The area of the region,enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x$,is

  • A
    $\frac{1}{3}(12 \pi-1)$
  • B
    $\frac{1}{6}(12 \pi-1)$
  • C
    $\frac{1}{6}(24 \pi-1)$
  • D
    $\frac{1}{3}(6 \pi-1)$

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Similar Questions

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 (in sq. units) enclosed between the curves $y=|x|$,$y=[x]$ and the ordinates $x=-1$,$x=0$,$x=1$ is

The area (in sq. units) bounded by the curves $x=y^2$ and $x=3-2y^2$ is

Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f(x)=2 x^{3}-3 x^{2}-12 x$. If $A$ is the total area of the region bounded by $y=f(x)$,the $x$-axis and the lines $x=a$ and $x=b$,then $4 A$ is equal to ..... .

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

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