The area of the region enclosed by y leq 4x^{ | vedclass

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(D) The region is bounded by $y = 4x^{2}$ (or $x^{2} = y/4$),$x^{2} = 9y$,and $y = 4$.From the graph,the area is symmetric about the $y$-axis.For a fi

Vedclass · Accepted Answer

$\frac{80}{3}$

Vedclass · Answer

(D) The region is bounded by $y = 4x^{2}$ (or $x^{2} = y/4$),$x^{2} = 9y$,and $y = 4$.From the graph,the area is symmetric about the $y$-axis.For a fixed $y$,the $x$-coordinates of the curves are $x = \pm \sqrt{y}/2$ and $x = \pm 3\sqrt{y}$.The width of the shaded region at height $y$ is $(3\sqrt{y} - \sqrt{y}/2) + (3\sqrt{y} - \sqrt{y}/2) = 2(5\sqrt{y}/2) = 5\sqrt{y}$.The total area $A$ is given by the integral with respect to $y$ from $0$ to $4$:$A = \int_{0}^{4} 5\sqrt{y} \, dy$$A = 5 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4}$$A = 5 \cdot \frac{2}{3} \cdot [4^{3/2} - 0]$$A = \frac{10}{3} \cdot 8 = \frac{80}{3}$.

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

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