3 x x+1. (4) One might be tempted to do what is done with fractions, i.e., combine them by means of a common denominator.

Transcription

1 Lecture 6 Techniques of integration (cont d) Integration of rational functions by partial fractions Relevant section from Stewart, Eighth Edition: 7.4 In this section, we consider the integration of rational functions, i.e., ratios of polynomials, f() = P() Q() = p m m +p m m + +p +p q n n +q n n + +q +q. () We ll let deg(p) = m and deg(q) = n denote the degrees of P and Q, respectively. The main idea is to epress the above rational function as a sum of a polynomial plus partial fractions. Eample : Suppose that we re given the following sum of two rational functions, (2) One might be tempted to do what is done with fractions, i.e., combine them by means of a common denominator. In this case, = 3(+)+( 2) ( 2)(+) = ( 2)( ) 4+ = 2 2. (3) In reverse, we have that = (4) The right side of the above equation is called the partial fraction decomposition of the rational function on the left side. As such, if we re faced with the problem of integrating the rational function on the left side, we can 59

2 use its partial fraction decomposition, since the components of the latter are easy to integrate, i.e., [ d = 2 + ] d + 3 = 2 d+ + d = 3ln 2 +ln + +C. (5) We now consider the general case of integrating a rational function of the form, f() = P() Q() = p m m +p m m + +p +p q n n +q n n + +q +q. (6) Case No. : If deg(p) < deg(q), then the rational function P()/Q() admits a partial fraction decomposition, which can then be integrated term by term. Case No. 2: If deg(p) deg(q), then the rational function P()/Q() may be epressed as follows, P() R() = S()+ Q() Q(), (7) where S() is a polynomial of degree m n and deg(r) < deg(q). The integration of the polynomial S() is straightforward. The rational function R()/Q() then falls under Case No. and therefore may be decomposed into partial fractions which can then be integrated. Eample 2: Consider the rational function, f() = (8) We can perform formal division as follows (it is difficult to typeset the usual long division scheme, so we ll present it in steps: = = = The result is that we have epressed the above rational function as a sum of a polynomial term plus a (proper) fraction, = (9) 6

3 As a result, the rational function may be easily integrated, [ +2 d = ] d = ln +C. () A more general discussion of partial fraction decompositions In what follows, we consider the problem of decomposing a rational function, where f() = R() Q() As we know from fractions, it is the denominator Q(), = r m m +r m m + +r +r q n n +q n n + +q +q, () deg(r) < deg(q), i.e., m < n. (2) Q() = q n n +q n n + +q +q, (3) which determines the decomposition. From the Fundamental Theorem of Algebra, Q() has n roots, i.e., i, i n, such that Q( i ) =, i n. (4) Even though the coefficients q i of Q() are real, it is possible that some roots are comple-valued. We ll eventually have to consider that case. Simplest case: Q() has n distinct, real roots This implies that Q() can be factored as follows, so that the roots are given by Q() = (a +b )(a 2 +b 2 ) (a n n +b n ), (5) i = b i a i. (6) In this case, the rational function R()/Q() can be epressed as a sum of partial fractions with linear denominators, i.e., R() Q() = A + A 2 A n + +. (7) a +b a 2 +b 2 a n +b n 6

4 Each of the terms on the right can easily be integrated. To illustrate, let us recall Eample from the beginning of this section: Here, with distinct, real roots = 2 and 2 = = (8) Q() = 2 2 = ( 2)(+), (9) The question now remains, Given a polynomial Q() with distinct real roots, how do we find the coefficients A i, a i and b i in the partial fraction epansion in Eq. (7)? The answer is, By using some basic algebra, as we show in the eample below. Eample 3: Consider the rational function, We may easily factor the denominator polynomial, f() = R() Q() = (2) Q() = 2 4 = (+2)( 2). (2) Since the degree of R() is less than the degree of Q(), we can assume that f() will admit an epansion of the form, = A +2 + B 2. (22) We must now solve for A and B. The first step is to basically undo the epansion on the right side, i.e., combine the two fractions by using the common denominator Q(), A = +2 + B 2 = A( 2)+B(+2) (+2)( 2) = (A+B)+( 2A+2B) (+2)( 2). (23) For this equation to be true for all, ecept = 2 and = 2, the numerators of both sides must be equal, i.e., + = (A+B)+( 2A+2B). (24) 62

