Step 1: Understanding the Concept:
Finding the area of a bounded region involves setting up a definite integral. Here, we are looking for the area between a parabola \(y^2 = 4ax\) and a vertical line \(x = 2a\) in the first quadrant. In the first quadrant, \(y\) is positive, so the curve is \(y = \sqrt{4ax} = 2\sqrt{a}\sqrt{x}\). The area is calculated by integrating the function \(y\) with respect to \(x\) from the starting boundary (origin, \(x=0\)) to the ending boundary (\(x=2a\)).
Step 2: Key Formula or Approach:
Area \(A = \int_{x_1}^{x_2} y \, dx\).
Given: \(y = 2\sqrt{a}x^{1/2}\), limits \(0 \to 2a\).
Step 3: Detailed Explanation:
Setting up the integral for the first quadrant:
\[ A = \int_{0}^{2a} 2\sqrt{a} \cdot \sqrt{x} \, dx \]
Factor out the constant \(2\sqrt{a}\):
\[ A = 2\sqrt{a} \int_{0}^{2a} x^{1/2} \, dx \]
Integrate using the power rule \(\int x^n dx = \frac{x^{n+1}}{n+1}\):
\[ A = 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2a} \]
\[ A = 2\sqrt{a} \cdot \frac{2}{3} \left[ x^{3/2} \right]_{0}^{2a} \]
\[ A = \frac{4\sqrt{a}}{3} \left[ (2a)^{3/2} - 0^{3/2} \right] \]
Simplifying \((2a)^{3/2}\):
\[ (2a)^{3/2} = (2a)^1 \cdot (2a)^{1/2} = 2a \sqrt{2} \sqrt{a} \]
Substitute this back:
\[ A = \frac{4\sqrt{a}}{3} \cdot (2a \sqrt{2} \sqrt{a}) \]
Multiply terms together:
\[ A = \frac{4 \cdot 2 \cdot \sqrt{2} \cdot a \cdot (\sqrt{a} \cdot \sqrt{a})}{3} \]
\[ A = \frac{8\sqrt{2} a \cdot a}{3} = \frac{8\sqrt{2}a^2}{3} \]
Step 4: Final Answer:
The area in the first quadrant is \(\frac{8\sqrt{2}a^2}{3}\) square units. This matches option (B).