Question

The area of the figure bounded by the parabola $x=-2y^2\text{and}x=1-3y^2$ is ?

• A. $1/6$
• B. $2/3$
• C. $3/2$
• D. $4/3$

Answer 1

Answer: (D)

$4/3$

$x=-2y^2\to \left(i\right)\&x=1-3y^2\to \left(ii\right)$

Equating equation (i) & (ii)

$\begin{array}{c}-2y^2=1-3y^2\\ -2y^2+3y^2=1\\ y^2=1\\ y\pm 1\end{array}$

So, area of region bounded by

$\begin{array}{c}A\Rightarrow 2{\int }_0^1\left[\left(1-3y^2\right)-\left(-2y^2\right)\right]dy\\ =2{\int }_0^1\left(1-y^2\right)dy\\ =2{\left[y-\frac{y^3}{3}\right]}_0^1\Rightarrow 2\left[\left(1-\frac{1}{3}\right)-\left(0-\frac{0}{3}\right)\right]\\ =2\left[\frac{1}{1}-\frac{1}{3}\right]\Rightarrow 2\left[\frac{2}{3}\right]=\frac{4}{3}sq.units\end{array}$