Area of Circle

Theorem
The area $$A$$ of a circle is given by the formula $$A=\pi r^2$$, where $$r$$ is the radius of the circle.

Proof
We start with the equation of a circle: $$x^2 + y^2 = r^2$$.

Thus $$y = \pm \sqrt{r^2 - x^2}$$, so from the geometric interpretation of the definite integral, $$A=\int_{-r}^r \left[ \sqrt{r^2 - x^2} - (-\sqrt{r^2 - x^2})\right] dx$$

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Let $$x = r \sin\theta$$ (note that we can do this because $$-r \leq x \leq r$$. Thus, $$\theta = \arcsin \left(\frac{x}{r}\right)$$ and $$dx = r\cos\theta d\theta$$.

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Proof by Shell Integration
The circle can be divided into a set of infintesimaly thin rings, each of which has area $$2\pi t dt$$, since the ring has length $2 \pi t$ and thickness $$dt$$.

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