MATH 5.4: Applications of integration	PPLATO @
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1 Opening items

1.1 Module introduction

You probably know that the volume of a sphere of radius r is $\frac43\pi r^3$, but do you know how to prove that this is the case? One way to proceed is to divide the sphere into a set of thin discs, to find an approximate expression for the volume of each disc, then add all the approximations, and so estimate the volume of the sphere. As we allow the discs to get thinner and thinner, the accuracy of the approximation improves and approaches a limiting value – the limit of a sum – which is usually known as a definite integral.

This module discusses several physical and geometrical applications of integration, all based on the fact that a definite integral is the limit of an appropriate sum. This idea is probably already familiar to you, since you may well have been introduced to definite integrals in the context of calculating areas under graphs, where such an area is approximated by a set of thin rectangles. However, the module starts with a review of the relation between a definite integral and an area, and discusses cases where the area actually corresponds to some physical quantity. It goes on to show how definite integrals can be used to find more complicated areas those enclosed by two intersecting graphs. Then it discusses some examples of solids (solids of revolution) whose volumes and surface areas can be written as definite integrals. (Here, you will find a derivation of the formula for the volume of a sphere.) Finally, it shows you how to express several other quantities masses of objects whose density is not constant, centres of mass and moments of inertia of solid objects, average values as definite integrals.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

Study comment In order to study this module, you will need to be familiar with the following terms: centre of mass, definite integral, improper integral, integrand, integration by parts, integration by substitution, limits of integration, modulus, moment of inertia, range of integration. If you are uncertain of any of these terms, you can review them now by referring to the Glossary which will indicate where in FLAP they are developed. In addition, you will need to be familiar with various trigonometric identities, and you should know how to find standard integrals (such as the integrals of xⁿ, or e^ax), and to evaluate definite integrals by the method of integration_by_substitutionsubstitution, or by integration by parts. You will also need to be able to sketch graphs of straight lines, quadratic_equationquadratic and cubic_equationcubic polynomial_expressionpolynomials, reciprocal functions, circles and ellipses; and know how to find the intersectpoints of intersection of two graphs. The following questions will allow you to establish whether you need to review some of these topics before embarking on this module.

If you feel unsure about any of the terms used in Questions R1 to R6, consult the Glossary.

2 Areas

2.1 Area under a graph

You already know that a definite integral $\Int_a^bf(x)\,dx$ i can be related to an ‘area under the graph’ of the function f(x). If f(x) is a function that is positive over the interval a ≤ x ≤ b, then the integral$\Int_a^bf(x)\,dx$ is equal to the magnitude of the area enclosed by the graph of the function f(x), the vertical lines x = a and x = b, and the x–axis (see Figure 1). The area of this region is known as the ‘area under the graph’ of f(x) over the interval a ≤ x ≤ b. It is worth recalling here the argument that relates the definite integral $\Int_a^bf(x)\,dx$ to the area A shown in Figure 1, since we shall be using the same line of reasoning in many different cases throughout this module.

The idea is that we can estimate a value for A by dividing the area up into a large number of thin rectangles. In Figure 2, the area under the graph of f(x) between x₁ = a and x_n+1 = b has been divided into n strips (although we only show six of them) which are then approximated by rectangles: the first is of height f(x₁) and width ∆x₁, the second is of height f(x₂) and width ∆x₂, and so on. The area of the i^th rectangle, covering the interval [x_i, x_i + ∆x_i] is f(x_i)∆x_i and the sum of the areas of all these rectangles provides a good approximation to A, in other words,

$\displaystyle A \approx \sum_{i=1}^n f(x_i)\,\Delta x_i$(1) i

As we allow the width of the rectangles to become smaller and smaller (while, as a consequence, n gets larger) the sum on the right–hand side of Equation 1 becomes an ever better approximation to A, and in the limit, the sum is actually equal to A.

This limit of a sum is defined to be the definite integral $\Int_a^bf(x)\,dx$, so that we have

$A = \Int_a^bf(x)\,dx$(2)

The ‘area under a graph’ and the ‘magnitude of the area bounded by a graph’

In the above discussion we assumed that the function f(x) is positive between a and b, and we will need to make a minor adjustment if f(x) is negative or changes sign in the interval. In FLAP, we use the following definition:

This has the consequence that in a region where f(x) is always negative, the area under the graph of f(x) is a negative quantity. We might, however, be interested instead in calculating the magnitude of the area enclosed by the graph of f(x), the vertical lines x = a and x = b, and the x–axis. Such a magnitude is, by definition, always positive.

You will find that some authors define ‘area under a graph’ in such a way that it always gives the magnitude of this area. However, we shall use the definition given above, and we will make it very clear if we want you to calculate the magnitude of an enclosed area.

The essential point for you to note is this: when calculating ‘the area under a graph’, the areas of the regions below the x–axis must be subtracted from the areas of the regions above the x–axis. On the other hand, when you are asked to find ‘the magnitude of the area bounded by the graph’ you must ensure that all the contributions to the area, from parts above or below the x–axis, are positive. This means that you have to consider separately the regions in which f(x) is positive and those in which it is negative, as in the following example.

In a question of this kind it is absolutely essential to be able to determine where the function changes sign; it will also help if you are able to sketch the graph of the function.

We can summarize the previous discussion very neatly in terms of the modulus of f(x):

Since |f(x)| = f(x) when f(x) is positive, and |f(x)| = −f(x) when f(x) is negative, the modulus sign takes care of any changes in sign that f(x) may undergo in the region of integration. The integral $\Int_a^b\lvert\,f(x)\rvert\, dx$ is not in general equal to $\Int_a^bf(x)\,dx$; the two are only equal if f(x) ≥ 0 throughout the interval a ≤ x ≤ b.

