Question

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

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{5}{3}$
• D. $\frac{1}{3}$

Answer 1

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

Solving the equations $x+2y^2=0$ and $x+3y^2=1$

we get points of intersection
$(-2,1)$ and $(-2,-1)$
The bounded region is show in figure.
The required area
$=2{\int }_0^1(1-{3y}^2)-(-{2y}^2)$
$=2{\int }_0^1(1-y^2)dy=2{[y-\frac{y^3}{3}]}_0^1=2\times \frac{2}{3}=\frac{4}{3}.$