]> Geometrical applications

### Chapter 2: Integration; Section 2: Definite Integrals; page 4

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# Geometrical Applications

## Length of an arc

The only geometrical application of definite integrals we saw so far, was the evaluation of an area. Any segment of a geometrical curve, commonly called an arc, has a length. As we already know the notion of length is relevant only in a metric space. We are going to limit ourselves first to curves described by continuous single valued functions.

Does the approximation we used for the evaluation of an area under a curve, give us also the length of the arc? The evaluation of an area, using the sum of the maximal or the minimal rectangles, is not appropriate for the length, as can be seen in Fig. Wrong length.

On the other hand the trapezoid method yields the correct length of an arc, as can be seen in Fig. Correct length.

Let's look closely, how the trapezoid method was used in order to approximate the length of an arc. We divided an interval of  x  of a function in  n  equal segments. The  kth  segment includes the values of  x  between  xk-1  and  xk . The approximated length of the arc corresponding to this segment is

 $\sqrt{{\left({x}_{k}-{x}_{k-1}\right)}^{2}+{\left({y}_{k}-{y}_{k-1}\right)}^{2}}=\sqrt{{\left(\Delta x\right)}_{k}^{2}+{\left(\Delta y\right)}_{k}^{2}}$ (2.2.4.1)

The approximated length of the curve is expressed by the sum of the lengths of all the segments

 ${L}_{n}=\sum _{k=1}^{n}\sqrt{{\left(\Delta x\right)}_{k}^{2}+{\left(\Delta y\right)}_{k}^{2}}=\sum _{k=1}^{n}\Delta {x}_{k}\sqrt{1+{\left(\frac{\Delta y}{\Delta x}\right)}_{k}^{2}}$ (2.2.4.2)

On the limit of  $n\to \infty$ , the sum becomes a definite integral

 $L=\underset{n\to \infty }{\mathrm{lim}}\sum _{k=1}^{n}\Delta {x}_{k}\sqrt{1+{\left(\frac{\Delta y}{\Delta x}\right)}_{k}^{2}}=\underset{a}{\overset{b}{\int }}\text{d}x\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}$ (2.2.4.3)

where  a  and  b  are the beginning and end points respectively, of the  x  interval.

As an example we are going to calculate the length of the arc used in the previous figures, namely the arc defined by the parabola

 $y=1-{\left(\frac{x}{3}\right)}^{2}\text{ }\text{in}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{interval}\text{ }0\le x\le 3$ (2.2.4.4)

According to (2.2.4.3), the length of the arc is

 $\begin{array}{l}L=\underset{0}{\overset{3}{\int }}\text{d}x\sqrt{1+{\left(\frac{2x}{9}\right)}^{2}}\text{ }\text{and}\text{\hspace{0.28em}}\text{by}\text{\hspace{0.28em}}\text{substituting}\text{ }u=\frac{2x}{9}\\ L=\frac{9}{2}\underset{0}{\overset{2}{3}}{\int }}\text{d}u\sqrt{1+{u}^{2}}=\frac{9}{2}\left[\frac{u\sqrt{1+{u}^{2}}}{2}+\frac{\mathrm{ln}\left(u+\sqrt{1+{u}^{2}}\right)}{2}\right]\underset{0}{\overset{\frac{2}{3}}{|}}=\\ =\frac{9}{4}\left[\frac{2}{3}\sqrt{1+{\left(\frac{2}{3}\right)}^{2}}+\mathrm{ln}\left(\frac{2}{3}+\sqrt{1+{\left(\frac{2}{3}\right)}^{2}}\right)-0\right]=\\ =\frac{9}{4}\left[\frac{2\sqrt{13}}{9}+\mathrm{ln}\left(\frac{2+\sqrt{13}}{3}\right)\right]\approx 3.209352\end{array}\right\}$ (2.2.4.5)

in accordance with the result obtained by the approximations of the trapezoid method.

By definition any length of an arc has a positive value. The integral (2.2.4.3) is positive only for  a<b . If one has to calculate the length of a curve which is not single valued, care should be taken to divide the curve in parts and perform the integration on each part separetely in the correct direction.