Although this description is quite neat, in practice we rarely try to integrate |f(x)| directly, and usually we consider separately regions where the function is positive, and regions where the function is negative, as in Question T1.

The physical significance of the definite integral

So far, we have simply interpreted f(x) geometrically, as the height of the graph y = f(x), in which case $\Int_a^b\lvert\,f(x)\rvert\, dx$ is indeed just the magnitude of the area bounded by the graph of y = f(x), measured in whatever scale units are used on the graph’s axes. However, if f(x) represents some physical quantity, then the definite integral will of course have a different physical significance. Here are two examples.

Velocity–time graphs

Figure 4 shows a graph of the velocity, v_x(t) of an object moving along the x–axis, against time t; note that v_x(t) changes sign twice. If we want to know the displacement s_x of the object from its initial position (at t = 0) after a given time T has elapsed, we can use the same reasoning that led from Equation 1 to Equation 2.

$\displaystyle A \approx \sum_{i=1}^n f(x_i)\,\Delta x_i$(Eqn 1)

$A = \Int_a^bf(x)\,dx$(Eqn 2)

We divide the time T into many short time intervals, each of duration ∆t. During any one of these time intervals, v_x(t) is approximately constant, so the corresponding displacement is approximately equal to v_x(t)∆t; then the total displacement is approximately equal to the sum $\sum v_x(t)\,\Delta t$. In the limit as ∆t tends to zero, we find that the displacement is given by the integral $\Int_0^Tv_x(t)\,dt$, i.e. the area under the graph of v_x(t) between t = 0 and t = T.

If, instead, we want to calculate the distance travelled by the object, we must recall that distance is the magnitude_of_a_real_quantitymagnitude of displacement. Thus the distance travelled in a short time interval ∆t is equal to |v_x(t)|∆t and, reasoning as before, we find that the distance travelled after time T is given by the integral $\Int_0^T\lvert\,v_x(t)\rvert\,dt$. So it is equal to the sum of the magnitudes of the areas enclosed by the graph of v_x(t) and the t–axis, between t = 0 and t = T.

Work

Suppose that an object is moving along the x–axis under the influence of a constant force F_x in the x–direction. The work done by the force in moving the object from x = a to x = b is W = F_xs_x where s_x is the displacement of the object, which in this case is b − a.

If the force, F_x(x) say, varies with x, then we can divide the interval a ≤ x ≤ b into many much smaller subintervals, of width ∆x, in each of which F_x(x) is approximately constant. The work done in moving the object through the small subinterval between x and x + ∆x is approximately F_x(x)∆x.

The total work done is approximately given by adding all these small amounts of work, so that

total work done $W \approx \sum F_x(x)\,\Delta x$

In the limit as ∆x decreases towards zero (and the number of subintervals increases) this approximation to W becomes increasingly accurate, and

$W = \Int_a^bF(x)\,dx$

It follows that we can interpret W to be the area under the graph of F_x(x) between x = a and x = b.

2.2 Area between two graphs

Suppose that we want to find the area A enclosed by the graphs of the two functions f(x) and g(x) shown in Figure 5. Proceeding as before, we divide the region up into thin slices, each of thickness ∆x, then approximate each slice by a thin rectangle. The height of the rectangle shown in Figure 5 is (f(x) − g(x)), and so its area is (f(x) − g(x))∆x. We now sum the areas of all the rectangles, and let ∆x tend to zero, so that in the limit the sum becomes an integral and we have

$A = \Int_a^b\left[\,f(x) - g(x)\right]\,dx$(3)

where a and b are the x–coordinates of the points of intersection of the two graphs. Notice that f(x) ≥ g(x) in the interval a ≤ x ≤ b, which ensures that f(x) − g(x) is positive, and therefore A is also positive.

It is quite possible to find such an area without drawing the graphs (as the above exercise shows), but if you are asked to calculate the area between the graphs of f(x) and g(x), you will find it much easier if you begin by sketching the two graphs on the same axes, and, of course, finding the points of intersection (which you need in order to be able to put in the correct limits in Equation 3).

The following example shows you how to proceed.

In all the examples so far, the functions f(x) and g(x) have been chosen so that the graph of f(x) lies above the graph of g(x). This ensures that the integrand in Equation 3,

$A = \Int_a^b\left[\,f(x) - g(x)\right]\,dx$(Eqn 3)

is positive, so that we obtain a positive answer for A. If we had chosen our two graphs the other way round, then the height of the rectangle in Figure 5 would have been equal to (g(x) − f(x)), and it is this function that we would have integrated. Thus, wherever f(x) > g(x), we integrate (f(x) − g(x)), to obtain the area; but in the case that g(x) > f(x), we integrate (g(x) − f(x)). In both these cases, we can write down the area of the region – known as the area_between_two_graphsarea between the graphs of f(x) and g(x) – as

It is important to use Equation 4 (not Equation 3) if the two graphs intersect more than twice, so that (f(x) − g(x)) is sometimes positive and sometimes negative, as in the next example. i

