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Question
Find the area of the circle x2 + y2 = 16
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Solution
By the symmetry of the circle, required area of the circle is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.
Given equation of the circle is x2 + y2 = 16
∴ y2 = 16 – x2
∴ y = `+- sqrt(16 - x^2)`
∴ y = `sqrt(16 - x^2)` ......[∵ In first quadrant, y > 0]
∴ Required area = 4(area of the region OPQO)
= `4 xx int_0^4 y*"d"x`
= `4 xx int_0^4 sqrt(16 - x^2) "d"x`
= `4int_0^4 sqrt((4)^2 - x^2) "d"x`
= `4[x/2 sqrt((4)^2 - x^2) + (4)^2/2 sin^-1 (x/4)]_0^4`
= `4{[4/2 sqrt((4)^2 - (4)^2) + 16/2 sin^-1 (4/4)] - [0/2 sqrt((4)^2 - (0)^2) + 16/2 sin^-1 (0/4)]}`
= `4{[0 + 8 sin^-1 (1)] - [0 + 0]}`
= `4(8 xx pi/2)`
= 16π sq.units
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