The area of the plane region bounded by the c | vedclass

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

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(D) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.To find the points of intersection,set $-2y^2 = 1 - 3y^2$,which gives $y^2 = 1$,so $y = \pm 1$.When $y = \pm 1$,$x = -2(1)^2 = -2$. Thus,the intersection points are $(-2, 1)$ and $(-2, -1)$.The area $A$ is given by the integral of the difference between the curves with respect to $y$ from $y = -1$ to $y = 1$:$A = \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) dy$$A = \int_{-1}^{1} (1 - y^2) dy$$A = [y - \frac{y^3}{3}]_{-1}^{1}$$A = (1 - \frac{1}{3}) - (-1 - \frac{-1}{3})$$A = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$ square units.

The area of the plane region bounded by the curves $x + 2y^2 = 0$ and $x + 3y^2 = 1$ is equal to

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