The area of a circle

Finally in this subsection, it is worth noting that we can use Equation 4,

$A =\Int_a^b\lvert\,f(x) - g(x)\rvert\,dx$(Eqn 4)

to find the areas of shapes such as circles and ellipses. For example, let us use it to prove that the area of a circle of radius r is πr². The equation of a circle of radius r centred on the origin is x² + y² = r²; so, for a given value of x (less than r), y can take the two values $+\sqrt{r^2-x^2}$ and $-\sqrt{r^2-x^2}$. Thus we can regard the region inside the circle as the area enclosed by the graphs of the two functions $f(x) = +\sqrt{r^2-x^2}$ and $g(x) = -\sqrt{r^2-x^2}$ (see Figure 9). The points of intersection of these two graphs are x = − r and x = + r. So Equation 4 gives, for the area A of the circle,

$A = \Int_{-r}^r\left[\sqrt{r^2-x^2}-\left(-\sqrt{r^2-x^2}\right)\right]\,dx = 2\Int_{-r}^r(\sqrt{r^2-x^2})\,dx$

The definite integral can be evaluated by making the substitution x = rsinu. Then $\sqrt{r^2-x^2} = \sqrt{r^2 (1 - \sin^2u)} = r\cos u$, and dx = rcosudu.

When x = r, sin(u) = 1, so u = π/2, and when x = −r, u = −π/2; thus the new limits of integration are π/2 and −π/2. So the integral becomes

$A = 2r^2 \Int_{-\pi/2}^{\pi/2}\cos^2 u\,du$

We can evaluate this integral by means of the trigonometric identity 2cos²u = 1 + cos(2u); substituting this into the integral, we find

$\displaystyle A = 2r^2 \int_{-/pi2/}^{\pi/2}\cos^2 u\,du = 2r^2\int_{-/pi2/}^{\pi/2}\dfrac12\{1+\cos(2u)\}\,du = r^2\left[u+\dfrac12\sin(2u)\right]_{-/pi2}^{\pi/2}$

$\phantom{A }= r^2\left[\dfrac{\pi}{2}+\dfrac12\sin(\pi) - \left(-\dfrac{\pi}{2}+\dfrac12\sin(-\pi)\right)\right] = \pi r^2$

3 Solids of revolution

3.1 Volume of revolution

Suppose that we want to find the volume of a cone, of height h and base radius r. Such a cone is shown in Figure 10; we have drawn it so that its vertex is at the origin and its axis of symmetry lies along the x–axis. To find the volume we can approximate the cone by a set of thin discs, each of thickness ∆x. One of these discs, D, is shown in Figure 10. The centre of the disc D has coordinates (x, 0). We calculate the volume of each disc, and then obtain an estimate for the volume of the cone by adding these volumes. We then allow ∆x to tend to zero, and in the limit the sum becomes a definite integral between the limits 0 and h, which gives the exact value of the volume of the cone. In order to be able to evaluate this integral, we must, of course, obtain an expression for the volume of a typical disc in terms of x.

To calculate the volume of a typical disc D, we need first to know its radius. This is equal to the y–coordinate of the point A, and to express this in terms of x, we need to find the equation of the line OP in Figure 10. This line passes through the origin and the point (h, r); its gradient is r/h and its intercept is zero. So the equation of the line is y = rx/h. Consequently, the radius of the disc D is rx/h.

Hence

cross-sectional area of $D = \pi\left(\dfrac rh\right)^2$

and

volume of D = area × thickness = $D = \pi\left(\dfrac rh\right)^2x^2\Delta x$

Hence the volume V of the cone is given by the integral

$V = \pi\left(\dfrac rh\right)^2\Int_0^hx^2\,dx = \pi\left(\dfrac rh\right)^2\left[\dfrac13x^3\right]_0^h = \frac13\pi r^2h$

an expression that you have probably seen before.

We will show shortly that we can generalize this strategy to find the volume of any solid which can be regarded, to a good approximation, as a series of discs all centred on the x–axis. But we first introduce some terminology.

Look at Figure 10; imagine rotating the triangle OPQ (which lies in the (x, y) plane) about the x–axis. As you do so, the line OP will sweep out the surface of the cone. We can say, therefore, that the cone is generated by rotating the area OPQ about the x–axis. This is an example of a solid of revolution a solid which can be obtained by rotating the area under a graph (or part of a graph) about some axis. The volume of a solid of revolution is known as a volume of revolution.

Figure 12 shows an arbitrary solid of revolution, obtained by rotating the area under the graph of the function f(x) between the points x = a and x = b about the x–axis. As we did with the cone, we can approximate this volume by a set of thin discs of thickness ∆x; we show one such disc, D, in Figure 12. The radius of D is equal to the y–coordinate of the point A; that is, it is equal to f(x).

So the cross–sectional area of D is π[f(x)]², and its volume is π[f(x)]²∆x.

Summing over the volumes of all the discs, and allowing ∆x to tend to zero, we obtain a definite integral for the volume V of the solid of revolution:

3.2 Surface of revolution

Suppose now that we want to find, not the volume but the surface area of the cone shown in Figure 10. By now, you may be thinking that you know what is coming. Perhaps what we must do is to approximate the cone by a set of thin discs (just as we did to calculate its volume), work out the surface area of a typical disc, add up all these surface areas, and let the thickness of the discs tend to zero, to obtain an integral giving the surface area of the cone. This process does NOT work, and an example should convince you that this is the case.

The disc D in Figure 10 has radius rx/h, its circumference is 2πrx/h, and so its surface area (= circumference × thickness) is 2πrx/h∆x. Thus the integral giving the total surface area of the discs as their thickness tends to zero is

$2\pi\dfrac rh\Int_0^hx\,dx = 2\pi\dfrac rh\left[\dfrac12x^2\right]_0^h = \pi rh$(6)

However, Equation 6 does not give the correct answer for the surface area of the cone, as we can easily verify!

