The area of the figure bounded by the parabol | vedclass

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Question

(A) Given the parabolas are $x = -2y^{2}$ and $x = 1 - 3y^{2}$.To find the points of intersection,set the equations equal:$-2y^{2} = 1 - 3y^{2}$$y^{2}

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$\frac{4}{3}$ sq units

Vedclass · Answer

(A) Given the parabolas are $x = -2y^{2}$ and $x = 1 - 3y^{2}$.To find the points of intersection,set the equations equal:$-2y^{2} = 1 - 3y^{2}$$y^{2} = 1$$y = \pm 1$When $y = \pm 1$,$x = -2(1)^{2} = -2$.The points of intersection are $(-2, 1)$ and $(-2, -1)$.The area is bounded by $x = 1 - 3y^{2}$ (right curve) and $x = -2y^{2}$ (left curve) from $y = -1$ to $y = 1$.Area $= \int_{-1}^{1} [(1 - 3y^{2}) - (-2y^{2})] dy$$= \int_{-1}^{1} (1 - y^{2}) dy$Since the function is even,Area $= 2 \int_{0}^{1} (1 - y^{2}) dy$$= 2 [y - \frac{y^{3}}{3}]_{0}^{1}$$= 2 [1 - \frac{1}{3}]$$= 2 [\frac{2}{3}] = \frac{4}{3}$ sq units.

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

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