The area bounded by the curve $y^2 = x$,the line $y = 4$,and the $y$-axis is

Area of the region bounded by the ellipse $9x^2 + 16y^2 = 1$ in the first quadrant is . . . . . . sq. units.

The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$. Find the value of $a$.

Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

If the line $x=\alpha$ divides the area of region $R=\{(x, y) \in \mathbb{R}^2: x^3 \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts,then which of the following is true?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$

The area of the region $\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}$ is