The equation of a circle centered at the origin is usually given like this x 2 + y 2 = r 2 . Recall that you

can rewrite this to get:

y=+/- sqrt (r^2-x^2); where the "+" is the top half of the circle, and the "-" is the bottom half of the circle.

a) Write the area of a full circle out as an integral of some function.
b) Derive the area of a circle, by evaluating the integral you wrote in part (a).

A = ∫ [−r to r] √(r² − x²) − (−√(r² − x²)) dx
A = ∫ [−r to r] 2√(r² − x²) dx
A = (x √(r² − x²) + r² sin⁻¹(x/r)) | [−r to r]
A = (r √(r² − r²) + r² sin⁻¹(r/r)) − (−r √(r² − r²) + r² sin⁻¹(−r/r))
A = (0 + r² sin⁻¹(1)) − (0 + r² sin⁻¹(−1))
A = r² (π/2) − r² (−π/2)
A = π r²

a.
A = 2∫(sqrt(r^2 - x^2))dx [-r,r]

b.
Let r be the hypotenuse of a right triangle, x be a leg, and θ be the angle opposite x.
Then r*sin(θ) = x, r*cos(θ)dθ = dx.
Additionally, sqrt(r^2 - x^2) = r*cos(θ).
A = (2r^2)∫(cos(θ)^2)dθ [-π/2,π/2]
= (2r^2)∫((1/2)cos(2θ) + 1/2)dθ [-π/2,π/2]
= (r^2)∫(cos(2θ) + 1)dθ [-π/2,π/2]
Let u = 2θ, du = 2dθ.
A = ((r^2)/2)∫(cos(u) + 1)du [-π,π]
= ((r^2)/2)(sin(u) + u) [-π,π]
= ((r^2)/2)(sin(π) + π - sin(-π) + π)
= ((r^2)/2)(2π)
= πr^2