1.

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))`

Answer» 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`
`= 2 int _(pi/4)^ (pi/2) (a cos theta)(a cos theta) d theta`
`= 2a^2 int_ (pi/4)^(pi/2) (cos 2 theta +1)d theta`
`= a^2[ int_(pi/2)^(pi/4)Cos 2 theta d theta + int 1 d theta]`
`= a^2[[sin 2 theta/2] + [ theta]]`
`= a^2[[-1/2] +[pi/4]]`
`= a^2[pi/4 - 1/2]`
answer


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