Question

The area of the region bounded by the parabola ( y^2 = 27x ) and the line ( x = 1 ) is ________ sq.units.

• A. ( 2\sqrt{3} )
• B. ( 3\sqrt{3} )
• C. ( 4\sqrt{3} )
• D. [suspicious link removed]

Hint:

Area of parabola $y^2 = 4ax$ from $x=0$ to $x=h$ is $\frac{8}{3}\sqrt{a}h^{3/2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the area between a rightward opening parabola and a vertical line.
Step 3: Detailed Explanation:
The equation of the parabola is \( y^2 = 27x \implies y = \pm \sqrt{27x} = \pm 3\sqrt{3x} \).
The region is bounded by \( x=0 \) to \( x=1 \). Due to symmetry about the x-axis:
\[ \text{Area} = 2 \int_0^1 y \, dx \]
\[ \text{Area} = 2 \int_0^1 3\sqrt{3} \sqrt{x} \, dx = 6\sqrt{3} \int_0^1 x^{1/2} \, dx \]
\[ \text{Area} = 6\sqrt{3} \left[ \frac{2}{3} x^{3/2} \right]_0^1 = 6\sqrt{3} \times \frac{2}{3} (1 - 0) \]
\[ \text{Area} = 4\sqrt{3} \text{ sq. units} \]
Step 4: Final Answer:
The area is \( 4\sqrt{3} \) sq. units.