Chapter 7

March 18 2017

Chapter 7 Applications of Integrals 7.1 The Intuitive Meaning of Integration - in chapter 6 we accomplished two major purposes - first, we approximated the area under a given curve by certain sums and found the exact area by forming the limit of these sums - second, we learned how to calculate the numerical value of this limit by using the much more powerful method provided by the Fundamental Theorem of Calculus - almost the whole content of chapter 6 can be compressed into the following statement - if f(x) is continuous on [a, b], then lim max dxk->0 E1n f(dxk) xk = Sab f(x) dx = F(x) ]ab = F(b) - F(a) - where F(x) is any indefinite integral of f(x) - there are many other quantities in geometry and physics that can be treated in essentially the same way - among these are volumes, arc lengths, surface areas, and such basic physical quantities as the work done by a variable force acting over a given distance - in each case the process is the same - an interval of the independent variable is divided into small subintervals, the quantity in question is approximated by certain corresponding sums - and the limit of these sums yields the exact value of the quantity in the form of a definite integral - which is then evaluated by means of the Fundamental Theorem - once we have seen the details of this limit of sums being carried out for the area under a curve - as was done in chapter 6, it is unnecessary and boring to think through these details over and over again for each new quantity that we meet - the notation needed for this is complicated and repetitive - and actually impedes the intuitive understanding that we wish to cultivate - in this spirit, we consider the easy, intuitive way of constructing the definite integral - we think of the area under the curve as composed of a great many think vertical rectangular strips - the typical strip has height y and width dx - therefore it has the area dA = y dx = f(x) dx - this area is called the differential element of area, or simply the element of area - it is located at an arbitrary position within the region - this position is specified by a value of x between a and b - we now think of the total area A of the region as the result of adding up these elements of area dA as our typical strip sweeps across the region - this act of addition or summation can be symbolized by writing - A = S dA - since the element of areas sweeps across the region as x increases from a to b - we can express the idea with greater precision by writing - A = S dA = S y dx = Sab f(x) dx - we reach a true definite integral only in the last step - where the variable of integration and the limits of integration become visibly present - in this way, we glide smoothly over the many details - and setup the definite integral for the area directly, without having to think about limits of sums at all - from this point of view - integration is the act of calculating the whole of a quantity by cutting it up into a great many convenient small pieces and adding up these pieces - it is this intuitive Leibnizian approach to the process of integration that we intend to illustrate and reinforce in the following sections 7.2 The Area Between Two Curves - suppose we are given two curves y = f(x) and y = g(x) - with points of intersection at x = a and x = b - with the first curve lying above the second on [a, b] - in setting up an integral for the area between these curves, it is natural to use thin vertical strips - the height of such a strip is the distance f(x) - g(x) - the element of area is therefore - dA = [f(x) - g(x)] dx - and the total area is - A = S dA = Sab [f(x) - g(x)] dx - we integrate from the smaller limit of integration a to the larger b so that the increment (or differential) dx will be positive - it should be pointed out that a and b are values of x for which the two functions yield the same y's - that is, they are the solutions to the equation f(x) = g(x) - we are given an outline of the steps that should be followed in finding an area by integration - step 1 sketch the region whose area is to be found - write down on the sketch the equations of the bounding curves and find their points of intersection - step 2 decide whether to use thin vertical strips that have width dx or thin horizontal strips that have width dy - draw a typical strip on the sketch - step 3 by looking at the sketch and using the equations of the bounding curves - write down the area dA of the typical strip as the product of its length and its width - express dA entirely in terms of the variable (x or y) appearing in the width - step 4 integrate dA between appropriate x or y limits - these limits are found by examining the sketch - use symmetry whenever possible, in order to simplify the calculation 7.3 Volumes: The Disk Method - if the region under a curve y = f(x) between x = a and x = b is revolved about the x-axis - it generates a three-dimensional figure called a solid of revolution - the symmetrical shape of this solid makes its volume easy to compute - when the region is revolved about the x-axis - the typical thin vertical strip of thickness dx generates a thin circular disk shaped like a coin - with radius y = f(x) and thickness dx - the volume of this disk is our element of volume dV - since the disk is a cylinder, its volume is clearly the area of the circular face times the thickness - dV = pi y2 dx = pi f(x)2 dx - we now imagine that the solid of revolution is filled with a very large number of thin disks like this - so the total volume is the sum of all the elements of volume as our typical disk sweeps through the solid from left to right - that is, as x increases from a to b - V = S dv = S pi f(x)2 dx = Sab pi f(x)2 dx - similarly we can confirm the volume formulas of other basic solids - for obvious reasons, the method of these examples is usually called the disk method - the same idea can be applied to solids