The area bounded by the curves $y=x^2$ and $y-6=-|x|$ is

Calculate the area enclosed by the curves $x^2 = 2 - y$ and $x^2 = y$.

The area of the smaller portion between the curves $x^2 + y^2 = 8$ and $y^2 = 2x$ is

$OABC$ is a unit square where $O$ is the origin and $B=(1,1)$. The curves $y^2=x$ and $x^2=y$ divide the area of the square into three parts $a_1, a_2, a_3$. If $a_1, a_2, a_3$ are the areas (in sq units) of these parts respectively,then $a_1+2a_2+3a_3=$

Let $C_{1}$ be the curve obtained by the solution of the differential equation $2xy \frac{dy}{dx} = y^{2} - x^{2}, x > 0$. Let the curve $C_{2}$ be the solution of $\frac{2xy}{x^{2} - y^{2}} = \frac{dy}{dx}$. If both the curves pass through $(1, 1)$,then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to:

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is