The area in the first quadrant bounded by y=9x^2,\ x=0,\ y=1 and y=4…

The correct answer is Option (4) → 14/9 Sq.Units

The region lies between the lines y = 1 and y = 4, so horizontal strips are natural. On the curve, in the first quadrant,

    \( \displaystyle x=\frac{\sqrt{y}}{3} \)

So

    \( \displaystyle \text{Area}=\int_{1}^{4}x\,dy \)

    \( \displaystyle =\frac{1}{3}\int_{1}^{4}y^{1/2}\,dy \)

    \( \displaystyle =\frac{1}{3}\cdot\frac{2}{3}\left[y^{3/2}\right]_{1}^{4} \)

    \( \displaystyle =\frac{2}{9}(8-1) \)

    \( \displaystyle =\frac{14}{9} \)