Mathematics · Ch 11 — Applications of the Integrals

Area of a Circle (Standard Form)

The standard equation of a circle of radius $r$ centred at the origin is $x^2+y^2=r^2$ . Solving for $y$ , the upper half of the circle is the curve

y = \sqrt{r^2 - x^2}, \qquad -r \leq x \leq r,

and the full circle's area is derived by integrating this expression and then using the circle's symmetry.

Setting up the quarter-circle integral. By symmetry about both axes, the area of the whole circle is exactly $4$ times the area of the part lying in the first quadrant alone. The first-quadrant portion is bounded by the arc $y=\sqrt{r^2-x^2}$ , the $x$ -axis, and the ordinates $x=0$ and $x=r$ , so its area is

A_{1} = \int_0^r \sqrt{r^2 - x^2}\,dx.

Evaluating by the substitution $x = r\sin\theta$ . Let $x = r\sin\theta$ , so $dx = r\cos\theta\,d\theta$ . When $x=0$ , $\sin\theta=0$ so $\theta=0$ ; when $x=r$ , $\sin\theta=1$ so $\theta=\pi/2$ . Also,

\sqrt{r^2-x^2} = \sqrt{r^2 - r^2\sin^2\theta} = r\sqrt{1-\sin^2\theta} = r\cos\theta \quad (\cos\theta \geq 0 \text{ on } [0,\pi/2]).

Substituting,

A_1 = \int_0^{\pi/2} (r\cos\theta)(r\cos\theta)\,d\theta = r^2\int_0^{\pi/2}\cos^2\theta\,d\theta.

Using the double-angle identity $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$ ,

A_1 = r^2 \int_0^{\pi/2} \frac{1+\cos 2\theta}{2}\,d\theta = \frac{r^2}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/2} = \frac{r^2}{2}\left(\frac{\pi}{2} + 0 - 0 - 0\right) = \frac{\pi r^2}{4}.

So the first-quadrant quarter-circle has area $\dfrac{\pi r^2}{4}$ .

Scaling up to the full circle. Multiplying by $4$ (one factor for each of the four congruent quadrants) gives the area of the whole circle:

\text{Area of circle} = 4 \times \frac{\pi r^2}{4} = \pi r^2,

the familiar formula, now derived from the definition of area as a definite integral rather than simply quoted.

Getting a semicircle or a quadrant instead. The same antiderivative works for any portion of the circle -- only the limits of integration and the final multiplying factor change:

Upper semicircle (bounded by the arc and the $x$ -axis, from $x=-r$ to $x=r$ ): this is twice the first-quadrant piece by symmetry about the $y$ -axis, giving $2 \times \dfrac{\pi r^2}{4} = \dfrac{\pi r^2}{2}$ -- or equivalently, half of $\pi r^2$ , as expected for a half-circle.
Quarter circle in one quadrant: no scaling is needed at all -- it is exactly $A_1 = \dfrac{\pi r^2}{4}$ by itself, the result already found above.
Full circle: multiply the quarter-circle result by $4$ , as shown. …

Figure 1Circle, standard form, area by the quadrant method

What this figure shows. A single circular curve is drawn centred at the origin, with coordinate axes shown passing through the centre. The circle crosses the x-axis at the labelled points (-r,0) and (r,0), and crosses the y-axis at (0,r) and (0,-r). The portion of the circle's interior lying in the first quadrant only (bounded by the arc, the positive x-axis, and the positive y-axis) is shaded, distinct in shading from the unshaded remaining three quadrants of the circle's interior, to show that the shaded quarter is one of four congruent quarters making up the whole circle. A thin vertical strip of width dx is drawn inside the shaded quarter at a representative x between 0 and r, with height reaching from the x-axis up to the arc, labelled y = sqrt(r^2 - x^2), illustrating the rep …

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