CBSE Class 12 · 2009 · Area Under CurvesLong Answer

Find the area of the region enclosed between the two circles

x^{2} + y^{2} = 9

and

(x - 3)^{2} + y^{2} = 9

Answer: $(6 π - \frac{9 3}{2})$ square units

Solution

Both circles have radius

3

, centres

O (0, 0)

and

C (3, 0)

.
Eliminating

y

9 - x^{2} = 9 - (x - 3)^{2} \Rightarrow x = \frac{3}{2}

, so the circles meet at

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

.
By symmetry about the

x

-axis,

A = 2 [\int_{0}^{3/2} 9 - (x - 3)^{2} d x + \int_{3/2}^{3} 9 - x^{2} d x] .

Using the standard antiderivative with

a = 3

and simplifying,

A = 2 [- \frac{9 3}{8} + 3 π - \frac{9 3}{8}] = 2 [3 π - \frac{9 3}{4}] = 6 π - \frac{9 3}{2} .

Hence the enclosed area is

(6 π - \frac{9 3}{2})

square units.

CBSE Class 12 Board · 2009 · Subjective