The circumference of a circle can be used as an example:

 $\begin{array}{l}{x}^{2}+{y}^{2}={r}^{2}\text{ }\text{where}\text{ }r>0\text{ }\text{is}\text{\hspace{0.28em}}\text{a}\text{\hspace{0.28em}}\text{constant}\\ y=±\sqrt{{r}^{2}-{x}^{2}}\text{\hspace{0.28em}}\text{;}\text{ }\frac{\text{d}y}{\text{d}x}=\mp \frac{x}{\sqrt{{r}^{2}-{x}^{2}}}\\ L\left(\text{for}\text{\hspace{0.28em}}y>0\right)=\underset{-r}{\overset{r}{\int }}\frac{r\text{d}x}{\sqrt{{r}^{2}-{x}^{2}}}=r\underset{-1}{\overset{1}{\int }}\frac{\text{d}u}{\sqrt{1-{u}^{2}}}=\pi r\\ L\left(\text{for}\text{\hspace{0.28em}}y<0\right)\ne \underset{r}{\overset{-r}{\int }}\frac{r\text{d}x}{\sqrt{{r}^{2}-{x}^{2}}}=-\pi r\end{array}\right\}$ (2.2.4.6)

If a curve is given in a parametric form, then we divide the interval of the parameter  p  in  n  sub-intevals, the same way it was done with  x  for the explicit form:

 $\begin{array}{l}\left\{\begin{array}{l}y=y\left(p\right)\\ x=x\left(p\right)\end{array}\right\}p=\text{parameter}\\ L=\underset{n\to \infty }{\mathrm{lim}}\sum _{k=1}^{n}\sqrt{{\left(\Delta x\right)}_{k}^{2}+{\left(\Delta y\right)}_{k}^{2}}=\\ =\underset{n\to \infty }{\mathrm{lim}}\sum _{k=1}^{n}{\left(\Delta p\right)}_{k}\sqrt{{\left(\frac{\Delta x}{\Delta p}\right)}_{k}^{2}+{\left(\frac{\Delta y}{\Delta p}\right)}_{k}^{2}}=\\ =\underset{a}{\overset{b}{\int }}\text{d}p\sqrt{{\left(\frac{\text{d}x}{\text{d}p}\right)}^{2}+{\left(\frac{\text{d}y}{\text{d}p}\right)}^{2}}\end{array}\right\}$ (2.2.4.7)

where  a<b  are the end points of the interval. Here one also has to pay attention that the integration is achieved in an increasing sense of the parameter, in order to obtain only positive values.

For example we will repeat the calculation of the circumference  L  of a circle (2.2.4.6), expressed this time in its parametric form. Since the parameter is an angle (θ), we can integrate once within the limits from  0  to  2π .

 $\begin{array}{l}\left\{\begin{array}{l}x=r\mathrm{cos}\theta \\ y=r\mathrm{sin}\theta \end{array}\right\}\text{\hspace{0.28em}}r>0\\ L=\underset{0}{\overset{2\pi }{\int }}\text{d}\theta \sqrt{{r}^{2}{\mathrm{sin}}^{2}\theta +{r}^{2}{\mathrm{cos}}^{2}\theta }=\\ =r\underset{0}{\overset{2\pi }{\int }}\text{d}\theta =2\pi r\end{array}\right\}$ (2.2.4.8)

## Volume of revolution around x

Any continuous single valued function  y(x)  in an interval of  x  [a,b] , where  a<b  defines a surface in the  (x,y)  plane, which is limited by the  x  axis and by the curve  y(x) , and is bounded by the interval of  x . When rotated by an angle of  2π  in space about the  x  axis, the surface covers a volume called volume of revolution.

By rotation, any point of the function makes a circular motion with the radius equal to the absolute value of the function:  |y(x)| . The volume of rotation can be expressed approximately as a sum of discs, each one with a circular surface of  πy²(x)  and with a width of  Δx . For  Δx→0  the volume converges to

 $V=\pi \underset{a}{\overset{b}{\int }}{y}^{2}\left(x\right)\text{d}x$ (2.2.4.9)

This way of expressing the volume of revolution around the x axis by discs, is named the disc method.