We have another way of finding this surface area: we can imagine cutting the cone along a straight line running from its base to its vertex and spreading the cone out flat. This gives us the sector of a circle shown in Figure 14; the radius of the circle is equal to the slant height l of the cone (i.e. the length of the line OP in Figures 10 and 14), and the length of the arc PR is equal to the circumference of the base of the cone, 2πr. (If you do not believe that this is the figure obtained, try doing th eexperiment in reverse. Cut out a sector of a circle; you will find that you can roll it up into a cone.)

The surface area of the cone is equal to the area of the sector shown in Figure 14. This sector is a fraction θ/2π of the complete circle of radius l, where θ is the angle $\rm{P\hat{O}R}$ (measured in radians).

So its area must be equal to (θ/2π) × the area of this circle (πl²), i.e. to $\frac12\theta l^2$.

The angle θ is equal to the length of the arc PR divided by the radius l of the circle, θ = 2πr/l, so we find

surface area S of cone = $\dfrac12\left(\dfrac{2\pi r}{l}\right)l^2 = \pi rl$(7)

– NOT the same as the result obtained in Equation 6.

The reason why Equation 6 gives the wrong answer is simply that, while the volume of a thin slice of the cone is well approximated by the volume of a thin disc of the same thickness and with radius equal to the average radius of the slice, the surface area of such a slice is not equal, even approximately, to the surface area of the corresponding disc. However, we can arrive at an integral giving the surface area of the cone if we consider carefully what the surface area of a thin slice of it actually is.

Figure 15a shows such a slice Ω, whose thickness (equal to the length AB) we will call ∆s. We can think of this slice as being generated by rotating the line segment AB about the x–axis; in which case we see that the point B travels a total distance equal to 2π × BC. So if we imagine cutting the slice along AB, and laying it flat, we will obtain the shape shown in Figure 15b part of a ring. The area of this region is, to a good approximation, given by the product of its thickness ∆s and the length of the outer arc bounding it, 2π × BC:

area of slice Ω = 2π × BC × ∆s(8)

Before we can use Equation 8 to give us an integral equal to the surface area of the cone, we must express all quantities in it in terms of x. The length BC is equal to the y–coordinate of the point B, which is equal to rx/h. We now need to relate ∆s to ∆x, and this can be done using Pythagoras’s theorem.

From Figure 15c, we see that (∆s)² = (∆x)² + (∆y)² = (∆x)² (1 + (∆y)²/(∆x)²)

i.e.$\Delta s = \Delta x\sqrt{1+(\Delta y)^2/(\Delta x)^2}$

The ratio ∆y/∆x is simply equal to the gradient of the line OP, which is r/h. So we have

$\Delta s = \Delta x\sqrt{1 + \dfrac{r^2}{h^2}}$

and substituting this, and BC = rx/h in Equation 8, we find

area of the slice Ω = $2\pi~~\underbrace{\dfrac rhx}_{\color{purple}{\large{\rm BC}}}~~\underbrace{\left(\sqrt{1 + \dfrac{r^2}{h^2}}\right)\Delta x}_{\color{purple}{\large{\Delta x}}}$

We now add up all such areas, allow the thickness of the slices to tend to zero, and so obtain a definite integral for the surface area S of the cone:

$\displaystyle S = 2\pi\int_0^h\left(\dfrac rhx\sqrt{1 + \dfrac{r^2}{h^2}}\right)\,dx = 2\pi\underbrace{\dfrac rh\sqrt{1 + \dfrac{r^2}{h^2}}}_{\color{purple}{\large{\substack{\text{everything here}\\[2pt]\text{is constant}}}}}{\small\int}_0^hx\,dx$

The answer you have found does not yet look quite the same as the answer obtained in Equation 7, which was πrl. However, it is not hard to show that it is in fact the same. Figure 16 shows that the slant height l, the height h and the radius r of the cone are related by Pythagoras’s theorem: l² = h² + r², which can be rearranged to give $l = h\sqrt{ 1 + r^2/h^2}$. So the right–hand sides of Equation 9,

$S = 2\pi\dfrac rh\sqrt{1+\dfrac{r^2}{h^2}}\left[\dfrac12x^2\right]_0^h = \pi rh\sqrt{1+\dfrac{r^2}{h^2}}$(Eqn 9)

and Equation 7,

surface area S of cone = $\dfrac12\left(\dfrac{2\pi r}{l}\right)l^2 = \pi rl$(Eqn 7)

are identical.

Now that we have seen how to write the surface area of a cone as an integral, we can generalize the method to find the area of any surface of revolution that is, any surface produced by rotating a graph about an axis.

Figure 17a shows the graph of the function f(x) and its surface of revolution between the points x = a and x = b. To find the surface of revolution S, we approximate the graph by a set of line segments of equal length ∆s, as shown in Figure 17b. When one of these is rotated about the x–axis, it generates a surface Ω whose surface area is approximately equal to that of a slice of the solid of revolution. The area of the surface Ω is approximately equal to its thickness ∆s multiplied by the distance travelled by one end of the line segment as it is rotated about the x–axis, 2πf(x). So we have

area of Ω ≈ 2πf(x)∆s(10)

Figure 17c shows that in this general case, we have the approximate relation

$\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2} = \Delta x \sqrt{1+(\Delta y/\Delta x)^2}$(11)

This approximation will become better as ∆x and ∆s get smaller and smaller. We can also write

