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Question
Using integration, find the area of the region `{(x, y): 0 ≤ y ≤ sqrt(3)x, x^2 + y^2 ≤ 4}`
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Solution
Solving `y = sqrt(3)x` and `x^2 + y^2` = 4, we get the points of intersection as `(1, sqrt(3))` and `(-1, - sqrt(3))`
The required area = the shaded area = `int_0^1 sqrt(3)x dx + int_1^2 sqrt(4 - x^2) dx`
= `sqrt(3)/2 [x^2]_0^1 + 1/2 [xsqrt(4 - x^2) + 4 sin^-1 x/2]_1^2`
= `sqrt(3)/2 + 1/2 [2pi - sqrt(3) - 2 pi/3]`
= `(2pi)/3` square units
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