Home
Class 12
MATHS
Prove that the area in the first quadran...

Prove that the area in the first quadrant enclosed by the axis, the line `x=sqrt(3)y` and the circle `x^2+y^2=4\ i s\ pi//3.`

Text Solution

Verified by Experts

`x^(2)+y^(2)=4` is the equation of a circle whose centre is (0, 0) and radius is 2. On solving the equation of circle to the straight line `x=sqrt(3)y` we obtained the points of intersection `A(sqrt(3), 1)` and `D(-sqrt(3), -1)`.

Now required area
`=` area of OACO + area of CABC
`=int_(0)^(sqrt(3))(x)/sqrt(3)dx+int_(sqrt(3))^(2)sqrt(4-x^(2))dx`
`=[(x^(2))/(2sqrt(3))]_(0)^(sqrt(3))+[(x)/(2)sqrt(4-x^(2))+(4)/(2)"sin"^(-1)(x)/(2)]_(sqrt(3))^(2)`
`=(sqrt(3))/(2) +[(0+2"sin"^(-1)1)-((sqrt(3))/(2)+2"sin"^(-1)(sqrt(3))/(2))]`
`=(sqrt(3))/(2)+2*(pi)/(2)-(sqrt(3))/(2)-2*(pi)/(3)`
`=(pi)/(3)` sq. units.
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF INTEGRALS

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 8.2|7 Videos

  • APPLICATIONS OF INTEGRALS

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|19 Videos

  • APPLICATIONS OF INTEGRALS

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 8 C Questions For Competitive Examinations|10 Videos

  • APPLICATIONS OF DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|24 Videos

  • Continuity and Differentiability

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|23 Videos

Similar Questions

Explore conceptually related problems

Find the area of the region in the first quadrant enclosed by x-axis, line x=sqrt(3)y and the circle x^2+y^2=4 .

Find the area of the region in the first quadrant enclosed by x-axis, the line y=sqrt(3\ )x and the circle x^2+y^2=16

Area of the region in the first quadrant exclosed by the X-axis, the line y=x and the circle x^(2)+y^(2)=32 is

Find the area of the region in the first quadrant enclosed by the y-axis, the line y=x and the circle x^2+y^2=32 , using integration.

Find the area of the region in the first quadrant enclosed by the x-axis, the line y" "=" x" , and the circle x^2+y^2=32 .

Area lying in the first quadrant and bounded by the circle x^(2)+y^(2)=4 the line x=sqrt(3)y and x-axis , is

Find the area enclosed by the curves y= sqrtx and x= -sqrty and the circle x^(2) + y^(2) =2 above the x-axis.

Find the area lying in the first quadrant and bounded by the curve y=x^3 and the line y=4xdot

Area lying in the first quadrant and bounded by the circle x^2+y^2=4 and the lines x= 0 and x= 2 is:

The radius of the larger circle lying in the first quadrant and touching the line 4x+3y-12=0 and the coordinate axes, is