$\Delta y \approx \dfrac{dy}{dx}\Delta x$   i.e.   ∆y ≈ f ′(x)∆x(12)

and again, this approximation will improve as ∆x tends to zero. Substituting Equations 11 and 12 into Equation 10 gives

area of the thin strip Ω ≈ $\underbrace{2\pi\,f(x)}_{\color{purple}{\large\substack{\text{length of}\\[2pt]\text{the strip }\Omega}}}\underbrace{\sqrt{1+\left[\,f'(x)\right]^2}\Delta x}_{\color{purple}{\large\substack{\Delta s\\[2pt]\text{width of the strip }\Omega}}}$

and when we add up the area of all the slices, and allow ∆x to tend to zero, we obtain an integral for the surface S of revolution:

4 Totals and averages

4.1 The mass of an object of variable density

Suppose that we want to find the mass of a rod of length L and uniform cross–sectional area A, whose density ρ(x) varies with distance x from one end of the rod (see Figure 18). We proceed by dividing the rod into thin slices of thickness ∆x. The volume of one of these slices is A∆x, and since the density within this slice is approximately constant, and equal to ρ(x), the mass of the rod between x and x + ∆x is approximately ρ(x)A∆x. The total mass M of the rod is approximately given by the sum of the masses of all the thin slices. In the limit as ∆x tends to zero, this sum becomes a definite integral, exactly equal to the mass of the rod:

$M = A\Int_0^L\rho(x)\,dx$(14)

The method used to derive Equation 14 can be generalized to find the mass of some solids which are not in the shape of rods. For example, suppose we want to find the total mass of a disc, of radius R and depth h whose density ρ(r) varies with the distance r from the axis of the disc (see Figure 19a). Since the density depends only on r, and on neither the angular displacement round the disc nor the depth, we can easily write down an approximate expression for the mass of a thin ring-shaped portion of the disc (shown in Figure 19b) if we know the volume of such a portion. We therefore divide the disc into thin concentric rings of radius r and thickness ∆r, calculate the mass of one such ring, and add up the masses of all the rings; then, as usual, as we let ∆r tend to zero, the resulting integral gives us the mass of the disc.

To find the mass of the ring, note that its volume is approximately equal to the product of its inner circumference 2πr, its thickness ∆r and its depth h. So that

mass of the ring = $\overbrace{\underbrace{2\pi r}_{\color{purple}{\large\substack{\text{circumference}\\[2pt]\text{of ring}}}} \underbrace{h}_{\color{purple}{\large\substack{\text{depth}\\[2pt]\text{of ring}}}}\underbrace{\Delta r}_{\color{purple}{\large\substack{\text{thickness}\\[2pt]\text{of ring}}}}}^{\color{purple}{\large\text{volume of ring}}}\underbrace{\rho(r)}_{\color{purple}{\large\substack{\text{density}\\[2pt]\text{of ring}}}}$

and the mass M of the whole disc is given by:

$M = 2\pi h\Int_0^Rr\rho(r)\,dr$(15)

The same strategy can be used to find the mass of a sphere of radius R whose density ρ(r) depends only on the distance r from the centre of the sphere. Here, we divide the sphere into thin spherical shells of thickness ∆r, concentric with the sphere. The volume of such a shell is approximately equal to the surface area of its inner surface, 4πr², multiplied by its thickness ∆r.

4.2 Centre of mass

If we have a set of particles of masses mⁱ distributed at positions x0ⁱ along the x–axis, the position xc of the centre of mass of this set of particles is given by

$x_{\rm c} = \dfrac{{\large\sum}_im_ix_i}{{\large\sum}_im_i}$(17)

Suppose for example that we have three small spheres of lead of mass 0.1 kg, 0.15 kg and 0.2 kg, attached to a thin rod, made of aluminium, at distances 5 cm, 10 cm and 75 cm, respectively, from one end of the rod. The centre of mass is the point about which the rod will balance, i) and according to Equation 17 this point will be approximately

$\dfrac{5 \times 0.1 + 10 \times 0.15 + 75 \times 0.2}{0.1 + 0.15 + 0.2}\,{\rm cm} \approx 37.78\,{\rm cm}$

from one end. We are ignoring the mass of the aluminium rod in this calculation since it is presumed to be small in comparison to the other masses. i

If we remove the lead weights then we can no longer ignore the mass of the rod, in which case we are dealing with a mass distributed uniformly instead of a number of discrete masses. In the case of a uniform rod i, made of aluminium say, it is clear that it will balance about its centre point, but generally we may need to use integration to find the position of a centre of mass.

Suppose, for example, that we want to find the centre of mass of the non–uniform rod shown in Figure 18, of length L, cross–sectional area A and density ρ(x) (which varies along the length of the rod). We divide the rod into slices of thickness ∆x, and now we may treat it as a number of discrete masses and apply Equation 17 to obtain an approximate expression for x_c.

The mass of the slice between x and x + ∆x is approximately ρ(x)A∆x; and now we multiply the mass of this slice by its x–coordinate, sum over all slices, and divide this sum by the sum of the masses of the slices:

$x_{\rm c} = \dfrac{{\large\sum}_ix\rho(x)A\,\Delta x}{{\large\sum}_i\rho(x)A\,\Delta x}$

As ∆x tends to zero, this ratio of sums becomes a ratio of integrals, and we obtain an exact value:

The integral in the denominator here is (from Equation 14) equal to the mass M of the rod; so an alternative way to write Equation 18a is

4.3 Moment of inertia

When you try to push a heavy object, the difficulty increases with the object’s mass. On the other hand, if you try to rotate an object about an axis, i the difficulty increases with a quantity known as its moment of inertia about that axis. You may have seen pictures of someone having difficulty opening the massive doors of a bank vault; this is not usually because there is resistance in the hinges, but because the doors have a large moment of inertia about the axis of the hinges (and once you get the door started it is just as difficult to stop it). The mass of an object is simply one of its intrinsic properties; but while the moment of inertia is related to an object’s mass, it also depends crucially on the choice of axis. The moment of inertia of a telegraph pole about the axis of circular symmetry of the pole is relatively small, but its moment of inertia about an axis through one end of the pole, and perpendicular to the pole, is quite considerable.

