Question

The area of the figure bounded by the parabolas $x=-2y^2$ and $x=1-3y^2$ is

• A. $\frac{4}{3}\text{square units}$
• B. $\frac{2}{3}\text{square units}$
• C. $\frac{3}{7}\text{square units}$
• D. $\frac{6}{7}\text{square units}$

Answer 1

The correct option is A $\frac{4}{3}\text{square units}$

The curves intersects at
$2y^2=3y^2-1\Rightarrow y=\pm 1\Rightarrow x=-2$

$I=2\underset{0}{\overset{1}{\int }}(1-3y^2-(-2y^2\left)\right)dy=2\underset{0}{\overset{1}{\int }}(1-y^2)dy=2(1-\frac{1}{3})=\frac{4}{3}$