The example of a revolved parabola

 $\begin{array}{l}y=\sqrt{\frac{x}{2}}\text{ }\text{for}\text{ }0\le x\le 2\\ V=\pi \underset{0}{\overset{2}{\int }}\frac{x}{2}\text{d}x=\pi \left(\frac{{x}^{2}}{4}\right)\underset{0}{\overset{2}{|}}=\pi \end{array}\right\}$ (2.2.4.10)

is explained graphically in Fig. Volume of revolution around x.

In another example we will take the cigar shaped ellipsoid of revolution, obtained by the rotation of the positive branch of the canonical form of an ellipse about the x axis.

 $\begin{array}{l}y=b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}\text{ }\text{where}\text{ }a>b>0\\ V=\pi \underset{-a}{\overset{a}{\int }}{y}^{2}\text{d}x=\pi {b}^{2}\underset{-a}{\overset{a}{\int }}\left(1-\frac{{x}^{2}}{{a}^{2}}\right)\text{d}x=\\ =\pi {b}^{2}\left(x-\frac{{x}^{3}}{3{a}^{2}}\right)\underset{-a}{\overset{a}{|}}=\frac{4\pi }{3}a{b}^{2}\end{array}\right\}$ (2.2.4.11)

This result is consistent with the volume of a sphere for  a = b , as can be expected.

Of course the same method (2.2.4.9) can be used for functions presented in parametric forms. We will repeat this example (2.2.4.11), but this time applying the parametric presentation of the ellipse.

 $\begin{array}{l}\left\{\begin{array}{l}x=a\mathrm{cos}\beta \\ y=b\mathrm{sin}\beta \end{array}\right\}\text{ }\text{where}\text{ }\left\{\begin{array}{l}a>b>0\\ \beta =\text{parameter}\end{array}\right\}\\ V=\pi \underset{-a}{\overset{a}{\int }}{y}^{2}\text{d}x=-\pi a{b}^{2}\underset{\pi }{\overset{0}{\int }}{\mathrm{sin}}^{3}\beta \text{d}\beta =\\ =\pi a{b}^{2}\underset{0}{\overset{\pi }{\int }}{\mathrm{sin}}^{3}\beta \text{d}\beta =\\ =\pi a{b}^{2}\left(-\mathrm{cos}\beta +\frac{{\mathrm{cos}}^{3}\beta }{3}\right)\underset{0}{\overset{\pi }{|}}=\frac{4\pi }{3}a{b}^{2}\end{array}\right\}$ (2.2.4.12)

## Volume of revolution around y

Since one can look at  y  as the independent variable and  x  as the inverse function, one can apply the revolution about the  y  axis, the same way we did previously with the revolution about  x . However there is a method for obtaining the volume of revolution about the  y  axis without inverting the function and still using the independent variable  x  for the integration.

A continuous single valued function  y(x)  in an interval of  x  [a,b] , where  $0\le a , defines a surface in the  (x,y)  plane between the  x  axis and the curve  y(x) , and is bounded by the interval of  x . When rotated by an angle of  2π  about the  y  axis, the surface covers a volume of revolution.

In order to obtain the integral expression of this volume, let's look at a straight-line segment parallel to the  y  axis, between the points  (x,0)  on the  x  axis and the corresponding point of the function  (x,y) , with  x  within the interval. This segment is of length  |y| , and its distance from the  y  axis remains  x  (constant) during the rotation. Therefore by the rotation, it covers a cylindrical shell with radius  x , and has a surface of  2πx|y| . If this shell has a width of  Δx , its product with this surface gives the volume of the shell. This finally yields the expression of the total volume

 $V=2\pi \underset{a}{\overset{b}{\int }}|y\left(x\right)|x\text{d}x$ (2.2.4.13)

This way of expressing the volume of revolution around the  y  axis by cylindrical shells is named the shell method.

Let's look at the following example using the shell method.