of other types - in which the element of volume is not necessarily a circular disk - a minor variation of the disk method is often useful - suppose the strip being revolved about an axis is separated from the axis of rotation by a certain distance - in this case the element of volume generated by the strip is a disk with a hole in it - what might be described as a washer - the volume of this washer is the volume of the disk minus the volume of the hole 7.4 Volumes: The Shell Method - there is another method of finding volumes which is often more convenient - the disk method rotated the typical strip in the region under a curve about the x-axis - however, if the region is rotated about the y-axis - then we get an entirely different solid of revolution and the vertical strip generates a thin-walled cylindrical shell - this shell can be thought of as resembling a soup can whose top and bottom have been removed - or perhaps a thin-walled cardboard mailing tube - its volume dV is essentially the area of the inner surface (2pixy) times the thickness of the wall (dx) - dV = 2pixy dx - as the radius x of this shell increases from x = 0 to x = b - we can see that the resulting series of concentric shells fills the solid of revolution from the axis outward - in much the same way as the concentric layers of an onion fill the onion from the centre outward - the total volume of this solid is therefore the sum, or integral, of the elements of volume dV - V = S dV = S 2pixy dx = S0b 2pix f(x) dx - since y = f(x) - in principle, this volume V can also be found by using horizontal disks generated by thin horizontal strips - however, this could turn out to be difficult - since the given equation y = f(x) would have to be solved for x in terms of y - just as in our other applications of integration - these formulas give brief expression to a complex process involving limits of sums - and as usual, we omit the details of the process in the interests of clarity 7.5 Arc Length - an arc is the part of a curve that lies between two specific points A and B - physically, the length of an arc is a very simple concept - mathematically, it is somewhat more complicated - from a physical point of view, we merely bend a piece of string to fit the curve from A to B - mark the points corresponding to A and B - straighten out the string, and measure its length with a ruler - this process can be carried out by means of an approximation procedure that lends itself to mathematical treatment - divide the arc AB into n parts by using points P0 = A, P1 ... Pn = B - place pins at these points - and let the string stretch in short straight-line paths from each pin to the next - if we take large and larger values of n - and at the same time require that the pins be placed closer and closer together - then the length of the string should approximate the length of the arc - we now express this idea in mathematical language and derive a practical method of calculating arc length by integration - let us assume that the arc under discussion is the graph of a continuous function y = f(x) for a <= x <= b - we partition the interval [a, b] into n subintervals by using points x0 = a ... xn = b - we let Pk be the point (xk, yk) where yk = f(xk) - the total length of the polygonal path P0 ... Pn is the sum of the lengths of the chords joining each point to the next - if we write dxk = xk - xk-1 and dyk = yk - yk-1 for k = 1 ... n - then it is clear that by the Pythagorean theorem we have - length of kth chord = sqrt dxk2 + dyk2 = sqrt 1 + (dyk/dxk)2 dxk - we now assume that y = f(x) is not only continuous but also differentiable - this permits us to replace the ratio inside the radical - which is the slope of the chord joining Pk-1 to Pk - by the value of the derivative at some point xsk between xk-1 and xk - dyk/dxk = f'(xsk) where xk-1 < xsk < xk - the justification for this step lies in the fact that the chord is parallel to the tangent at some point on the curve between Pk-1 and Pk (mean value theorem) - so the length of the kth chord is sqrt 1 + [f'(xsk)]2 dxk - so the total length of the polygonal path is - Ek1n sqrt 1 + [f'(xsk)]2 dxk - we now obtain our conclusion by forming the limit of these sums as n approaches infinity and the length of the longest subinterval approaches zero - length of arc AB = lim maxdxk > 0 Ek1n sqrt 1 + [f'(xsk)]2 dxk = Sab sqrt 1 + [f'(x)]2 dx - at first, this formula may appear to be rather hard to keep in mind - however, if we use the Leibniz notation dy/dx instead of f'(x) - then the following intuitive approach makes this formula much easier to grasp and remember - let the letter s denote the variable arc length from A to a variable point on the curve - let s be allowed to increase by a small amount ds - so that ds is the differential element of arch length - let dx and dy be the corresponding changes in x and y - we think of dds as so small that this part of the curve is virtually straight - so ds is the hypotenuse of a tiny right triangle called the differential triangle - for this triangle the Pythagorean theorem says that ds2 = dx2 + dy2 - and this simple equation is the source of all wisdom in the calculation of arc lengths - if we solve for ds, then factor dx and rewrite it from the radical - we clearly get - ds = sqrt 1 + (dy/dx)2 dx - we now touch again on the basic theme of this chapter and point out that the total length of the arc AB can be thought of as the sum, or integral, of all the elements of arc ds as ds sweeps along the curve from A to B - so the length of arc AB = S ds = Sab sqrt 1 + (dy/dx)2 dx - this formula tells us that x is the variable of integration and that y is to be treated as a function of x - however, it is sometimes more convenient to treat x as a function of y - most mathematicians do not memorize these functions as they stand - instead, they start with ds2 = dx2 + dy2 - and mentally perform as needed the simple manipulations