Class 12 Maths Application of Integrals The area of the region bounded by a curve and a line

The area of the region bounded by a curve and a line

In this section, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse when their standard forms are given.

Example: Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle  x2 + y2 = 4

Solution:

The area of the region bounded by the circle x2 + y2 = 4, x = √3y, and the x-axis is the  area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1).

Area OAB = Area ∆OCA + Area ACB

Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2 Area ACB = ʃ32 y dx

= ʃ32 √(4 – x2) dx

= [x√(4 – x2)/2 + (4/2) * sin-1 x/2]√32

= [2 * π/2 - (√3/2) * √(4 - 3) - 2sin-1 √3/2]

= [π - √3/2 – 2(π/3)]

= [π/3 - √3/2]

Therefore, area enclosed by x-axis, x = √3y and the circle x2 + y2 = 4 in the first quadrant

= √3/2 + [π/3 - √3/2] = π/3 units

Example: Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2

Solution:

The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is the area ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis. So, Area ABCD = 2 * Area ABC

Now, Area of ABC = ʃa/√2a y dx

= ʃa/√2a √(a2 – x2) dx

= [x√( a2 – x2)/2 + (a2/2) * sin-1 x/a]a/√2a

= [a2/2 * π/2 - (a/2√2) * √( a2 - a2/2) - a2/2 * sin-1 1/√2]

= a2/2 * π/2 - (a/2√2) * (a/√2) - a2/2 * (π/4)

= a2π/4 - a2/4 - a2π/8

= (a2/4)(π - 1 - π/2)

= (a2/4)(π/2 - 1)

So, area of ABCD = 2[(a2/4)(π/2 - 1)] = (a2/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x2 + y2 = a2 cut off by the line x = a/√2 is

(a2/2)(π/2 - 1) units.

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