The area (in square units) of the region boun | vedclass

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(D) Given curves are $x=y^2$ and $x=3-2y^2$.To find the points of intersection,set $y^2 = 3-2y^2$.$3y^2 = 3 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$

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$4$

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(D) Given curves are $x=y^2$ and $x=3-2y^2$.To find the points of intersection,set $y^2 = 3-2y^2$.$3y^2 = 3 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$.Thus,the points of intersection are $(1, 1)$ and $(1, -1)$.The region is symmetric about the $x$-axis.Required area $= 2 \int_0^1 (x_{right} - x_{left}) dy = 2 \int_0^1 ((3-2y^2) - y^2) dy$.$= 2 \int_0^1 (3-3y^2) dy = 6 \int_0^1 (1-y^2) dy$.$= 6 [y - \frac{y^3}{3}]_0^1 = 6 (1 - \frac{1}{3}) = 6 (\frac{2}{3}) = 4$ square units.

The area (in square units) of the region bounded by the curves $x=y^2$ and $x=3-2y^2$ is

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