The part of the circle $x^2 + y^2 = 9$ between $y = 0$ and $y = 2$ is revolved about the $y$-axis. The volume of the generating solid will be:

Let $y = g(x)$ be the inverse of a bijective mapping $f : R \rightarrow R$ defined by $f(x) = 3x^3 + 2x$. The area bounded by the graph of $g(x)$,the $x-$axis,and the ordinate at $x = 5$ is:

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is:

The area of the region bounded by the curve $y = x^2 - x - 6$ and the $X$-axis is . . . . . . sq. units.

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 area (in sq unit) of the region bounded by the curves $2x = y^2 - 1$ and $x = 0$ is