 $\begin{array}{l}y=2\left({x}^{2}-1\right)\text{ }\text{for}\text{ }0\le x\le 1\\ V=2\pi \underset{0}{\overset{1}{\int }}|y|x\text{d}x=4\pi \underset{0}{\overset{1}{\int }}\left(1-{x}^{2}\right)x\text{d}x=\\ =4\pi \left(\frac{{x}^{2}}{2}-\frac{{x}^{4}}{4}\right)\underset{0}{\overset{1}{|}}=\pi \end{array}\right\}$ (2.2.4.14)

This example is shown and explained graphically in Fig. Volume of revolution around y.

The figure also shows a volume of revolution of the same shape, as the one obtained previously, by rotating (2.2.4.10) about the  x  axis. As a matter of fact, the geometrical curves used for both rotations are the same, but just inverted, translated and rotated in order to obtain the same result. On the other hand if the same function within the same interval is used for revolution about the  x  axis, or about the  y  axis, the obtained shapes of the volumes of revolution will in general be different. Take for instance the function (2.2.4.14) and apply a revolution about the  x  axis. You can qualitatively figure out the differences, just by a simple drawing, without doing any calculation.

This is well illustrated by the example of the revolution about the  y  axis of an ellipse presented in canonical form, which results in a lentil shaped ellipsoid of revolution. In order to do the revolution, we'll restrict ourselves by the positive branch of the function, and the non-negative interval of  x . This will result in half of the required volume and should be doubled. The application of (2.2.4.13) on the parametric form of the ellipse gives

 $\begin{array}{l}\left\{\begin{array}{l}x=a\mathrm{cos}\beta \\ y=b\mathrm{sin}\beta \end{array}\right\}\text{ }\text{where}\text{ }\left\{\begin{array}{l}a>b>0\\ 0\le \beta \le \frac{\pi }{2}\end{array}\right\}\\ V=-4\pi {a}^{2}b\underset{\frac{\pi }{2}}{\overset{0}{\int }}{\mathrm{sin}}^{2}\beta \mathrm{cos}\beta \text{d}\beta =\\ =4\pi {a}^{2}b\frac{\left({\mathrm{sin}}^{3}\beta \right)\underset{0}{\overset{\frac{\pi }{2}}{|}}}{3}=\frac{4\pi }{3}{a}^{2}b\end{array}\right\}$ (2.2.4.15)

that should be compared to (2.2.4.12).

## Surface of revolution

Any continuous single valued function  y(x)  in an interval of  x  [a,b] , where  a<b , rotated (in three dimensions) by an angle  2π  about the  x  axis, defines a surface of revolution. We are going to learn how to evaluate the surface area, by the use of definite integrals.

Let's look at a small element of a curve with length of  ΔL . By revolving the curve about the  x  axis, this element forms a surface in a form a circular collar with radius  |y|  and width  ΔL . The area of this surface is therefore  2π|y|ΔL . By adding up all the areas formed by all the elements, we'll obtain the area of the surface of revolution. From (2.2.4.3) we know how to express the infinitisimal element of length  dL

 $\text{d}L=\text{d}x\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}$ (2.2.4.16)

which yields for the area of the surface of revolution

 $S=2\pi \underset{\Delta L\to 0}{\mathrm{lim}}\sum |y|\Delta L=2\pi \underset{a}{\overset{b}{\int }}\text{d}x|y\left(x\right)|\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}$ (2.2.4.17)

As an example, we'll use the parabola from (2.2.4.10), which yields a surface called paraboloid of revolution

 $\begin{array}{l}y=\sqrt{\frac{x}{2}}\text{ }\text{interval:}\text{\hspace{0.28em}}0\le x\le 2\text{ }\frac{\text{d}y}{\text{d}x}=\frac{1}{\sqrt{8x}}\\ S=2\pi \underset{0}{\overset{2}{\int }}\text{d}x|y\left(x\right)|\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}=\\ =2\pi \underset{0}{\overset{2}{\int }}\text{d}x\sqrt{\frac{x}{2}\left(1+\frac{1}{8x}\right)}=\pi \sqrt{2}\underset{0}{\overset{2}{\int }}\text{d}x\sqrt{x+\frac{1}{8}}=\\ =\pi \sqrt{2}\frac{2{\left(x+\frac{1}{8}\right)}^{\frac{3}{2}}}{3}\underset{0}{\overset{2}{|}}\approx 9.1751\end{array}\right\}$ (2.2.4.18)

This example is also shown and explained graphically , in Fig. Surface of revolution around x.

