If the area of the region $\{(x, y): 0 \leq y \leq \min \{2x, 6x-x^2\}\}$ is $A$,then $12A$ is equal to

Let $T$ and $C$ respectively be the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$. Then the area of the region above the parabola $x^2 = y + 4$,below the transverse axis $T$ and on the right of the conjugate axis $C$ is:

The area of the smaller region enclosed by the curves $y^{2}=8x+4$ and $x^{2}+y^{2}+4\sqrt{3}x-4=0$ is equal to.

The area bounded by the parabolas $y=x^2$ and $y=1-x^2$ is equal to

The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4$,$y = 4$ and the coordinate axes. If $S_1, S_2, S_3$ are respectively the areas of these parts numbered from top to bottom,then $S_1:S_2:S_3$ is

The area included between the two curves $y^2 = 4ax$ and $x^2 = 4ay$ is