Some objects are designed so as to have very large moments of inertia. For example a flywheel is constructed to be as massive and with as large a diameter as is convenient, with most of its mass as far from the axis as possible. We will see why such a design is sensible shortly.

Finding the moment of inertia of a number of point masses is relatively easy. When the mass is distributed throughout an object we generally need to employ integration, but not in the following case.

The following example is based on a similar idea.

The previous cases were easy because the object could be divided into a number of small masses which were (approximately) the same distance from the axis; when the distance from the axis varies we will need to employ a more sophisticated summation process, which will lead to a definite integral. Here are some examples:

Thin rods

This approach can easily be adapted to the case where the density of the thin rod is not constant, but varies along its length. Suppose that the rod in Example 7 has density ρ(x) at a distance x from one end, and constant cross–sectional area A.

Solid cylinders and discs

Equation 20,

I = MR²(Eqn 20)

gives us the moment of inertia of a hollow cylinder about its axis of circular symmetry. We can use this result to calculate the moment of inertia of a solid cylinder of uniform density about its axis of circular symmetry (see Figure 22). We simply divide the cylinder into a large number of concentric thin–walled hollow cylinders, i use Equation 20 to write down the moment of inertia of a typical one of these, add up all such moments of inertia, and so arrive at an integral.

We first need to calculate the mass of a typical cylindrical shell of thickness ∆r, and, since we are assuming that the density of the cylinder is uniform, this just means finding the volume of the shell. Its volume is approximately equal to the product of its circumference (2πr), its thickness (∆r) and its height (h); so if the cylinder has density ρ,

∆M = the mass of a cylindrical shell = $\underbrace{~2\pi r~}_{\color{purple}{\large\text{circumference}}} \overbrace{~~h~~}^{\color{purple}{\large\text{height}}} \underbrace{~\Delta r~}_{\color{purple}{\large\text{thickness}}} \overbrace{~~\rho~~}^{\color{purple}{\large\substack{\text{constant}\\[2pt]\text{density}}}}$

Its radius is r; so (replacing M by ∆M and R by r in Equation 20) we find that the moment of inertia of this hollow cylinder is

2πrhρ∆r × r² = 2πr³ρh∆r

The total moment of inertia of the cylinder is therefore approximately given by:

$\sum 2\pi r^3\rho h\Delta r$.

As ∆r tends to zero, the sum becomes an integral, giving the moment of inertia I exactly:

$I = 2\pi \rho h\Int_0^Rr^3\,dr$

We evaluate the integral, to find

$I = 2\pi \rho h\left[\frac14r^4\right]_0^R = \frac12\pi\rho hR^4$(22)

It is usually more convenient to have an expression for I in terms of the mass M of the cylinder, rather than the density ρ. We know that the volume of the cylinder is πR²h so ρ = M/πR²h. Substituting for ρ in Equation 22 gives

$I = \frac12\pi hR^4 \times \underbrace{\left(\dfrac{M}{\pi R^2h}\right)}_{\color{purple}{\large{\text{density} \rho}}} = \frac12MR^2$

Again, this result does not depend on the height of the cylinder (as in Equation 20),

I = MR²(Eqn 20)

and because of this, Equation 23 applies equally well to a thin disc as to a long cylinder.

If the density ρ(r) of the cylinder varies with distance r from its axis, the moment of inertia about the axis can also easily be written down as a definite integral. It is:

$I = 2\pi h\Int_0^Rr^3\rho(r)\,dr$(24)

where R and h are the radius and height of the cylinder.

Solids of revolution

Since the height of the cylinder does not appear in Equation 23,

$I = \frac12MR^2$(Eqn 23)

that formula applies equally well to a thin disc as to a long cylinder. This means that we may use it to find the moment of inertia, about the x–axis, of any solid of revolution. We simply approximate the solid of revolution by a set of thin discs (as we did to find its volume), calculate the moment of inertia of one such disc using Equation 23, and, in the usual way, arrive at an integral.

4.4 Function averages

In everyday language the word ‘average’ is a much abused term. Of course, we know roughly what we mean by saying ‘an average man’ or ‘an average day for the time of year’. We mean that the ‘man’ or ‘day’ is in some way a good representative of all men or days. In this subsection and the next we will discuss two forms of ‘average’ that are quite distinct, the point being that our choice of meaning for the word ‘average’ depends on the context. Our first illustration concerns average velocity.

You are probably familiar with the definition of average velocity between two times t₁ and t₂ as

$\dfrac{\text{total displacement between } t = t_1 \text{ and } t = t_2}{\text{total time } (t_2 - t_1)}$

In the case of an object travelling along the x–axis, we saw in Subsection 2.1 that if we know the velocity v_x(t) as a function of time, we can find the total displacement between two times as an integral, $\Int_{t_1}^{t_2}v_x(t)\,dt$.

