Question

Find the area bounded by the curves $x=y^2$ and $x=3-2y^2$

• A. $2$ sq. units
• B. $4$ sq. units
• C. $6$ sq. units
• D. $8$ sq. units

Answer 1

The correct option is B $4$ sq. units

The two curves represent parabolas with vertices at $(0,0)$ and $(3,0)$
They intersect at $(1,1)$ and $(1,-1)$
So the required area is
area of OPMQO $=2$( area of OPMO)
$=2({\int }_0^1\sqrt{x}dx+{\int }_1^3\sqrt{\frac{3-x}{2}}dx)$
$=2{\left(\frac{2}{3}{x}^{\frac{3}{2}}\right)}_0^1-{[2\frac{1}{\sqrt{2}}.\frac{2}{3}{(3-x)}^{\frac{3}{2}}]}_1^3$
$=2[\frac{2}{3}-(0-\frac{1}{\sqrt{2}}\frac{2}{3}{2}^{\frac{3}{2}})]=2(\frac{2}{3}+\frac{4}{3})=4$