Area of a circle

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The area of a circle with radius r is [\pi r^2].

Proof

The following is a proof for the area of a circle and an ellipse using calculus:

A circle is a form of ellipse, denoted by the equation

[\frac + \frac = 1].

Solving the equation of the ellipse for [y], the following is derived:

[\frac = 1 - \frac = \frac \ or \ y = \pm\frac\sqrt]

Because the ellipse is symmetric with respect to both axes, the total area [A] is four times the area in the first quadrant. The part of the ellipse in the first quadrant is given by the function
[y = \frac\sqrt \ 0 \le x \le a] and so[:\ \frac A = \int_0^a \frac \sqrt \,dx]

To evaluate this integral we substitute [x= a\sin\theta]. Then [dx = a \cos \theta d\theta]. To change the limits of integration we note that when [x = 0], [\sin \theta = 0], so [\theta = 0]; when [x = a], [\sin \theta = 1], so [\theta = \frac].

Also

[\sqrt \ = \ \sqrt \ = \ \sqrt \ = \ a |\cos \theta| \ = \ a \cos \theta]

since [0 \le \theta \le \frac]

Therefore

[A = 4 \frac \int_0^a \sqrt \,dx \ = 4 \frac \int_^ a \cos \theta \cdot a \cos \theta \,d\theta]

: [ = 4ab \int_^ \cos^2 \theta \,d\theta \ = 4ab \int_^ \frac (1 + \cos 2\theta)]

: [ = 2ab[theta + frac sin 2theta]_^ \ = 2ab (\frac + 0 - 0)]

: [\ \ \ = \pi ab]

This shows that the area of an ellipse with semiaxes [a] and [b] is [\pi ab]. In particular, taking [a = b = r], the formula for the area of a circle with radius [r], [A = \pi r^2], has been proven.

 

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