So, if we introduce the notation v_av for average velocity, then

$v_{\rm {av}} = \dfrac{\Int_{t_1}^{t_2}v_x(t)\,dt}{t_2 - t_1}$ i

We can use the same method to calculate average values of other functions. For example, consider the electrical power used by a domestic appliance. The power P supplied to a particular appliance by the mains in the United Kingdom is designed so that it varies with time according to the formula

P(t) = P₀sin²(ω₀t)

where P₀ is a constant depending on the power rating of the particular appliance.

A sketch of P against t is shown in Figure 23. The power rating quoted for any domestic appliance is actually defined to be the average power consumed. We can find an expression for this average power P_av in terms of P₀ by calculating the average over just one cycle, since the symmetry of the graph means that the average over many cycles is the same as the average over one cycle.

So we will calculate the average power consumed between t = 0 and t = π/ω, which is given by the integral of P(t) between these times, divided by the time interval:

$P_{\rm av} = \dfrac{{\large\int_0^{\pi/\omega}}P(t)\,dt}{\pi/\omega - 0} = \dfrac{\omega}{\pi}{\large\int_0^{\pi/\omega}}P_0\sin^2(\omega t)\,dt$

To evaluate the integral, we use the trigonometric identity sin²(ωt) = ½ [1 − cos(2ωt)], so that

$P_{\rm av} = \dfrac{\omega P_0}{2\pi}{\large\int_0^{\pi/\omega}}\left[1-\cos(2\omega t)\right]\,dt = \dfrac{\omega P_0}{2\pi}\left[t-\dfrac{1}{2\omega}\sin(2\omega t)\right]_0^{\omega/\pi}$

and since sin2π = sin0 = 0, we finally find

$P_{\rm av} = \dfrac{\omega P_0}{2\pi} \times \dfrac{\pi}{\omega} = \dfrac{P_0}{2}$

So the average power is half the peak power P₀.

 

So far in this subsection we have taken the integration variable to be time t, so that the averages in question were time averages. However, the notion of the average value of a function can be defined quite generally: for any function f(x), the average value f_av over the interval a ≤ x ≤ b is defined as

4.5 Mean value of a distribution

The word ‘average’ is often used in a completely different sense to the one introduced in Subsection 4.4. For example, if you sat three exams and scored 81% on the first, 52% on the second and 74% on the third, you might say that your ‘average score’ on all three was $\dfrac{\text{81% + 52% + 74%}}{3} = \text{69%}$. But it would be more correct to call this your mean score. Generally speaking, the mean of N numbers is defined as the sum of the numbers, divided by N. i

There are several different ways of writing an expression for a mean value. To illustrate the point, let us suppose that we have several, say N, particles moving with different constant speeds and we want to know the mean value $\langle v\rangle$ of their speeds. We measure the speed of each one, and find that N₁ of them have speed v₁, N₂ have speed v₂, and so on up to N_n having speed v_n (so that N₁ + N₂ + ... + N_n = N, where n is the number of different speed groups). Instead of adding up the measured values one by one to obtain the mean value of all our measurements, we can multiply each of the n values obtained by the number of times it occurs, add the results, and divide by N. What we obtain is the mean_of_a_distributionmean value of the distribution:

$\displaystyle \langle v\rangle = \dfrac1N\sum_{i=1}^nN_iv_i$(27)

The fraction f_i of times that the result v_i occurs is simply N_i/N so we can rewrite Equation 27 in the form

$\displaystyle \langle v\rangle = \sum_{i=1}^n\,f_iv_i$(28)

(Notice that it follows that $\displaystyle \sum_{i=1}^n\,f_i = 1$ because N₁ + N₂ + ... + N_n = N.)

If we are interested, not in the mean speed, but in the mean value of some other physical quantity x, then this is given by an equation just like Equation 28:

$\displaystyle \langle x\rangle = \sum_{i=1}^n\,f_ix_i$(29)

where f_i is the fraction of particles possessing the value x_i of x, and i = 1, 2, 3, ... n.

Now suppose we are interested in finding the mean speed of molecules in a gas_phasegas. There are so many molecules that Equation 28, as it stands, is not going to be very useful to us; only a very small fraction of molecules will have any given speed, and we would have a very large number of speeds to sum over. Instead, what we do is divide up the range of all speeds available to the molecules into small intervals ∆v, and concentrate, not on the fraction f_i of molecules with a particular speed v_i but on the fraction of molecules with speeds lying within the small interval between v and v + ∆v. We expect this fraction to be proportional to the size ∆v of the interval, and we write it as f(v)∆v. The function f(v) is known as a speed distributiondistribution function: it tells us what fraction of molecules have speed close to the value v.

We can then write Equation 28 as

$ \langle v\rangle = {\large\sum}v\underbrace{~\,f\,\Delta v~}_{\color{purple}{\large\substack{\text{fraction with}\\[2pt]\text{speed between }\\[1pt]v \text{ and } v+∆v}}}$

and, as we have done so often before in this module, we allow ∆v to tend to zero, so that the sum becomes an integral, and since there will be an upper bound, V say, to the speed of the molecules, we can write

$ \langle v\rangle = \Int_0^Vv\,f(v)\,dv$(30) i

An integral of this sort is far easier to perform than a sum over many different speeds, particularly when we know the specific form for the function f(v).

Remember that f(v)∆v represents the fraction of molecules with speeds lying between v and v + ∆v, and all these fractions must sum to 1, in other words

$\Int_0^Vf(v)\,dv = 1$

The following function provides an appropriate mathematical model for f(v) molecules in a gas, and is known as the Maxwell–Boltzmann speed distribution:

$f(v) = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{3/2}v^2\exp(-mv^2/2kT)$(31) i

where m is the mass of one of the molecules, T is the temperature of the gas, and k is Boltzmann’s constant. This speed distribution is sketched in Figure 25.

