Question

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

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

Answer 1

Answer: (C)

$\frac{4}{3}$

Solving the equation $x+2y^2=0$ & $3y^2+x=1$
we get $A(-2,1)$ & $B(-2,-1)$
Required Area $=2[{\int }_0^{1}xdy-{\int }_0^{1}xdy]$
$=2{\int }_0^1(1-3y^2+2y^2)dy$
$=2{\int }_0^1(1-y^2)dy$
$=2{[y-\frac{y^3}{3}]}_0^1=2(1-\frac{1}{3})=\frac{4}{3}$ sq.units