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)

$\frac{4}{3}$ square units

Given curves are $x=-2y^2$ and $x=1-3y^2$

$-2y^2=1-3y^2$

$⟹y^2=1⟹y=\pm 1$

$\therefore$ Curves intersect at $(-2,\pm 1)$
Area $={\int }_{-1}^1(1-3y^2-(-2y^2\left)\right)dy$

$={\int }_{-1}^1(1-y^2)dy$

$={\int }_{-1}^0(1-y^2)dy+{\int }_0^1(1-y^2)dy$

Substituting $y=-y$ in first integral we get,

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

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

$=2{|y-\frac{y^3}{3}|}_0^1$

$=2(1-\frac{1}{3})=\frac{4}{3}$

Hence, area bounded by the given curves is $\frac{4}{3}$ square units.