Step 1: Understanding the Question:
The question asks for the area of the region enclosed by the parabola \(y^2 = 4x\) (which opens to the right) and the vertical line \(x = 3\).
Step 2: Key Formula or Approach:
To find the area bounded by curves, we can integrate. Since the parabola is given as \(y^2 = 4x\), it's easier to express \(x\) in terms of \(y\) and integrate with respect to \(y\). The area \(A\) between two curves \(x_{right}\) and \(x_{left}\) from \(y=c\) to \(y=d\) is given by:
\[
A = \int_{c}^{d} (x_{right} - x_{left}) \, dy
\]
Step 3: Detailed Explanation:
(i) Find the points of intersection:
Substitute \(x = 3\) into the equation of the parabola to find the limits of integration:
\[
y^2 = 4(3) = 12
\]
\[
y = \pm\sqrt{12} = \pm 2\sqrt{3}
\]
So, the region extends from \(y = -2\sqrt{3}\) to \(y = 2\sqrt{3}\).
(ii) Set up the integral:
In the bounded region, the right boundary is the line \(x = 3\) (\(x_{right} = 3\)), and the left boundary is the parabola \(y^2 = 4x\), or \(x = \frac{y^2}{4}\) (\(x_{left} = \frac{y^2}{4}\)).
The area is:
\[
A = \int_{-2\sqrt{3}}^{2\sqrt{3}} \left(3 - \frac{y^2}{4}\right) dy
\]
Since the integrand \( (3 - \frac{y^2}{4}) \) is an even function and the limits are symmetric about the origin, we can simplify the integral:
\[
A = 2 \int_{0}^{2\sqrt{3}} \left(3 - \frac{y^2}{4}\right) dy
\]
(iii) Evaluate the integral:
\[
A = 2 \left[ 3y - \frac{y^3}{4 \cdot 3} \right]_{0}^{2\sqrt{3}} = 2 \left[ 3y - \frac{y^3}{12} \right]_{0}^{2\sqrt{3}}
\]
Now, substitute the upper limit \(y = 2\sqrt{3}\):
\[
A = 2 \left[ \left(3(2\sqrt{3}) - \frac{(2\sqrt{3})^3}{12}\right) - (0) \right]
\]
\[
A = 2 \left[ 6\sqrt{3} - \frac{8 \cdot 3\sqrt{3}}{12} \right] = 2 \left[ 6\sqrt{3} - \frac{24\sqrt{3}}{12} \right]
\]
\[
A = 2 [6\sqrt{3} - 2\sqrt{3}] = 2[4\sqrt{3}] = 8\sqrt{3}
\]
Step 4: Final Answer:
The area of the region is \(8\sqrt{3}\) square units.