Question

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

• A. $-\frac{4}{3}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is B $\frac{4}{3}$

The given curves are $x+2y^2=0$ and $x+3y^2=1$.

The first curve is $x+2y^2=0$ which implies $2y^2=-x$ and hence

$y^2=-x/2$, which is a parabola.

Second curve $x+3y^2=1$which is again a parabola as $y^2=-1/3(x-1)$.

Hence we are required to find the area between the two parabolas.

On solving the equations of the two parabolas, we get the points of intersection as $(-2,1)$and $(-2,-1)$.

Hence we set$x=-2$ and $y=1,-1$.

Hence the required area is$\int (-2y^2-1+3y^2)dy$ , where the integral runs from 0 to 1.

= $\int (y^2-1)dy$ , where the integral runs from 0 to 1.

= $(1/3-1)$

= $\frac{2}{3}$.

Hence the area of the region bounded by the curves is $\frac{4}{3}$.