Question

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

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

Hint:

For curves of the form \(y^2 = ax\), remember the graph is symmetric about the \(x\)-axis. So the total area between the curve and a vertical boundary can be computed using \[ \text{Area} = 2\int y\,dx. \] This avoids calculating upper and lower parts separately.

Answer 1

The Correct Option is C

Concept: For curves given in the form \(y^2=f(x)\), the area enclosed between the curve and a vertical line is often found using symmetry: \[ \text{Area} = 2 \int_{x_1}^{x_2} y\,dx \] because the curve is symmetric about the \(x\)-axis.

Step 1: Express \(y\) from the equation of the curve. \[ y^2=4x \] \[ y=\pm \sqrt{4x} \] \[ y=\pm 2\sqrt{x} \]

Step 2: Set up the area integral. The curve intersects the line \(x=3\), so the limits are \(x=0\) to \(x=3\). \[ \text{Area} = 2\int_{0}^{3} 2\sqrt{x}\,dx \] \[ =4\int_{0}^{3} x^{1/2}\,dx \]

Step 3: Evaluate the integral. \[ 4\left[\frac{2}{3}x^{3/2}\right]_{0}^{3} \] \[ =\frac{8}{3}(3^{3/2}) \] \[ =\frac{8}{3}(3\sqrt{3}) \] \[ =8\sqrt{3} \]