to derive the equations - this way, the whole package of ideas is almost impossible to forget - the example arc length calculations should serve as a warning - for if we try to find the length of an arc on almost any familiar curve - then the resulting integral will probably be impossible for us to work out - at this stage we must choose our problems very carefully in order for the integrals to be computable - we should also be aware of our urgent need for more integration techniques - filling this need is the main purpose of the next three chapters 7.6 The Area of a Surface of Revolution - let us consider a smooth curve lying above the x-axis - when this curve is revolved about the x-axis - it generates a surface of revolution - we now set ourselves the problem of calculating the area of such a surface - for reasons that will become clear - we begin by considering a very simple surface of revolution - the curved lateral part of a cone whose base has radius r and whose slant height is L - if this cone is cut down the side from the vertex to the base, that is, along a generator, and laid out flat - then we get a sector of a circle of radius L whose curved edge has length 2pir - and the lateral area A of the cone equals the area of this sector - it is geometrically clear that the ratio of the area of the sector to the complete area of the circle equals the ratio of the curved edge to the complete circumference of the circle - A / pi L2 = 2pir / 2piL, A = pi r L - the lateral surface of the cone can evidently be thought of as the surface of revolution swept out by a generator as it revolves around the axis - if the formula is written as A = L 2 pi (1/2 r) - then we see that the lateral area of a cone equals the product of the length of a generator and the distance travelled by the midpoint on its journey around the axis - next, we generalize slightly and find the area of the surface of revolution generated when a line segment of length L is revolved about an axis at a distance r from its midpoint - this area is the lateral area of a frustum of a cone - if we denote this area by A, then A is the difference between the lateral areas of the two cones - so A = pi r1 L1 - pi r2 L2 = pi (r1 L1 - r2 L2) - by similar triangles, it is clear that L2 / r2 = L1 / r1 or r1 L2 = r2 L1 - this enables us to write A in the form - A = pi (r1 L1 - r1 L2 + r2 L1 - r2 L2) = pi[ r1 (L1 - L2) + r2 (L1 - L2) ] = pi (L1 - L2) (r1 + r2) = (L1 - L2) 2 pi (r1 + r2 / 2) = L 2 pi r - we therefore conclude that in tihs case as well - the area of the surface of revolution equals the product of the length of the segment and the distance travelled by the midpoint on its journey around the axis - we now apply these ideas to the general area problem stated at the beginning of this section - our approach will be intuitive and geometric - we begin by approximating the smooth curve y = f(x) by a polygonal path consisting of many short line segments connecting nearby points on the curve - the surface generated by revolving the curve about the x-axis will have approximately the same area as the surface generated by revolving this polygonal path about the x-axis - the latter surface is evidently made up of a number of pieces - each of which is shaped like a frustum of a cone - if the element of arc length ds is revolved about the x-axis - then it generates a thin ribbon-like element of area dA - and if the midpoint of ds is at a distance y from the x-axis, then the above discussion tells us that - dA = 2 pi y ds = 2 pi y sqrt 1 + (dy/dx)2 dx - we obtain the total area A of the surface by forming the sum, or integral, of all the elements of area dA as dA sweeps along the complete surface - A = S dA = S 2 pi y ds = Sab 2 pi y sqrt 1 + (dy/dx)2 dx - where y is assumed to be known as a function of x, y = f(x) - if we choose instead to revolve our curve about the y-axis - and thereby to generate an entirely different surface of revolution - then in the same way its area is given by - A = S 2 pi x ds - the underlying idea in both of these formulas can be expressed by writing - A = S 2 pi (radius of revolution) ds - in using this formula to perform an actual calculation - the element of arc length ds must be written in terms of a convenient variable of integration and appropriate limits of integration must be provided 7.7 Hydrostatic Force - in the previous sections of the chapter - we have seen how integration can be used to answer many natural questions that arise in geometry - in the next two sections we consider several applications to physics and engineering - applications of this type usually require, in addition to the pertinent mathematical knowledge - a reasonable grasp of the basic scientific principles involved - in the problems we study, these scientific principles are very simple - so simple, indeed, that students with no previous experience in science should be able to understand them easily - the main theme of our work continues to be the idea that the whole of a quantity can be calculated by dividing it into many convenient small pieces and adding up these pieces by means of integration - a flat plate of unspecified shape is submerged vertically in a body of water - to find the total force exerted by the water against one face of this plate - we imagine this face to be divided into a large number of horizontal strips - the typical strip is at a depth h below the surface - its width dh is so small compared with h that the pressure is essentially constant over the entire strip - and has the value p = wh - the area of the strip is dA = x dh - so the element of force dF acting against the strip is given by - dF = p dA = w h x dh - the total force F acting against the whole face of the plate is now obtained by integrating these elements of force as our typical strip sweeps across the plate from top to bottom - F = S dF = Sab w h x dh 7.8 Work and Energy - examples in physics