The area of the given bounded by the curves x2 = 16y, y = 1, y = 4 and the Y-axis, lying in the first quadrant:

Question

The area of the given bounded by the curves $x^{2} = 16 y$ , $y = 1$ , $y = 4$ and the $Y$ -axis, lying in the first quadrant:

Smart Achievers · Accepted Answer

(3)

Ans. $\frac{56}{3}$ sq. units

Explanation:

$Let A be the required area.$

$A = \int_{1}^{4} x d y$

$= \int_{1}^{4} \sqrt{16 y} d y = \int_{1}^{4} 4 \sqrt{y} d y$

$= 4 {[\frac{2}{3} y^{3 / 2}]}_{1}^{4}$

$= \frac{8}{3} \times [8 - 1] = \frac{8}{3} \times 7$

$∴$ $A = \frac{56}{3} sq. units.$

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