Question

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

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

Answer 1

The correct option is A

$\frac{4}{3}\mathrm{sq}\mathrm{units}$

Given, $\mathrm{x}+2{\mathrm{y}}^2=0$ and $\mathrm{x}+3{\mathrm{y}}^2=1$

Equate $\mathrm{x}$ of both the equation,

$$\begin{gathered} 1-3{\mathrm{y}}^2=-2{\mathrm{y}}^2 \\ \Rightarrow {\mathrm{y}}^2=1 \\ \Rightarrow \mathrm{y}=\pm 1 \end{gathered}$$

Find the corresponding value of $\mathrm{x}$

$$\begin{gathered} \mathrm{y}=-1\Rightarrow \mathrm{x}=-2 \\ \mathrm{y}=1\Rightarrow \mathrm{x}=-2 \end{gathered}$$

We have two points of intersection,

$\left(-2,1\right)\mathrm{and}\left(-2,-1\right)$

The desired area,

$\begin{array}{rcl}A& =& 2{\int }_0^1\left[\left(1-3{\mathrm{y}}^2\right)-\left(-2{\mathrm{y}}^2\right)\right]d\mathrm{y}\\ & =& 2{\int }_0^1\left(1-{\mathrm{y}}^2\right)d\mathrm{y}\\ & =& 2{\left[\mathrm{y}-\frac{{\mathrm{y}}^3}{3}\right]}_0^1\\ & =& 2\times \frac{2}{3}\\ & =& \frac{4}{3}\mathrm{sq}\mathrm{units}\end{array}$

Hence option(A) is correct.