After learning how to calculate the area under a curve by using the definite integral, we'll now see how the definite integral can be applied, for **calculating an area enclosed by a loop**.

We'll begin with the **case of an enclosed area between two single valued continuous functions that intersect at two points**, defining in this way an interval of the independent variable *x* . In order to use definite integrals for calculating the enclosed area, we also require that the functions are integrable in the interval. In order to use the notion of an area under a function, we'll start by also requiring that both functions are non-negative in this interval.

For any point in this interval (except the end points), **one of the functions, say y_{1}(x) , should be greater than the second y_{2}(x) **; otherwise we'll have intersection points inside the interval. From the definition of area under a curve, it follows that

$$\begin{array}{l}\text{for}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{interval}\text{\hspace{1em}}\left[a,b\right]\\ \{\begin{array}{l}{y}_{1}\left(a\right)={y}_{2}\left(a\right)\\ {y}_{1}\left(b\right)={y}_{2}\left(b\right)\end{array}\\ \text{for}\text{\hspace{0.17em}}x\ne a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\ne b:\\ {y}_{1}\left(x\right)>{y}_{2}\left(x\right)\ge 0\\ S={\displaystyle \underset{a}{\overset{b}{\int}}{y}_{1}}\text{d}x-{\displaystyle \underset{a}{\overset{b}{\int}}{y}_{2}}\text{d}x=\\ ={\displaystyle \underset{a}{\overset{b}{\int}}{y}_{1}}\text{d}x+{\displaystyle \underset{b}{\overset{a}{\int}}{y}_{2}}\text{d}x\end{array}\}$$ | (1) |
---|

The last row of (1) can be interpreted as the integral, whose path is clockwise along the loop, defined by the appropriate functions.

Let's take the following *example*

$\begin{array}{l}\{\begin{array}{l}{y}_{1}={x}^{2}+1\text{\hspace{1em}}\text{and}\text{\hspace{1em}}{y}_{2}=2{x}^{2}\\ a=-1\text{\hspace{1em}}\text{and}\text{\hspace{1em}}b=1\end{array}\\ S=\underset{-1}{\overset{1}{\int}}\left({x}^{2}+1\right)\text{d}x+2\underset{1}{\overset{-1}{\int}}{x}^{2}\text{d}x=\\ =\left(\frac{{x}^{3}}{3}+x\right)\underset{-1}{\overset{1}{|}}+2\frac{{x}^{3}}{3}\underset{1}{\overset{-1}{|}}=\\ =\left(\frac{1}{3}+1\right)-\left(-\frac{1}{3}-1\right)+2\left(-\frac{1}{3}-\frac{1}{3}\right)=\\ =-\frac{2}{3}+2=\frac{4}{3}\end{array}\}$ | (2) |
---|

which is shown in **Fig. Enclosed area of parabolas(1) **.

This example uses functions *y* that are non-negative for all the points of the loop enclosing the area. The notion of the clockwise sense of the integration was graphically emphasized in the figure. The question that should now be asked is, what happens if there are points of the loop with negative *y* values? As we'll see it does not matter: **the result of the integration gives the enclosed area with a positive sign, as far as the clockwise sense is observed**.

In order **to prove** this, let's use (1) and translate both functions in *y* by the same constant value *Δy* . By doing so, the enclosed area remains unchanged, but on the other hand the values of *y* change and can also be negative. The clockwise sense of integration becomes

$\begin{array}{l}\underset{a}{\overset{b}{\int}}\left({y}_{1}+\Delta y\right)\text{d}x+\underset{b}{\overset{a}{\int}}\left({y}_{2}+\Delta y\right)\text{d}x=\\ =\underset{a}{\overset{b}{\int}}{y}_{1}\text{d}x+\underset{b}{\overset{a}{\int}}{y}_{2}\text{d}x+\Delta y\underset{a}{\overset{b}{\int}}\text{d}x+\Delta y\underset{b}{\overset{a}{\int}}\text{d}x=\\ =S+\Delta y\left[\left(b-a\right)+\left(a-b\right)\right]=S\end{array}\}$ | (3) |
---|

After the proof, we can modify the ** example** (2) by adding Δ

$\begin{array}{l}{y}_{1}={x}^{2}\\ {y}_{2}=2{x}^{2}-1\end{array}\}$ | (4) |
---|

The graphical presentation this time can be seen in **Fig. Enclosed area of parabolas(2) **

We can now talk about any area enclosed in a loop not intersecting itself. **It does not matter if this loop is formed from two functions, or it is defined by many segments of different curves**, as long as the integration is done in a clockwise sense, it will result in a positive value. The result of a counter-clockwise sense of integration will give also the area but with a negative sign.

**The enclosing loop** does not have to be defined by explicit functions, but **could be given also by a parametric representation**. There are cases, where the whole loop can be defined by just one parametric presentation, whose area can be obtained by one single integration.

In the following ** example** we'll use the area enclosed by the loop formed by the Folium of Descartes:

${x}^{3}-3axy+{y}^{3}=0$ |
---|

Its parametric presentation is:

$\left\{\begin{array}{l}x=\frac{3at}{1+{t}^{3}}\\ y=\frac{3a{t}^{2}}{1+{t}^{3}}\end{array}\right\}$ |
---|

From previous analysis it is known that **the function y(x) has a minimum at the origin (x,y) = (0,0)** , and that the point

As explained in this figure, the area *S* of the loop can be calculated from

$S=2\left(\underset{0}{\overset{\frac{3a}{2}}{\int}}x\text{d}x+\underset{\frac{3a}{2}}{\overset{0}{\int}}y\text{d}x\right)$ | (5) |
---|

where *y* corresponds to the curve (the Folium of Descartes) for the interval of the parameter $0\le t\le 1$
and the integral can be solved by substituting *u* = 1+*t*³

$\begin{array}{l}\underset{\frac{3a}{2}}{\overset{0}{\int}}y\text{d}x=\underset{1}{\overset{0}{\int}}y\frac{\text{d}x}{\text{d}t}\text{d}t=\\ =\underset{1}{\overset{0}{\int}}\frac{3a{t}^{2}}{1+{t}^{3}}\frac{3a\left(1+{t}^{3}\right)-3{t}^{2}3at}{{\left(1+{t}^{3}\right)}^{2}}\text{d}t=\\ ={\left(3a\right)}^{2}\underset{1}{\overset{0}{\int}}{t}^{2}\frac{1-2{t}^{3}}{{\left(1+{t}^{3}\right)}^{3}}\text{d}t=\\ =3{a}^{2}\underset{2}{\overset{1}{\int}}\frac{3-2u}{{u}^{3}}\text{d}u=\\ =3{a}^{2}\left(-\frac{3}{2{u}^{2}}+\frac{2}{u}\right)\underset{2}{\overset{1}{|}}=-\frac{3{a}^{2}}{8}\end{array}\}$ | (6) |
---|

and finally, from (5) and (6) one obtains the area *S*

$S=2\left(\frac{9{a}^{2}}{8}-\frac{3{a}^{2}}{8}\right)=\frac{3{a}^{2}}{2}$ |
---|