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Question
Find the area of the region bounded by the curve y = `sqrt(36 - x^2)`, the X-axis lying in the first quadrant and the lines x = 0 and x = 6
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Solution
Let A be the required area.
Given equation of the curve is y = `sqrt(36 - x^2)`
∴ A = `int_0^6 y "d"x`
= `int_0^6 sqrt(36 - x^2) "d"x`
= `int_0^6 sqrt((6)^2 + x^2) "d"x`
= `[x/2 sqrt((6)^2 - x^2) + (6)^2/2 sin^-1 (x/6)]_0^6`
= `[6/2 sqrt((6)^2 - (6)^2) + (6)^2/2 sin^-1 (6/6)] - [0/2 sqrt((6)^2 - 0) + (6)^2/2 sin^-1 (0/6)]`
= `0 + 36/2 sin^-1 (1) - 0`
= `36/2 (pi/2)`
= `(36pi)/4` sq.units
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