A continuous single valued function  y(x)  in an interval of  x  [a,b] , where  $0\le a , rotated (in space) by an angle of  2π  about the  y  axis, also defines a surface of revolution.

The evaluation of the area is done in a similar way, as the case of rotation about the  x  axis, except that the radius of an elementary collar is  x  this time. The expression for the area is therefore

 $S=2\pi \underset{\Delta L\to 0}{\mathrm{lim}}\sum x\Delta L=2\pi \underset{a}{\overset{b}{\int }}x\text{d}x\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}$ (2.2.4.19)

As an example we will use the parabola from (2.2.4.14). From the examples of the volumes of revolution, we know that this paraboloid is of the same shape as that from (2.2.4.18), and therefore we expect the same area:

 $\begin{array}{l}y=2\left({x}^{2}-1\right)\text{ }\text{interval:}\text{\hspace{0.28em}}0\le x\le 1\text{ }\frac{\text{d}y}{\text{d}x}=4x\\ S=2\pi \underset{0}{\overset{1}{\int }}x\text{d}x\sqrt{1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^{2}}=\\ =2\pi \underset{0}{\overset{1}{\int }}x\text{d}x\sqrt{1+16{x}^{2}}=8\pi \underset{0}{\overset{1}{\int }}x\text{d}x\sqrt{\frac{1}{16}+{x}^{2}}=\\ =8\pi \frac{{\left(\frac{1}{16}+{x}^{2}\right)}^{\frac{3}{2}}}{3}\underset{0}{\overset{1}{|}}\approx 9.1751\end{array}\right\}$ (2.2.4.20)

For another example we will use the torus. The torus (or toroid) is the geometrical form of a doughnut. It is obtained by the revolution about the  y  axis, of a circle with radius  r , and its center situated at a distance  R  from the  y  axis where  R > r . Its surface area is

 $S=4{\pi }^{2}rR$ (2.2.4.21)

The torus is shown graphically and (2.2.4.21) is obtained in Fig. Surface of a torus.

## Exercises

Exercise 1. Calculate the length of the curve $y=b\mathrm{cosh}\left(\frac{x}{b}\right)\text{ }\text{for}\text{ }-b\le x\le b$ where b is a positive constant.

Exercise 2. A plane surface is limited by the x axis and the line  y = kx  for the x interval of  [0,a],

where a and k are positive constants.

1. Find the volume of revolution around the x axis!
2. Find the volume of revolution around the y axis!
3. For what value of k are the two volumes equal?
4. From the volume of revolution around the x axis, obtain the volume of a cone in terms of r (the radius of its base) and h (its height)!
5. Validate your result (of the volume of a cone you obtained) with the volume of revolution around the y axis!

Exercise 3. Use the parametrisation given in Fig. Surface of a torus and calculate the volume of a torus!

Exercise 4. The parametric representation of the cycloid (1.2.5.9) is

$\left\{\begin{array}{l}x=r\left(\phi -\mathrm{sin}\phi \right)\\ y=r\left(1-\mathrm{cos}\phi \right)\end{array}\right\}$

For the angular interval of [0,2π]

1. What is the length of the curve?
2. What is the area of the surface of revolution around the x axis?
Hint: $\mathrm{cos}\phi ={\mathrm{cos}}^{2}\left(\frac{\phi }{2}\right)-{\mathrm{sin}}^{2}\left(\frac{\phi }{2}\right)$

Exercise 5. The curve

$\left\{\begin{array}{l}x=r\left(\phi +\mathrm{sin}\phi \right)\\ y=r\left(1+\mathrm{cos}\phi \right)\end{array}\right\}$

represents the same cycloid as in exercise 4, but translated along the x axis, so that the maximum is at x=0. A plane surface is enclosed between the curve in the φ interval of  [0,π], and the coordinate axes. Calculate the volume of revolution

1. around the x axis.
2. around the y axis.

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