The area (in sq. units) of the region bounded by \( x^2=4y \), \( y=1,y=9 \) and \( x=0 \) in the first quadrant is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{104}{3} \)
The region lies between the y-axis (x = 0) and the parabola, from y = 1 to y = 9. In the first quadrant, \( x=2\sqrt{y} \), so integrate with respect to y:
\( \displaystyle \text{Area}=\int_{1}^{9}2\sqrt{y}\,dy \)
\( \displaystyle =2\left[\frac{2}{3}y^{3/2}\right]_{1}^{9} \)
\( \displaystyle =\frac{4}{3}(27-1) \)
\( \displaystyle =\frac{104}{3} \)
Tip: The region starts at the line y = 1, not at the origin. Taking the lower limit as 0 gives \( \frac{4}{3}\times 27=36 \), Option (2).