Advertisement Remove all ads

Find the area of the sector bounded by the circle x2+ y2 = 16, and the line y = x in the first quadrant - Mathematics and Statistics

Sum

Find the area of the sector bounded by the circle x2+ y2 = 16, and the line y = x in the first quadrant

Advertisement Remove all ads

Solution

Given equation of the circle is x2+ y2 = 16   ......(i)

and equation of the line is y = x   ......(ii)

From (i), w get

y2 = 16 − x2 

∴ y = `sqrt(16 - x^2)`    ......(iii) .....[∵ In first quadrant, y > 0]

Substituting (ii) in (i), we get

x2 + y2 = 16

∴ 2x2 = 16

∴ x2 = 16

∴ x2 = 8

∴ x = `2sqrt(2)`    .....[∵ In first quadrant, x > 0]

When x = `2sqrt(2)`, y= `2sqrt(2)`

∴ The point of intersection is `"B"(2sqrt(2), 2sqrt(2))`.

Required area = area of the region OCABO

= area of the region OCBO + area of the region ABCA

= `int_0^(2sqrt(2)) x  "d"x + int_(2sqrt(2))^4 sqrt(16 - x^2)  "d"x`    .....[From (iii) and (ii)]

= `1/2[(2sqrt(2))^2 - 0] + [4/2 sqrt(16 - 16) + 16/2 sin^-1 (1) - {(2sqrt(2))/2 sqrt(16 - 8) + 16/2 sin^-1 (1/sqrt(2))}]`

= `8/2 + 16/2(pi/2) - (2sqrt(2))/2 (2sqrt(2)) - 16/2(pi/4)`

= 4 + 4π − 4 − 2π

= 2π sq.units

Concept: Area Bounded by the Curve, Axis and Line
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×