The area of the region bounded between the line \( x=3 \) and the parabola \( y^{2}=6x \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 12\sqrt{2} \) Sq. units
The parabola \( y^{2}=6x \) is symmetric about the x-axis, and the line \( x=3 \) meets it at \( y=\pm3\sqrt{2} \). So the area is twice the part above the x-axis.
\( \displaystyle A=2\int_{0}^{3}\sqrt{6x}\,dx \)
\( \displaystyle =2\sqrt{6}\left[\frac{2}{3}x^{3/2}\right]_{0}^{3} \)
\( \displaystyle =\frac{4\sqrt{6}}{3}\cdot3\sqrt{3} \)
\( \displaystyle =4\sqrt{18} \)
\( \displaystyle =12\sqrt{2} \)
Shortcut: The region is a parabolic segment, whose area is
\( \displaystyle \frac{2}{3}\times\text{base}\times\text{height} \).
Here the base is \( 6\sqrt{2} \) and the height is 3, so the area is
\( \displaystyle \frac{2}{3}\times6\sqrt{2}\times3=12\sqrt{2} \).