5 But for this equation to be true, the coefficients of and = must be equal, which leads to a linear system of equations in the unknowns A and B, A+B = 2A+2B =. (25) There are many, many ways to solve such a system. For eample, we could use one equation to epress one unknown, say A, in terms of the other, i.e., B, then substitute that result into the second equation to solve for B. Another way is to multiply one equation, or both, by suitable numbers which will permit an elimination of one variable. For eample, if we multiply the first equation by 2, 2A+2B = 2 (26) then adding both sides of this equation to corresponding sides of the second equation yields 4B = 3, which implies that B = 3 4. From the first equation, we then have that A = 4. In any case, we have the result, = (27) We may now integrate both sides of this equation to obtain, = [ ] d 2 = 4 ln ln 2 +C. (28) 4 Net simplest case: Q() has n real roots but some are repeated Let s consider the simplest of these cases: Suppose that the first root of Q() is repeated, i.e., the first two roots are equal, and the rest of the roots are distinct. Then Q() can be factored as follows, Q() = (a +b ) 2 (a 3 +b 3 ) (a n +b n ). (29) Associated with each of the distinct roots, 3 to n will be a simple partial fraction of the form A i a i +b i, 3 i n. (3) But what about the situation regarding the first root? It turns out that neither a fraction of the form, A a +b, (3) 63

6 nor a fraction of the form, A (a +b ) 2, (32) will, in general, suffice. What is generally needed is a combination of these two terms, i.e., It is best to illustrate with an eample. A A 2 + a +b (a +b ) 2. (33) Eample 4: The rational function, Here, ( ) 2. (34) Q() = ( ) 2, (35) with roots = and 2 = 3 =. We shall have to assume an epansion of the form, ( ) 2 = A + B + C ( ) 2, (36) where A, B and C are to be determined. We proceed in a straightforward reverse manner, i.e., putting everything over a common denominator, to find these unknowns: ( ) 2 = A + B + C ( ) 2 = A( )2 +B( )+C ( ) 2 = A(2 2+)+B( 2 )+C ( ) 2 = (A+B)2 +( 2A B +C)+A ( ) 2. (37) Equating the numerators on both sides of the equation, which implies equating the respective coefficients of 2, and =, we obtain the following linear system in the unknowns, A+B = 2A B +C = 3 A = 4. (38) From the third equation and the first, we find that B = 3 and then from the second that C = 8. So the net result is that ( ) 2 = ( ) 2. (39) 64

7 This result should show that both of the last two terms are needed in the decomposition. We may now easily integrate both sides of the equation to obtain ( ) 2 d = [ 4 3 ] + 8 ( ) 2 d = 4ln 3ln A more complicated case: Q() has comple roots 8 +C. (4) ( ) The first important point is that if Q() is a polynomial in, and i is a comple root of Q(), then its comple conjugate i is also a root. In other words, the roots of a polynomial with comple coefficients occur in comple conjugate pairs. This will simplify the procedure. But its perhaps best to consider a particular eample: Eample 5: The rational function, f() = (4) We claim that nothing further can be done with this rational function if we wish to work with real numbers. The roots of the polynomial, Q() = , (42) are = 2± where i 2 =. As such, Q() could be factored as = ±2i, (43) Q() = (+ 2i)(++2i), (44) so that the partial fraction decomposition of the original rational function would have the form, = A + 2i + B ++2i. (45) The problem is that A and B will be comple-valued. In principle, this can be done, but it makes matters quite complicated. The solution is to work only with real numbers and acknowledge that the rational function f() in (4) is in its simplest form. 65

8 The integration of rational functions with quadratic denominators, such as f() in (4) is perhaps not trivial, but it is quite doable. In this case, we ll perform the following complete the square operation on Q(), Q() = = ( 2 +2+)+4 = (+) (46) We ll then employ this result in the integral of f() as follows, We ll now use the result, Then d = (+) 2 +4 d = u 2 du, where u = + = du = d. (47) +4 Resubstituing for u = + yields the result, 2 +a 2 d = a Tan ( ) +C. (48) a u 2 +4 = 2 Tan ( u ) +C. (49) d = 2 Tan ( ) +. (5) 2 Let s check this result: Correct! [ ( )] d + Tan d 2 = [ 2 + ( + 2 = = = [ 4 + (+)2 4 ) 2 ] 2 ] 4+(+) (5) In general, suppose that we have the rational function, f() = R() Q(), (52) 66

