Question

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

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

Answer 1

Answer: (D)

$x+2y^2=0\Rightarrow y^2=-\frac{x}{2}$
[Left handed parabola with vertex at (0, 0)]
$x+3y^2=1\Rightarrow y^2=-\frac{1}{3}\left(x-1\right)$
[Left handed parabola with vertex at (1, 0)]
Solving the two equations we get the points of intersection as (-2, 1), (-2, -1)

The required area is ACBDA, given by
$=\left|\underset{-1}{\overset{1}{\int }}\left(1-3y^2-2y^2\right)dy\right|={\left|y-\frac{5y^3}{3}\right|}_{-1}^1$
$=\left|\left(1-\frac{5}{3}\right)-\left(-1+\frac{5}{3}\right)\right|=2\times \frac{2}{3}=\frac{4}{3}$ sq. units.