \left( \frac{9}{5} \right)^{1/2} \).