9 where deg(r) < deg(q). Furthermore suppose that the nth degree polynomial Q() has n 2 distinct real roots and one pair of comple conjugate roots 2 =. Then the appropriate partial fraction decomposition of f() has the form, R() Q() = A +B a 2 + A 3 A n + +, (53) +b +c a 3 +b 3 a n +b n where the comple conjugate roots 2 = are roots of the quadratic polynomial, a 2 +b +c. (54) If the polynomial Q() has more than one set of comple conjugate roots, but they are all distinct, then the partial fraction decomposition of f() will have to a term of the form, A+B a 2 +b+c, (55) for each comple conjugate pair See Case III in Stewart s tet, Section 7.4, Page 497. Q() contains a repeated set of comple conjugate roots. See Case IV in Stewart s tet, Section 7.4, Page499. Foreample, ifthecompleconjugateroots 2 = appeartwice, i.e., 4 = 2, 3 =, and all other roots are real and distinct, then Q() has the form, Q() = (a 2 +b +c ) 2. (56) In this case, its partial fraction decomposition will have the form, R() Q() = A +B a 2 +b +c + Wow! That s complicated! A 2 +B 2 (a 2 +b +c ) 2 + A 5 a 5 +b A n a n +b n. (57) 67

10 Appendi: Material covered in the Monday, January 6 tutorial In last Friday s lecture (Lecture 4), in the section on integrals which contain the term a 2 2, we obtained the following result quite easily by means of the simple substitution method, a 2 2 d = a 2 2 +C, (58) Here, we show that the method of trigonometic substitution can also be used, as one might epect. As discussed in the lecture, the appropriate substitution is = asinθ = d = acosθdθ. (59) Notingthat wemustalsowritetheinthedenominator intermsof θ, substitutionyields thefollowing, asinθ a 2 d = 2 acosθ (acosθ)dθ = asinθdθ = cosθ+c a = a 2 2 +C a = a 2 2 +C, (6) in agreement with our earlier result. The method of completing the square In the section of the course dealing with trigonometric substitution to date, we have been concerned with integrals which contain terms of the form a 2 2, a and 2 a 2. The epresssions in the square roots contain only two terms: a constant and a quadratic in. What happens if a linear term in also appears in the argument, e.g. a 2 +b 2? Let s consider an eample. Eample: The integral, d. (6) We can t use the usual substitution method at least in a simple form since the numerator is not the derivative, or a multiple of the derivative, of the term in square brackets. The trick here is to 68

11 complete the square in the argument, , of the square root: The original integral becomes, Now use the following substitution, = 25 ( 2 +2) = The integral on the right then becomes, u du = 26 u 2 = 25 ( 2 +2+)+ = 26 (+) 2. (62) d. (63) 26 (+) 2 u = + = = u and du = d. (64) u du 26 u 2 du. (65) 26 u 2 The first integral on the right side can be handled with the simple method of substitution to give u du = 26 u 2 +C. (66) 26 u 2 The second integral has the form a 2 u du = Sin ( u ) +C. (67) 2 a (This was Eample 2 in Friday s lecture.) Here, a = 26. Combining these results, ( ) u u du = 26 u 2 Sin. (68) 26 u 2 26 We finally resubstitute for, i.e, u = + to yield the final result, ( + d = 26 (+) 2 Sin )+C Let s check this result by differentiation: ( + = Sin )+C. (69) 26 [ d ( )] d 2 Sin 26 Correct! = = = ( ) 2 ( 2 2) (+) (7)

