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The area of the smaller segment cut off ...

The area of the smaller segment cut off from the circle `x^(2)+y^(2)=9` by x = 1 is

A

`(1)/(2)(9 sec^(-1)3-sqrt(8))` sq unit

B

`(1)/(2)(9 sec^(-1)3-sqrt(8))` sq unit

C

`(sqrt(8)-9sec^(-1)3)` sq unit

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
B
Required area `=2int_(1)^(3)ydx=2int_(1)^(3)sqrt(9-x^(2))dx`
`=2.(1)/(2)[xsqrt(9-x^(2))+9sin^(-1).(x)/(3)]_(1)^(3)`

`=[9sin^(-1)(1)-sqrt(8)-9sin^(-1)((1)/(3))]=[9.(pi)/(2)-9sin^(-1)((1)/(3))-sqrt(8)]`
`=[9{cos^(-1)((1)/(3))}-sqrt(8)]" "[becausecos^(-1)theta=(pi)/(2)-sin^(-1)theta]`
`=[9sec^(-1)3-sqrt(8)]` sq unit
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