how to get the formula of area of a circle using integration
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How can we derive the familiar area formula A equals pi r squared using integration? Integration is a powerful mathematical tool for finding areas of curved regions. Today we'll explore how to use the disk method by integrating circular cross-sections to prove this fundamental formula.
Let's set up the integration. We start with the circle equation x squared plus y squared equals r squared. Solving for y gives us y equals plus or minus square root of r squared minus x squared. We can visualize vertical strips within the circle, where each strip has width dx and height 2 times square root of r squared minus x squared. This leads us to the integral: A equals the integral from negative r to r of 2 square root of r squared minus x squared dx.
Now we'll solve this integral using trigonometric substitution. We substitute x equals r sine theta, so dx becomes r cosine theta d theta. The limits change from x in the interval negative r to r, to theta in the interval negative pi over 2 to pi over 2. The square root term transforms to r cosine theta. This gives us the integral: A equals the integral from negative pi over 2 to pi over 2 of 2 r squared cosine squared theta d theta.
Now we evaluate the integral using the double angle identity. Cosine squared theta equals one plus cosine two theta over two. Applying this identity, our integral becomes r squared times the integral of one plus cosine two theta. We can split this into two separate integrals. The first integral gives us r squared times pi, while the second integral of cosine two theta equals zero. Therefore, our final result is A equals pi r squared, confirming our familiar formula through integration.
Let's explore an alternative approach using polar coordinates. In polar coordinates, a circle of radius R is described by r equals R, where r is constant. The area element in polar coordinates is r dr d theta. We set up the double integral: A equals the integral from 0 to 2 pi, integral from 0 to R of r dr d theta. Evaluating the inner integral first gives us R squared over 2. Then the outer integral gives us pi R squared. This elegant method also yields A equals pi R squared, demonstrating the power and consistency of integration techniques.