12 Techniques of integration - final comments In the past few lectures we have discussed a number of important methods of integration. In many integrals encountered in applications, the straightforward use of one or perhaps two of these methods will often be sufficient to obtain an indefinite integral or to compute a definite integral. That being said, there are many integrals which are much harder to handle. Sometimes, one will have to eperiment with a number of methods or combination of methods before the problem is solved if, indeed, the problem can be solved (more on this below). It is beyond the scope of this course to eamine these more difficult cases in detail. That being said, it is recommended that you read Section 7.3 of Stewart s tetbook entitled, Strategy in Integration. Finally, not all functions can be antidifferentiated in closed form, i.e., their antiderivatives can be epressed in terms of standard functions. An eample is the function f() = e 2. If such functions occur in definite integrals, e.g., e 2 d, (7) then it may be sufficient to produce good numerical approimations. Later in ther course, there may be some time to eamine a few methods of producing numerical approimations to definite integrals. Final integration eample for this section: The area of an ellipsoidal region Before we begin a new section, namely, the use of integrals to compute volumes of regions in R 3, we shall compute one final definite integral the area of the region in the plane enclosed by the elliptical curve given by 2 a 2 + y2 =. (72) b2 This may be the first time that you are seeing the equation of an ellipse. In the special case that a = b = R, the above equation becomes, after multiplication by R 2, 2 +y 2 = R 2, (73) i.e., the equation of a circle of radius R and with center at (,). In the case that a b, an ellipse may be considered as a kind of squashed circle. From Eq. (72), the -intercepts of the ellipse, found by setting y =, are = a and = a. Similarily, the 7

13 y-intercepts are y = b and y = b. A generic case for which a > b is sketched in the figure below. y b y = b 2 a 2 a a y = b 2 a 2 b Ellipse To find the area enclosed by the ellipse, we may consider it as being formed by two curves which are defined for [ a,a]: y = f() = b y = g() = b 2 a 2 the upper curve 2 a 2 the lower curve. (74) As you will recall from MATH 37, the area enclosed by these two curves is A = = a a a = 2b [f() g()]d a a 2b a 2 a 2 d 2 d. (75) a2 The final integral may look a little strange since it doesn t have the form a 2 2. We could epress the integrand as 2 a 2 = a a 2 2, (76) in which case we would immediately make the change of variable = asintheta. But we can make this change of variable in the last integral in (75): = asinθ = d = acosθdθ. (77) The square root term becomes 2 a 2 = sin 2 θ = cosθ. (78) 7

14 We also need to change the limits of integration over to limits of integration over θ. The limit = a corresponds to θ = π/2. And the limit = a corresponds to theta = π/2. Substituting all of these results into (75) gives a A = 2b = 2ab = 2ab = 2ab = 2ab a π/2 π/2 π/2 π/2 π/2 π/2 2 a 2 d cos 2 θdθ [ 2 + ] 2 cos2θ dθ [ θ 2 + ] 4 sin2θ dθ ( π )] 4 + [( π 4 + ) (double angle formula) = 2ab π 2 = πab. (79) As a check: In the special case, a = b = R, the circle of radius R, the area is A = πr 2. 72

15 Using integrals to compute volumes of regions in R 3 Relevant section from Stewart: 6.2 The lecture followed very closely the treatment in the tetbook by Stewart in Section 6.2, so all details and figures won t be presented here you can simply study Section 6.2. In all cases, we are concerned with the problem of finding the volume V of a solid S R 3. Some very simple special cases are:. Rectangular bo with dimensions l, w and h. V = lwh. 2. (Right) circular cylinder with base radius r and height h. V = πr 2 h. 3. A cylindrical-like shape with base area A and height h. V = Ah. 4. A sphere of radius r. V = 4 3 πr3. (You most probably know this result but may never have seen a derivation. We ll derive the result very shortly.) Of course, what we ll be doing in this course is finding the volumes of more complicated solids using, yes, the Spirit of Calculus. Volume of a solid obtained by slicing Here, we assume that the solid S has a principal ais, as is shown in Figure 2 of Section 6.2 of Stewart. An eample would be a long loaf of bread, or a French baguette. We ll assume, as is done in Figure 2, that the ais is parallel to the -ais. Furthermore, we assume that it is completely contained in the 3D region a b. We shall also assume that for each [a,b], the cross-sectional area A( ) of solid S is known. This is the area of the region of intersection of the solid S with the plane, P = {(,y,z) R 3 = }. (8) The plane P is perpendicular to the -ais (and parallel to the yz-plane). Thenetstepis, asalways, toconstructapartitionofthe-ais: Foragivenn >, let = b a n and define, as before, the partition points, k = a+k, k n, (8) 73

