Q. The area of the plane region bounded by the curves $x + 2\ y^{2} = 0\ and\ x + 3y^{2} = 1$ is equal to
A) $\frac{4}{3}\ sq\ uints$
B) $\frac{5}{3}\ sq\ units$
C) $\frac{1}{3}\ sq\ units\$
D) $\frac{2}{3}\ sq\ units$
Explanation
(a)
On solving the given curves, we get
$y = \pm 1\ and\ x = - 2$
$\therefore Required\ aerea = \left| \int_{- 1}^{1}{\left( x_{1} - x_{2} \right)dy} \right|$
$= \left| \int_{- 1}^{1}{\left( 1 - 3y^{2} + 2y^{2} \right)dy} \right|$
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$= \left| 2\int_{0}^{1}{\left( 1 - y^{2} \right)dy} \right|$
$= \left| 2\left\lbrack y - \frac{y^{3}}{3} \right\rbrack_{0}^{1} \right|$
$= \frac{4}{3}\ sq\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.