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Question
Find the area of the circle x2 + y2 = 62
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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 = 6.
Given equation of the circle is x2 + y2 = 62
∴ y2 = 62 – x2
∴ y = `+- sqrt(6^2 - x^2)`
∴ y = `sqrt(6^2 - x^2)` ......[∵ In first quadrant, y > 0]
∴ Required area = 4 (area of the region OPQO)
= `4 xx int_0^6 y*"d"x`
= `4 xx int_0^6 sqrt(6^2 - x^2) "d"x`
= `4[x/2 sqrt((6)^2 - x^2) + (6)^2/2 sin^-1 (x/6)]_0^6`
= `4{[6/2 sqrt((6)^2 - (6)^2) + (6)^2/2 sin^-1 (6/6)] - [0/2 sqrt((6)^2 - (0)^2) + (6)^2/2 sin^-1 (0/6)]}`
= `4{[0 + 36/2 sin^-1 (1)] - [0 + 0]}`
= `4(36/2 xx pi/2)`
= 36π sq.units
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