Area Under Curves — CBSE Class 12 · 2008 Board Question with Solution

Using integration, find the area of the region enclosed between the 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)

.
Eliminating

y

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

, giving intersection points

(1, \pm 3)

.
By symmetry about the

x

-axis the required (lens) area is

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

Using

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

with

a = 2

A = 2 [(- \frac{3}{2} - 2 \cdot \frac{π}{6} + 2 \cdot \frac{π}{2}) + (2 \cdot \frac{π}{2} - \frac{3}{2} - 2 \cdot \frac{π}{6})] = \frac{8 π}{3} - 23 .

Hence the enclosed area is

(\frac{8 π}{3} - 23)

square units.

CBSE Class 12 Board · 2008 · Subjective