16 so that = a and n = b. Then define the planes P k which are defined by P k = {(,y,z) R 3 = k } k n. (82) The intersection of these planes with the solid region S produces a set of n slices of S which we ll call S k. These slices S k could be viewed as slices of bread after a loaf has been sliced. As was done in the case of the Riemann integral, we select, from each interval I k = [ k, k ], a sample point k I k. We then approimate the volume V k of each slice S k as follows, V k A( k ). (83) The right hand side is the volume of a cylinder with constant cross-sectional area A( k ) and height. This is illustrated in Figure 3 of Section 6.2 of Stewart. The total volume V of solid S is then approimated by the sum V n k= A( k ). (84) As n increases, we epect this approimation to become better and better. In fact, the above sum is a Riemann sum which, in the limit n, converges to the volume V, i.e., n b V = lim A( k ) = A() d. (85) n k= In the special case that A() = A, i.e., constant cross-sectional area, then the solid S is a cylindricallike region with volume V = Ah, where h = (b a), which is in agreement with the above formula, V = b a a Ad = A(b a). (86) Eample : A right circular cone with base radius r and height h. For convenience, we let the ape of the cone be situated at the origin (,,) and the positive -ais define the ais of the cone so that the center of the base of the cone is the point (h,,). A side view is shown below. The boundary lines of the side view of the cone, y = ± r h, (87) h correspond to the radii of the circular cross sections of the cone when it is intersected with planes that are perpendicular to the -ais. At each [,] the cross sectional area A() is A() = πy 2 = π r2 h 22. (88) 74

17 y y = r h h Right circular cone Using the volume integral derived earlier, the volume V of the cone may then be computed as follows, V = h = πr2 h 2 A() d h = πr2 h 2 3 h3 2 d = 3 πr2 h. (89) This result may have been known to you. The interesting feature of this result is that the volume of the cone is one-third the volume of the right circular cylinder with the same values of base radius r and height h. Eample 2: A pyramid with rectangular base having sides of length l and w and height h, as sketched below. h w l Pyramid This problem is a slight generalization of Eample 8 in Stewart, Section 6.2, p We won t work 75

18 it out here, but the result is V = lwh. (9) 3 Thevolume of the pyramid is one-third the volume of a rectangular bo with sides of length l, w and h. Eample 3: The cone and pyramid problems can be generalized to the case of a conical solid with a base that has a nonstandard shape and area A, as sketched below. h A Generalized cone From the ape of this solid, situated at a height h above the base (it can be above any point of the base in fact, it can lie outside the area directly above the base) the surface of the solid is formed from straight lines that emanate from the ape to every point on the base. Ifwenow placetheapeat theorigin (,,) andorient thesolidsothat thebaseareais perpendicular to the -ais, then the -ais will serve as the ais along which to perform the slicing, as shown below. With a little bit of work, it can be shown that at a point [,h], the cross sectional A() is given by A() = 2 h2a. (9) (Note that A() is not a linear function of, even though the straight lines in the diagram might suggest that it is. We must remember that the cross-sectional slice is a two-dimensional object.) The volume integral then becomes V = = h h A() d 2 h 2Ad = A j h 2 2 d = Ah. (92) 3 76

19 The volume V is one-third the volume of a generalized cylinder with base area A and height h. Eample No. 4: Volume of a sphere of radius r. If the sphere is placed at the origin, then its boundary is composed of the set of all points satisfying, 2 +y 2 +z 3 = R 2. (93) This problem is worked out in detail in Eample of Stewart, Section 6.2, p The final answer is well known to you, V = 4 3 πr3. (94) Eample No. 5: The cross section of a football-shaped solid which is obtained by revolving the ellipse, 2 a 2 + y2 =, (95) b2 about the -ais. This ellipse was studied earlier and is shown again below. y b y = b 2 a 2 a a y = b 2 a 2 b Side view of football-shaped solid obtained by revolving ellipse about -ais. The tiny slice of width and centered at, shown in the figure, is now the side view of a cylindrical slice of the solid, the volume of which is well approimated by V A()d = πy 2 ( ) = πb 2 2 a 2. (96) 77

