The area (is sq. units) bounded by the curves $x=y^2$ and $x=3-2 y^2$ is

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Aicharya · Accepted Answer

$x=y^2 	ext { and } x=3-2 y^2$ Solving for $y, y^2=3-2 y^2 \Rightarrow y= \pm 1$ $\begin{aligned} & 	ext { Required Area } \ & =2\left(\int_0^1 \sqrt{x} d x+\int_1^3 \sqrt{\frac{3-x}{2}} d xight) \ & =\left[\frac{4}{3} x^{3 / 2}ight]_0^1-\frac{2 \sqrt{2}}{3}\left[(3-x)^{3 / 2}ight]_1^3 \ & = \ & \frac{4}{3}-\left(0-\frac{2 \sqrt{2}}{3} 2^{3 / 2}ight)=\left(\frac{4}{3}+\frac{8}{3}ight)=4\end{aligned}$

The area (is sq. units) bounded by the curves $x=y^2$ and $x=3-2 y^2$ is

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