The area bounded by the parabola x^{2}=4y and | vedclass

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(A) The parabola is $x^2 = 4y$,which implies $x = \pm 2\sqrt{y}$.Since the area is bounded by the $Y$-axis and the parabola in the first quadrant,we c

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$\frac{4}{3}(8-2 \sqrt{2})$ sq. units

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(A) The parabola is $x^2 = 4y$,which implies $x = \pm 2\sqrt{y}$.Since the area is bounded by the $Y$-axis and the parabola in the first quadrant,we consider $x = 2\sqrt{y}$.The area $A$ bounded by the curve $x = f(y)$,the $Y$-axis,and the lines $y = 2$ and $y = 4$ is given by:$A = \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy$$A = 2 \int_{2}^{4} y^{1/2} \, dy$$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = 2 \cdot \frac{2}{3} \left[ y^{3/2} \right]_{2}^{4}$$A = \frac{4}{3} [4^{3/2} - 2^{3/2}]$$A = \frac{4}{3} [8 - 2\sqrt{2}]$Thus,the area is $\frac{4}{3}(8 - 2\sqrt{2})$ sq. units.

The area bounded by the parabola $x^{2}=4y$ and the lines $y=2$,$y=4$ and the $Y$-axis is

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