20 The volume V of the ellipsoid of revolution may now be computed as follows, V = a a A() d a ) = πb ( 2 2 d a a 2 a ) = 2πb ( 2 2 d (integrand is an even function) = 2πb 2 [ 3 3 a 2 = 2πb 2 ( a 3 a ) a 2 ] a = 4 3 πab2. (97) (The use of the fact that the integrand is an even function to reduce the integration range to [,a] was not done in class.) In the special case that a = b = r, then V = 4 3 πr3, the volume of a sphere, as epected. Note the fact that it is the length b that is squared in the above result, as opposed to a. If the solid were obtained by revolving the ellipse about the y-ais (resulting in something looking like a flattened soccer ball), then the volume would be V = 4 3 πa2 b. 78

21 Volumes (cont d) Computation of volumes by cylindrical shells and washers Relevant sections from Stewart, Eighth Edition: 6.2, 6.3 Eample No. : We wish to compute the volume of the solid of revolution which is formed when the region in the first quadrant enclosed by the curves y = and y = 2,, is revolved about the -ais. This solid is sketched below. The two curves and the region they enclose are sketched in the net figure. y y = y u y = 2 y l We shall perform the integration over the -ais. Once again, we ll construct the usual partition of the -ais into n subintervals of length = n. Consider a thin horizontal strip of width that sits above one of these subintervals, say I k = [ k, k ]. Vertically, the strip is located between two 79

22 values which we shall denote as y u and y l so that y l y y u. (98) (Of course, y u and y l are determined by the two curves that enclose the region.) Now imaginethat this stripofwidth isrevolved aboutthe-ais toproducea washer assketched beside the graph. The volume V k of this washer is given by V k = π(yu 2 y2 l ). (99) As stated earlier y l and y u are determined by the two curves enclosing the region. We ll choose a sample point k in the subinterval of width and define y l = ( k )2, y u = k. () The volume of the washer is then approimated by V k π(( k )2 ( k ) 4 ). () We now sum over all n washers that are produced by the partitioning and revolving, V n V k = k= n π(( k )2 ( k )4 ). (2) k= In the limit n, the Riemann sum converges to the integral V = π ( 2 4 )d ( = π 3 ) 5 = 2 π. (3) 5 Note: In the future, we ll avoid all of the cumbersome notation involving Riemann sums and simply state that the volume V of the washer of width and centered at an [,] is given by V π( 2 4 )d. (4) In fact, we can go one further step and state that the differential volume dv of the washer situated at is given by dv = π( 2 4 )d. (5) 8

23 From here, we can proceed as follows: The total volume V is given by V = dv = π ( 2 4 )d = = 2 π. (6) 5 Eample No. 2: We now compute the volume of the solid of revolution which is formed when the region in the first quadrant enclosed by the curves y = and y = 2,, is revolved about the y-ais. This solid is sketched below. The two curves and the region they enclose are sketched in the net figure. y y = y u y = 2 y l As before, we shall perform the integration over the -ais. And once again, consider a thin horizontal strip of width that sits above one of these subintervals, say I k = [ k, k ]. Vertically, the strip is located between the two values which we employed before, y l y y u. (7) Now imagine that this strip of width is revolved about the y-ais to produce a cylindrical shell as sketched below the graph. The volume V k of this washer is determined by cutting it along a 8

24 vertical line and opening it up to form a rectangular bo with dimensions l = 2π, h = y u y l w =. (8) As such, its volume is approimated as V 2π(y u y l ) = 2π( 2 ) = 2π( 2 3 ). (9) where is a sample point taken from the subinterval. We ll use this result to write that the differential volume element at is dv = 2π( 2 3 )d. () The total volume V of the solid is obtained by integrating over all differential volume elements from = to = : V = = 2π = 2π dv ( 2 3 )d ] [ 3 4 = π 6. () Eample No 3: We revisit Eample No., i.e., the solid obtained by revolving region D about the -ais. But we ll compute the volume V of this solid by means of an integration over y, which means that cylindrical shells will be generated instead of washers. The situation is sketched below. Since the integration is being performed over y, we must epress the lower and upper values of which define the strip, i.e., l and u, respectively, in terms of y. Since l is defined by the line y =, we have l = y. Since u is defined by the line y = 2, or = y, we have u = y. 82

