Question

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

• A. $\frac{4}{3}$ square units
• B. $\frac{2}{3}$ square units
• C. $\frac{3}{7}$ square units
• D. $\frac{6}{7}$ square units

Answer 1

Answer: (A)

We have,
$x=-2y^2$
$\Rightarrow y^2=-\frac{1}{2}x...(i)$
and $x=1-3y^2$
$\Rightarrow y^2=-\frac{1}{3}(x-1)...(ii)$
From Eqs. (i) and (ii), we get
$-\frac{1}{2}x=-\frac{1}{3}(x-1)$
$\Rightarrow 3x=2(x-1)$
$\Rightarrow 3x=2x-2$
$\Rightarrow x=-2$
$\therefore y^2=-\frac{1}{2}(-1)=1$
$\Rightarrow y=\pm 1$
$\therefore$ Point of intersection of two curves is $(-2,\pm 1)$

$\therefore$ Required Area $=2\underset{0}{\overset{1}{\int }}\sin xdx$
$=2{\left[y-\frac{y^3}{3}\right]}_0^1$
$=2\left[1-\frac{1}{3}\right]$
$=\frac{4}{3}$ sq units