The area of the figure bounded by the parabol | vedclass

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(B) To find the area bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$, we first find their points of intersection by setting the two equati

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$\frac{4}{3}$ square unit

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(B) To find the area bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$, we first find their points of intersection by setting the two equations equal to each other:$-2y^{2} = 1 - 3y^{2}$$y^{2} = 1$$y = \pm 1$Thus, the curves intersect at $y = -1$ and $y = 1$.The area $A$ is given by the integral of the right curve minus the left curve with respect to $y$:$A = \int_{-1}^{1} [(1 - 3y^{2}) - (-2y^{2})] dy$$A = \int_{-1}^{1} (1 - y^{2}) dy$Since the integrand $(1 - y^{2})$ is an even function, we can simplify the integral:$A = 2 \int_{0}^{1} (1 - y^{2}) dy$$A = 2 [y - \frac{y^{3}}{3}]_{0}^{1}$$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

The area of the figure bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$ is

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