Area lying in the first quadrant between the curves $x^2 + y^2 = \pi^2$ and $y = \sin x$ is equal to :-

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

Let $f:[0,1] \rightarrow[0,1]$ be the function defined by $f(x)=\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36}$. Consider the square region $S=[0,1] \times [0,1]$. Let $G=\{(x, y) \in S: y>f(x)\}$ be called the green region and $R=\{(x, y) \in S: y(A)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.
$(B)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(C)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(D)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.

The area of the region given by $A = \{(x, y) : x^{2} \leq y \leq \min \{x+2, 4-3x\}\}$ is:

Let $A_{1}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and the $y$-axis in the first quadrant. Also,let $A_{2}$ be the area of the region bounded by the curves $y = \sin x$,$y = \cos x$,the $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then ..... .

The area of the region bounded by $x = 0, y = 0, x = 2, y = 2, y \leq e^x$ and $y \geq \log x$ is