CBSE Class 12 · 2013 · Area Under CurvesLong Answer

Using integration, find the area of the region enclosed between the two circles

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

and

(x - 2)^{2} + y^{2} = 4

Answer: $(\frac{8 π}{3} - 23)$ square units

Solution

Both circles have radius

2

, centres

O (0, 0)

and

C (2, 0)

; they intersect where

x^{2} = (x - 2)^{2} \Rightarrow x = 1

, i.e. at

(1, \pm 3)

.
By symmetry about the

x

-axis the lens area is

A = 2 [\int_{0}^{1} 4 - (x - 2)^{2} d x + \int_{1}^{2} 4 - x^{2} d x] .

Evaluating with

\int a^{2} - u^{2} d u = \frac{u}{2} a^{2} - u^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{u}{a}

(

a = 2

A = \frac{8 π}{3} - 23 square units.

CBSE Class 12 Board · 2013 · Subjective