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(A) The area of the region bounded by the curve $x^{2}=4y$,the lines $y=2$ and $y=4$,and the $y$-axis in the first quadrant is given by the integral o

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$\left(\frac{32-8 \sqrt{2}}{3}\right)$

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(A) The area of the region bounded by the curve $x^{2}=4y$,the lines $y=2$ and $y=4$,and the $y$-axis in the first quadrant is given by the integral of $x$ with respect to $y$ from $y=2$ to $y=4$.Given $x^{2}=4y$,we have $x = \sqrt{4y} = 2\sqrt{y}$ (since it is in the first quadrant,$x > 0$).Area $= \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy$$= 2 \int_{2}^{4} y^{1/2} \, dy$$= 2 \left[ \frac{y^{3/2}}{3/2} ight]_{2}^{4} = 2 	imes \frac{2}{3} \left[ y^{3/2} ight]_{2}^{4}$$= \frac{4}{3} \left[ 4^{3/2} - 2^{3/2} ight]$$= \frac{4}{3} \left[ 8 - 2\sqrt{2} ight]$$= \left( \frac{32 - 8\sqrt{2}}{3} ight) 	ext{ square units}$

Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

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