Submitted to HEJ Manuscript no.: ANM-010201-A

The area as an integral

The idea of area was our starting-point for the definition of the integral; but the connection between definite integral and area is still incomplete. The areas with which we are concerned in geometry are bounded by given closed curves. On the other hand, the area measured by the integral $I = \int _{a}^{b}f(x)\,dx$ is bounded only in part by the given curve $y=f(x)$, the rest of the boundary consisting of lines which depend on the choice of the coordinate system. If we introduce $t$ formally as a new independent variable in the above integral writing $x=x(t)$, $y=y(t)=f(x(t))$, we have

$$\int _{a}^{b}f(x)\,dx = \int _{{\it t_0}}^{{\it t_1}}{\rm y}(t)\,({\frac {\partial }{\partial t}}\,{\rm x}(t))\,dt$$

where $t_0$ and $t_1$ are the values of the parameter corresponding to the abscissa $x_0=a$ and $x_1=b$ respectively. Here we suppose that every point of the branch of the curve $y=f(x)$ corresponds to a single value of $t$ in the interval $t_0\leq t\leq t_1$, and conversely; furthermore $f(x)$ is everywhere positive and ${\frac {\partial }{\partial t}}\,{\rm x}(t)$ never vanishes in this interval.

We can express our formula for the area in a more elegant symmetrical form if we first transform the integral by integration by parts:

$$\int _{{\it t0}}^{{\it t1}}{\rm y}(t)\,({\frac {\partial }{ \... ...}{\partial t}}\,{\rm y}(t))\,dt + x*y\ \bigg\vert _{t_0}^{t_1}.$$

Since the curve is closed,

$$x(t_0)=x(t_1),\ y(t_0)=y(t_1),$$

and therefore

$$I = -\int _{{\it t0}}^{{\it t1}}{\rm y}(t)\,({\frac {\partial... ...\rm x}(t)\,({\frac {\partial }{ \partial t}}\,{\rm y}(t))\,dt .$$

If we form the arithmetic mean of the two expressions we obtain the symmetrical form

$$I = \frac{1}{2}\,\int _{{\it t0}}^{{\it t1}}{\rm y}(t)\,({\fr... ...{\rm x}(t)\,({\frac { \partial }{\partial t}}\,{\rm y}(t))\,dt.$$

EXAMPLE 7.1.

As an example of the application of our formula for the area consider the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. So we can define the function

$>$ f:=x-> b/a*sqrt(a^2-x^2);

$$f := x\rightarrow {\displaystyle \frac {b\,\sqrt{a^{2} - x^{2}}}{ a}}$$

In order to find its area we take the upper and lower halves of the ellipse separately and in this way we can express the area by the integral

$>$ 2*Int(f(x),x=-a..a)=value(2*int(f(x),x=-a..a));

$$2\,{\displaystyle \int _{ - a}^{a}} {\displaystyle \frac {b\... ...^{a}} {\displaystyle \frac {b\, \sqrt{a^{2} - x^{2}}}{a}} \,dx$$

If, however, we use the parametric representation $x=a*\cos(t),\ y=b*\sin(t)$, we find that

$>$ -1/2*Int(b*sin(t)*diff(a*cos(t),t)-

$>$ a*cos(t)*diff(b*sin(t),t),t=0..2*Pi);

$$- {\displaystyle \frac {1}{2}} \,{\displaystyle \int _{0}^{2\, \pi }} - b\,{\rm sin}(t)^{2}\,a - a\,{\rm cos}(t)^{2}\,b\,dt$$

which has the value

$>$ value(%);

$$b\,a\,\pi$$

In this subsection we have based the definition of the area on the concept of integral and have shown that this analytical definition has a truly geometrical character, since it yields a quantity independent of the coordinate system. It is, however, easy to give a direct geometrical definition of the area bounded by a closed curve which does not intersect itself, as follows: the area is the upper bound of the areas of all polygons lying interior to the curve. The proof that the two definitions are equivalent is quite simple, but will not be given here.

Submitted to HEJ Manuscript no.: ANM-010201-A