25 y y = y y = 2 y l u As a result, the volume V of the cylindrical shell is V = 2πy( u l ) y = 2πy( y y) y The infinitesimal volume element dv of this washer will be The total volume of the solid will be = 2π(y 3/2 y 2 ) y. (2) dv = 2π(y 3/2 y 2 )dy. (3) V = = 2π = 2π dv (y 3/2 y 2 )dy ) ( which is in agreement with our result from Eample No.. = 2π 5, (4) Eample No 4: We revisit Eample No. 2, i.e., the solid obtained by revolving region D about the y-ais. But we ll compute the volume V of this solid by means of an integration over y, which means that washers instead of cylindrical shells will be generated. The situation is sketched below. 83

26 y y = y y = 2 l u y Since the integration is being performed over y, we must epress the lower and upper values of which define the strip, i.e., l and u, respectively, in terms of y. Since l is defined by the line y =, we have l = y. Since u is defined by the line y = 2, or = y, we have u = y. As a result, the volume V of the washer is V = π( 2 u 2 l ) y The infinitesimal volume element dv of this washer will be The total volume of the solid will be = π(y y 2 ) y. (5) dv = π(y y 2 )dy. (6) V = = π = π dv (y y 2 )dy ) ( 2 3 which is in agreement with our result from Eample No. 2. = π 6, (7) 84

27 A note on conservative forces: Defining the potential energy in terms of definite integrals (The following material is taken from ERV s MATH 38P Fall 22 lecture notes, Week 2) Recall the definition of a conservative force in one dimension, F() = f()i: There eists a potential energy function U() such that U () = f() or f() = U (). (8) It follows that any force that is dependent only on the position coordinate is a conservative force. From Eq. (8), it follows that U() is the negative antiderivative of f(), i.e., U() = f()d. (9) As we ll see below, this formulation introduces the need for an arbitrary constant. A definite integral formulation will avoid the arbitrary constant. Eample: In the case of free fall near the earth, the force function is given by f() = mg. (2) From Eq. (9), the associated potential energy function is U() = ( mg)d = mgd = mg+c, (2) a well-known result. If we now impose the condition that U() =, we have that U() = mg+c = C = U() = mg. (22) In this case, all potential energy is measured with respect to the reference point =, at which U() =. You may recall that the definition of the potential energy function according to Eq. (8) turned out to be very convenient. From this definition, we were able to prove that the total mechanical energy function E(t), which is a sum of kinetic and potential energies, is constant in time. But the 85

28 formulation of the potential energy function in terms of definite integrals turns out to be even more useful, from both computational as well as physical viewpoints. Given a force function f() and a convenient reference point = a, we define the potential energy U() associated with f() as follows, U() = There are two noteworthy points regarding this definition: a f(s)ds. (23). From FTC I, we have that Therefore, Eq. (8) is satisfied. U () = d d a f(s)ds = f(). (24) 2. Since the lower limit of the definite integral is a, we have that U(a) = a a f(s)ds =. (25) In other words, the definite integral isolates a particular negative antiderivative of the force function f(). Let us now eamine the physical interpretation of the definite integral definition in Eq. (8). The integral a f(s) ds (26) represents the work done by the force F = f()i in moving the mass m from the position a to the position. As such, the function U(), which includes a negative sign in front of the integral, represents the work done against the force F = f()i in moving the mass m from position a to position. If U() is positive, it means that we have had to eert a force against the force F in order to displace the mass from a to. This causes a buildup of potential energy in the system. Eample: We return to the free-fall problem eamined earlier, in which f() = mg. We ll also let a = be the reference point. Then U() = ( mg)ds = 86 mgds = mg ds = mg, (27)

29 with U() =, in agreement with our earlier treatment. But note that no arbitrary constant was needed in this derivation. We simply built the initial condition into the function U() in terms of the lower limit of integration. Note that U() increases with. If the -coordinate of the mass is increased, i.e., it is elevated, then its potential energy increases. Eample: Recall the linear mass-spring problem, where the restorative force was given as f() = k, k >. (28) The equilibrium point of this force, i.e., the point at which the net force is zero, is =. We choose this to be our reference point, i.e., a =. The associated potential energy function is then U() = ( ks)ds = k sds = 2 k2, (29) a familiar result. Once again, there is no need for the arbitrary constant, which will eventually be fied according to a specified condition, e.g., U(a) =. Note that U() > for. If we move the mass from the equilibrium position =, then the spring will be either compressed ( < ) or etended ( > ) from its natural length. As a result, potential energy will be stored in the spring. 87