Question

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

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

Answer 1

The correct option is D $\frac{4}{3}sq.unit$

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

$1$st curve

$\Rightarrow x^2+2y^2=0$

$\Rightarrow x^2=-2y^2$

$\therefore y^2=-x/2$ (parabola)

$2nd$ curve

$\Rightarrow x+3y^2=1$

$\Rightarrow 3y^2=1-x$

$y^2=(1-x)/3$ (parabola)

So, area between parabolas is required

Solving

$-\frac{x}{2}=\frac{(1-x)}{3}$

$\Rightarrow -3x=2-2x$

$\Rightarrow -3x+2x=2$

$\Rightarrow -x=2$

$=x=-2$

$\therefore y=1,-1$

$(-2,-1)$ & $(-2,1)$ are points of intersection.

Required area

$=2|{\int }_0^1(-2y^2-1+3y^2)dy|$

$=2|{\int }_0^1(y^2-1)dy|=2{[y-\frac{y^3}{3}]}_0^1$

$=2|(\frac{1}{3}-1)|=2\times 2/3=4/3$ square units.