The area of a circle can be proven to be πr² by integrating the function y = √(r² - x²) from x = 0 to x = r, then multiplying by 4 to account for all four quadrants; this is achieved through trigonometric substitution (x = r sin θ) and applying the half-angle identity for cosine squared, ultimately yielding the familiar formula A = πr².
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Proof of Area of Circle using Calculus | Visualization | Application of integrals #mathsAjouté :
Have you ever wondered why the area of a circle is pi times r squared?
Let's not just accept it. Let's prove it.
A circle of radius r sits on a coordinate grid. A line from the center to the edge traces out the radius r. The angle it makes with the horizontal is called theta. The tip of that line has coordinates x and y, and the circle itself is described by the equation x squared plus y squared equals r squared.
To find the area, we begin by stacking thin vertical rectangles under the curve in the first quadrant. Each rectangle has a width of dx and a height equal to the y value on the circle, which is the square root of r squared minus x squared.
Watch as we keep adding more and more rectangles, thinner, more precise. They fill the quarter circle perfectly. Since all four quadrants are identical, the full area is exactly four times this quarter circle region. That means A equals four times the integral from zero to R of the square root of r squared minus x squared dx.
The square root makes this tricky, so we'll use a clever trigonometric substitution. We let x equal r sine theta, then dx becomes r cosine theta d theta. When x equals zero, theta is zero, and when x equals r, theta is pi over two.
Substituting into the square root, we get the square root of r squared minus r squared sine squared theta, which factors out to the square root of r squared times the quantity one minus sine squared theta.
Now, recall the Pythagorean identity, one minus sine squared theta equals cosine squared theta. So, the square root collapses cleanly to r cosine theta.
The integral now becomes four times the integral from zero to pi over two of r cosine theta times r cosine theta d theta, which is four r squared times the integral of cosine squared theta d theta.
To handle cosine squared, we apply the half angle identity. Cosine squared theta equals one plus cosine of two theta all divided by two.
Substituting that in, the integral becomes four r squared times the integral from zero to pi over two of the quantity one plus cosine two theta over two d theta, which simplifies to two r squared times the integral of one plus cosine of two theta.
Integrating term by term gives two r squared times the bracket theta plus sine two theta over two evaluated from zero to pi over two.
Plug in the upper limit, sine of pi vanishes to zero. Plug in the lower limit, everything is zero. We are left with two r squared times pi over two, and that equals pi r squared.
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