Question

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

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

Answer 1

Answer: (A)

$\frac{4}{3}$

$\begin{array}{c}1-3y^2=-2y^2\Rightarrow y^2=1\therefore y=\pm 1\\ y=-1\Rightarrow x=-2\\ y=1\Rightarrow x=-2\\ Theboundedregionisasunder\\ Thedesiredarea=2{\int }_0^1\left[\left(1-3y^2\right)-\left(-2y^2\right)\right]dy\\ =22{\int }_0^1\left[\left(1-y^2\right)\right]dy=2{\left[y-\frac{y^3}{3}\right]}_0^1\\ =2\times \frac{2}{3}=\frac{4}{3}sq.units\end{array}$

Hence, this is the answer.