Given X^2 + Y^2 = R^2, namely, a circle with center (0,0) and radius R, use integration to prove that the circle’s area equals πR2
To prove that the area of a circle with radius R is equal to πR^2 using integration, we can divide the circle into infinitesimally small strips and then integrate the area of each strip.
Consider an infinitesimally thin strip on the circle, located at a distance y from the x-axis. The width of this strip is dy. We can consider this strip as a portion of a rectangle, where the width is dy and the length is the circumference of the circle at that y-coordinate.
At any y-coordinate, the circumference of the circle is given by 2πr, where r is the distance from the y-axis to the circle at height y. Using the equation of the circle, we know that r = √(R^2 - y^2).
The length of the infinitesimally thin strip is 2π√(R^2 - y^2), and its width is dy. Therefore, the area of each strip is given by dA = 2π√(R^2 - y^2) dy.
To find the total area of the circle, we need to integrate the areas of all the infinitesimally thin strips from y = -R to y = R.
The integral for the area is given by:
A = ∫[from -R to R] 2π√(R^2 - y^2) dy
To solve this integral, we can make a substitution by letting y = Rsin(θ) and dy = Rcos(θ) dθ. The limits of integration also change accordingly: when y = -R, θ = -π/2, and when y = R, θ = π/2.
The integral transforms to:
A = ∫[from -π/2 to π/2] 2π√(R^2 - R^2sin^2θ) Rcos(θ) dθ
= 2πR^2 ∫[from -π/2 to π/2] √(1 - sin^2θ) cos(θ) dθ
Using the identity sin^2θ + cos^2θ = 1, we have:
A = 2πR^2 ∫[from -π/2 to π/2] cos^2(θ) dθ
Now, using the double-angle identity cos^2(θ) = (1 + cos(2θ))/2, the integral becomes:
A = 2πR^2 ∫[from -π/2 to π/2] (1 + cos(2θ))/2 dθ
Integrating term by term, we have:
A = πR^2 [θ/2 + (sin(2θ))/4] [from -π/2 to π/2]
= πR^2 [π/4 - (-π/4)]
= πR^2
Therefore, we have proved that the area of a circle with radius R is indeed equal to πR^2 using integration.