Area Under Curves — CBSE Class 12 · 2007 Board Question with Solution

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

x^{2} + y^{2} = 1

and

(x - 1)^{2} + y^{2} = 1

Answer: $(\frac{2 π}{3} - \frac{3}{2})$ square units

Solution

Both circles have radius

1

, with centres

O (0, 0)

and

C (1, 0)

.
Eliminating

y

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

, and then

y = \pm \frac{3}{2}

, so they meet at

(\frac{1}{2}, \pm \frac{3}{2})

.
The lens is symmetric about the

x

-axis. Its upper half is bounded on the left (

0 \leq x \leq \frac{1}{2}

) by the circle

(x - 1)^{2} + y^{2} = 1

and on the right (

\frac{1}{2} \leq x \leq 1

) by

x^{2} + y^{2} = 1

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

Each integral evaluates with

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

, giving

A = \frac{2 π}{3} - \frac{3}{2} square units.

CBSE Class 12 Board · 2007 · Subjective