The area of the region enclosed by $y \leq 4 x^{2}, x^{2} \leq 9 y and y \leq 4$ , is equal to :

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$\frac{40}{3}$

$\frac{80}{3}$

$\frac{112}{3}$

$\frac{56}{3}$

answer is D.

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Detailed Solution

$\begin{array}{l} Δ = 2 \cdot \int_{0}^{4} (3 \sqrt{y} - \frac{\sqrt{y}}{2}) dy \\ = 2 \cdot \int_{0}^{4} \frac{5}{2} \sqrt{y} dy \\ = 5 {(\frac{2}{3} y^{\frac{3}{2}})}_{0}^{4} = \frac{80}{3} \end{array}$

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