The area of the region that is common to the circle $x^2+y^2=16a^2$ and the parabola $y^2=6ax$ is

The area of the region bounded by the parabola $y=x^2$ and the curve $y=|x|$ is

Area bounded by the curve $y = \min \{\sin^2x, \cos^2x \}$ and $x-$ axis between the ordinates $x = 0$ and $x = \frac{5\pi}{4}$ is

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

Given: $f(x) = \begin{cases} x, & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1-x, & \frac{1}{2} < x \leq 1 \end{cases}$ and $g(x) = (x-\frac{1}{2})^2, x \in R$. Then the area (in sq. units) of the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt{3}$ is:

The area enclosed by $y^{2}=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$ is equal to