Question

Find the area of the smaller part of the circle x²+y²=a² cut off by the line x=a/√2

Answer 1

The area of the smaller part of the circle,x2+y2=a2,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.

∴Area ABCD=2×Area ABC

Area of ABC=∫aa/√2ydx

=∫aa/√2√a2-x2dx

=[x/2√a2-x2+a2/2sin-1x/a]aa/√2

=[a2/2(π/2)-2a/2√2√a2-a2/2-a2/2sin(1/√2)]

=a2π/4-a/2√2.a/√2-a2/2(π/4)

=a2π/4-a2/4-a2π/8

=a2/4[π-1-π/2]

=a2/4[π/2-1]

⇒Area ABCD=2[a2/4(π/2-1)]=a2/2(π/2-1)

Therefore,the area of smaller part of the circle,x2+y2=a2,cut off by the line,x=a/√2,

is a2/2(π/2-1)units.