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Find the area of the smaller part of ...

Find the area of the smaller part of the circle `x^2+y^2=a^2` cut off by the line `x=a/(sqrt(2))`

A

`a^4[pi/5 - 1/3]`

B

`a^2[pi/4 - 1/2]`

C

`a^2[pi/3 - 1/4]`

D

`a^3[pi/3 - 1/2]`

Text Solution

Verified by Experts

The correct Answer is:
B
require Area= `int ydx`
`= int_(a/sqrt2)^a 2 sqrt(a^2 - x^2) dx`
`=> x = aSin theta`
`dx= a cos theta d theta`
when `x=a/sqrt2, sin theta= 1/sqrt2`
`theta= pi/4`
when `x=a, sin theta = 1`
`theta= pi/2`
...
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