The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

The smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is

The volume of the solid formed by rotating the area enclosed between the curve $y^{2}=4x$, $x=4$, and $x=5$ about the $x$-axis is (in cubic units): (in $\pi$)

The area bounded by the curves $x = \sqrt{2 - y^2}$ and $|x| = |y|$ is -

The area of the region (in sq. units) enclosed by the curve $y=x^3-19x+30$ and the $x$-axis is

If $(a, 0); a > 0$ is the point where the curve $y = \sin 2x - \sqrt{3} \sin x$ cuts the $x$-axis first,and $A$ is the area bounded by this part of the curve,the origin,and the positive $x$-axis,then: