Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$.

The area bounded by the curve $x^2 = 8y$ and the straight line $x - 8y + 2 = 0$ is

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:

The area of the region bounded by the curve $y=|x-3|$,the $X$-axis,and the lines $x=0$ and $x=2$ is . . . . . . sq. units.

Let $f: R \rightarrow R$ be a twice differentiable function such that $f(x + y) = f(x) f(y)$ for all $x, y \in R$. If $f^{\prime}(0) = 4a$ and $f$ satisfies $f^{\prime \prime}(x) - 3a f^{\prime}(x) - f(x) = 0$,$a > 0$,then the area of the region $R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\}$ is:

The area of the region bounded by the curve $y^2 = 4x$,the $Y$-axis,and the line $y = 3$ is . . . . . . sq. units.