Question

The area enclosed between the curves, $x^2=y$ and $y^2=x$ is equal to:

• A. $\frac{1}{3}$ sq. unit
• B. $2{\int }_0^1(x-x^2)dx$
• C. Area enclosed by the region $\left\{\right(x,y):x^2\le y\le \sqrt{x}\}$
• D. Area enclosed by the region $\left\{\right(x,y):x^2\le y\le x\}$

Answer 1

Answer: (A), (B), (C)

$\frac{1}{3}$ sq. unit
$2{\int }_0^1(x-x^2)dx$
Area enclosed by the region $\left\{\right(x,y):x^2\le y\le \sqrt{x}\}$

Bothe the curves intersect at

$(0,0)$ and $(1,1)$
Now,
Area $={\int }_{x=0}^{x=1}(y_1-y_2)dx={\int }_{x=0}^1(\sqrt{x}-x^2)dx$
$={[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^3}{3}]}_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

Area can also be written as

$2{\int }_0^1(x-x^2)dx$

Due to its symmetry along $y=x$

Area enclosed by the region

$\left\{\right(x,y\left)\right\}:x^2\le y\le \sqrt{x}\}$

Twice the area enclosed by the region

$\left\{\right(x,y\left)\right\}:x^2\le y\le x\}$

Hence, options A, B, and C.