Question

Find the area included between the parabolas $y^2=x$ and $x=3-2y^2$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$A=(1,1)$ , $B=(0,\sqrt{\frac{3}{2}})$ , $C=(0,-\sqrt{\frac{3}{2}})$ , $D=(1,-1)$

Due to symmetry we can just calculate the area in the first quadrant and multiply by 2

$={\int }_0^{1}3-2y^{2}dy-{\int }_0^1{y}^{2}dy$

$=[3y-\frac{2y^3}{3}]-\frac{y^3}{3}=3-\frac{2}{3}-\frac{1}{3}$

Area of first quadrant $=2$

$Total\text{area}=4$