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

Let the area of the region $\{(x, y): 0 \leq x \leq 3, 0 \leq y \leq \min \{x^2+2, 2x+2\}\}$ be $A$. Then $12A$ is equal to

The area of the region (in sq. units) bounded by the curves $x^2+y^2=16$ and $y^2=6x$ is

Consider the functions $f, g: R \rightarrow R$ defined by
$f(x)=x^2+\frac{5}{12}$ and $g(x)=\begin{cases} 2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4} \end{cases}$
If $\alpha$ is the area of the region
$\{( x , y ) \in R \times R :| x | \leq \frac{3}{4}, 0 \leq y \leq \min \{f( x ), g( x )\}\}$,
then the value of $9 \alpha$ is.

The area of the region $\{(x, y): x^2 \leq y \leq 8-x^2, y \leq 7\}$ is

The area (in square units) of the region bounded by the curves $x=y^2$ and $x=3-2y^2$ is