Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be inverse of it, then area bounded by the curve $y = g(x)$ with $x$ axis between $x = 1, x = 2$ is (in square units)
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{3}{4}$
$1$
Let $A B$ be the la tusrectum of the parabola $y^2=4a x$ in the $X Y$-plane. Let $T$ be the region bounded by the finite arc $A B$ of the parabola and the line segment $A B$. A rectangle $P Q R S$ of maximum possible area is inscribed in $T$ with $P, Q$ on line $A B$, and $R, S$ on arc $A B$. Then, area $(P Q R S)$ area $(T)$ equals
If $A$ is the area in the first quadrant enclosed by the curve $C: 2 x^2-y+1=0$, the tangent to $C$ at the point $(1,3)$ and the line $x+y=1$, then the value of $60 A$ is
The area (in sq. units) of the region described by $\{(x,y):$${y^2} \le 2x \,and\,y \ge 4x - 1$$\}$ is
The volume of spherical cap of height $h$ cut off from a sphere of radius $a$ is equal to
Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt {1\, - \,{y^2}} $ is