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 ****x ^{2} + y^{2} = 4**

**Solution:**

The area of the region bounded by the circle x^{2} + y^{2} = 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 = ʃ_{√}_{3}^{2 }y dx

= ʃ_{√}_{3}^{2 }√(4 – x^{2}) dx

= [x√(4 – x^{2})/2 + (4/2) * sin^{-1} x/2]_{√3}^{2}

= [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 x^{2} + y^{2} = 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 x ^{2} + y^{2} = a^{2} cut off by the line x = a/**

**Solution:**

The area of the smaller part of the circle, x^{2} + y^{2} = a^{2} 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 = ʃ_{a/√}_{2}^{a }y dx

= ʃ_{a/√}_{2}^{a }√(a^{2} – x^{2}) dx

= [x√( a^{2} – x^{2})/2 + (a^{2}/2) * sin^{-1} x/a]_{a/√2}^{a}

= [a^{2}/2 * π/2 - (a/2√2) * √( a^{2} - a^{2}/2) - a^{2}/2 * sin^{-1} 1/√2]

= a^{2}/2 * π/2 - (a/2√2) * (a/√2) - a^{2}/2 * (π/4)

= a^{2}π/4 - a^{2}/4 - a^{2}π/8

= (a^{2}/4)(π - 1 - π/2)

= (a^{2}/4)(π/2 - 1)

So, area of ABCD = 2[(a^{2}/4)(π/2 - 1)] = (a^{2}/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x^{2} + y^{2} = a^{2} cut off by the line x = a/√2 is

(a^{2}/2)(π/2 - 1) units.

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