Question

Calculate the area of the region bounded by the curve \( y^2 = 4x \) and the line \( x = 3 \).

• A. \(6\sqrt{3}\)
• B. \(8\sqrt{3}\)
• C. \(12\sqrt{3}\)
• D. \(4\sqrt{3}\)

Hint:

When a parabola is given as \(y^2 = 4ax\), it is often easier to integrate with respect to \(y\) since \(x\) can be directly written as \( \frac{y^2}{4a} \).

Answer 1

The Correct Option is C

Concept: To find the area between a curve and a vertical line, we integrate with respect to \(y\) if the curve is expressed as \(x\) in terms of \(y\). For a region between two curves \(x = f(y)\) and \(x = g(y)\), the area is: \[ A = \int_{y_1}^{y_2} \left(x_{\text{right}} - x_{\text{left}}\right) dy \] Here, \[ y^2 = 4x \Rightarrow x = \frac{y^2}{4} \] The right boundary is \(x = 3\) and the left boundary is the parabola.
Step 1: {Find the points of intersection.} Substitute \(x = 3\) into the parabola: \[ y^2 = 4(3) = 12 \] \[ y = \pm 2\sqrt{3} \] Thus limits are \[ y = -2\sqrt{3} \quad \text{to} \quad y = 2\sqrt{3} \]
Step 2: {Set up the area integral.} \[ A = \int_{-2\sqrt{3}}^{2\sqrt{3}} \left(3 - \frac{y^2}{4}\right) dy \]
Step 3: {Evaluate the integral.} \[ A = \left[3y - \frac{y^3}{12}\right]_{-2\sqrt{3}}^{2\sqrt{3}} \] Substituting limits: \[ = \left(6\sqrt{3} - 2\sqrt{3}\right) - \left(-6\sqrt{3} + 2\sqrt{3}\right) \] \[ = 4\sqrt{3} + 4\sqrt{3} \] \[ = 12\sqrt{3} \]