The area (in square units) of the region bounded by the curves \(3y^2 = ax, y = a, a \gt 0\) and \(y\)-axis is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(a^2\)
The parabola \(3y^2 = ax\) gives \(x = \frac{3y^2}{a}\). It touches the y-axis at the origin, so the region lies between x = 0 and this curve, from y = 0 to the line y = a. Using horizontal strips:
\(\displaystyle \text{Area} = \int_{0}^{a} \frac{3y^2}{a}\,dy\)
\(\displaystyle = \frac{3}{a}\left[\frac{y^3}{3}\right]_{0}^{a}\)
\(\displaystyle = \frac{3}{a} \cdot \frac{a^3}{3} = a^2\)