In general, a distribution is a function which tells us what fraction of particles have values of a certain physical quantity lying in a particular small range: if we call the physical quantity X, then the distribution f(x) is defined so that

$f(x)\,∆x = \cases{\color{purple}{\text{fraction of particles with values of }X\\\text{in the interval between }x \text{ and }x + ∆x}\vphantom{p}}$(32)

The mean value x of X is then given by Equation 29,

$\displaystyle \langle x\rangle = \sum_{i=1}^n\,f_ix_i$(Eqn 29)

to be ${\large\sum}x\,f(x)\,\Delta x$, and, on allowing ∆x to tend to zero, the sum becomes an integral:

(where the limits of integration depend on the range of values that it is possible for x to take).

Mean values also arise in quantum mechanics. Consider the case of a particle that can move along the x–axis. Unlike classical mechanics, quantum mechanics does not allow us to calculate the position x(t) of this particle as a function of time. Instead, all it can tell us is the probability P(x)∆x of finding a particle on a small interval of the x–axis, between x and x + ∆x, at a given time. The function P(x) is known as a probability density.

Such functions must also have the property that $\Int_a^bP(x)\,dx =1$, since the probabilities must sum to 1.

If we know the probability density, we can use it to find the mean value which is defined as $\langle x\rangle$ of the position of the particle,

$\langle x\rangle = \Int_a^bxP(x)\,dx$(34)

where the limits of integration a and b depend on the region of the x–axis over which the particle is free to move.

5 Closing items

5.1 Module summary

1

The Subsection 2.1area under the graph of f(x) between x = a and x = b is defined as being equal to the definite integral

$\Int_a^bf(x)\,dx$. This definition has the consequence that in any region where f(x) is always negative, so is the area under its graph. Sometimes we may be interested in the total size of the area enclosed by the graph of f(x) and the x–axis, between x = a and x = b (where a < b); this is given by the integral $\Int_a^b\lvert\,f(x)\rvert\,dx$

2

2 If the graphs of two functions f(x) and g(x) intersect at the points x = a and x = b (where a < b), then the Subsection 2.2area between the graphs is defined as

$\Int_a^b\lvert\,f(x) - g(x)\rvert\,dx$

This includes the case where other points of intersection lie between x = a and x = b.

3

A Section 3solid of revolution is a solid which can be obtained by rotating the area under a graph (or part of a graph) about some axis. If the solid is obtained by rotating the area under the graph of f(x) between x = a and x = b about the x–axis, then its volume (known as a volume of revolution) is given by the definite integral $\pi\Int_a^b\left[\,f(x)\right]^2\,dx$, and the area of its surface (known as a Subsection 3.2surface of revolution) is given by the definite integral

$S = 2\pi{\Large\int}_a^bf(x)\sqrt{1+\left[\,f'(x)\right]^2}\,dx$(13)

4

Integration may be used to find the mass M of a solid whose density ρ varies with position. In the case of a rod of constant cross–sectional area A and length L, lying along the x–axis between 0 and L, $M = A\Int_0^L\rho(x)\,dx$. For a disc of height h and radius R, whose density ρ(r) depends only on distance r from the axis, $M = 2\pi h\Int_0^Rr\rho(r)\,dr$. For a sphere of radius R, whose density ρ(r) depends only on distance r from the centre, $M = 4\pi\Int_0^Rr^2\rho(r)\,dr$.

5

The position of the Subsection 4.2centre of mass of a solid object may also be found by integration; for example, for a rod of constant cross–sectional area A and length L, lying along the x–axis between 0 and L, the x–coordinate of the Subsection 4.2centre of mass is $x_{\rm c} = M \Int_0^Lx\rho(x)\,dx$.

6

Subsection 4.3Moments of inertia of thin rods, solid cylinders, and solids of revolution can also be written as definite integrals.

The integral giving the Subsection 4.3moment of inertia I about the x–axis of the solid of revolution is

$I = \frac12\pi\rho\Int_a^b\left[\,f(x)\right]^4\,dx$(Eqn 25)

7

The Subsection 4.4average value of a function f(x) over the interval a ≤ x ≤ b is defined as

$f_{\rm av} = \dfrac{\Int_a^bf(x)\,dx}{b-a}$(Eqn 26)

8

A Subsection 4.5distribution is a function specifying the fraction of particles which have values of a certain physical quantity lying in a particular small range between x and x + ∆x; the distribution f(x) is defined so that this fraction is equal to f(x) ∆x where $\Int f(x)\,dx = 1$ and the mean of the distribution, $\langle x\rangle$, is given by a definite integral: $\langle x\rangle = \Int x\,f(x)\,dx$, where a and b are the largest and smallest allowed values of x, respectively.

5.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Interpret the definite integral of a given function between given limits in terms of the area under the graph of that function between those limits; and (paying due attention to signs) calculate the magnitude of the total area enclosed by the graph of the function and the horizontal axis.

A3

Calculate the area enclosed between two intersecting graphs.

A4

Write down definite integrals giving the volume and surface area of a solid of revolution.

A5

Write down definite integrals giving the mass of a rod, disc or sphere whose density is not uniform.

A6

Express the position of the centre of mass of a solid as a definite integral.

A7

Find the moments of inertia of thin rods, cylinders, and solids of revolution as definite integrals.

A8

Calculate the average value of a given function over a given interval.

A9

Find the mean value of a distribution.

A10

Use the idea of a definite integral as the limit of a sum in order to set up an integral that gives the value of some required physical quantity.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

5.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.

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