JEE MAIN - Mathematics (2014 (Offline) - No. 5)
The area of the region described by
$$A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right.$$ and $$\left. {{y^2} \le 1 - x} \right\}$$ is :
$$A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right.$$ and $$\left. {{y^2} \le 1 - x} \right\}$$ is :
$${\pi \over 2} - {2 \over 3}$$
$${\pi \over 2} + {2 \over 3}$$
$${\pi \over 2} + {4 \over 3}$$
$${\pi \over 2} - {4 \over 3}$$
Explanation
Given curves are $${x^2} + {y^2} = 1$$ and $${y^2} = 1 - x.$$
Intersection points are $$x = 0,1$$
Area of shaded portion is the required area.
So, Required Area $$=$$ Area of semi-circle $$+$$ Area bounded by parabola
$$ = {{\pi {r^2}} \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} dx} $$
$$ = {\pi \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} $$
( As radius of circle $$=1$$ )
$$ = {\pi \over 2} + 2\left[ {{{{{\left( {1 - x} \right)}^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over { - {\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \right]_0^1$$
$$ = {\pi \over 2} - {4 \over 3}\left( { - 1} \right) = {\pi \over 2} + {4 \over 3}$$ Sq. unit
Intersection points are $$x = 0,1$$
Area of shaded portion is the required area.
So, Required Area $$=$$ Area of semi-circle $$+$$ Area bounded by parabola
$$ = {{\pi {r^2}} \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} dx} $$
$$ = {\pi \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} $$
( As radius of circle $$=1$$ )
$$ = {\pi \over 2} + 2\left[ {{{{{\left( {1 - x} \right)}^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over { - {\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \right]_0^1$$
$$ = {\pi \over 2} - {4 \over 3}\left( { - 1} \right) = {\pi \over 2} + {4 \over 3}$$ Sq. unit
Comments (0)
