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6.2Example 4 Let us find the length of the ellipse defined by the canonical equationx2 y2+= 1 (a ≥ b > 0).a 2 b2(6.56)Taking the parametrization x = a sin ψ , y = b cos ψ, 0 ≤ ψ ≤ 2π , we obtain2πl=(a cos ψ)2 + (−b sin ψ)2 dψ =0= 4a1−0where k 2 = 1 −The integrala 2 − a 2 − b2 sin2 ψ dψ =0'π/22πb2a2a 2 − b2 2sin ψ dψ = 4aa2π/2 1 − k 2 sin2 ψ dψ,0is the square of the eccentricity of the ellipse.ϕE(k, ϕ) =1 − k 2 sin2 ψ dψ0cannot be expressed in elementary functions, and is called an elliptic integral because of the connection with the ellipse just discussed.
More precisely, E(k, ϕ) isthe elliptic integral of second kind in the Legendre form. The value that it assumesfor ϕ = π/2 depends only on k, is denoted E(k), and is called the complete ellipticintegral of second kind. Thus E(k) = E(k, π/2), so that the length of an ellipse hasthe form l = 4aE(k) in this notation.6.4.3 The Area of a Curvilinear TrapezoidConsider the figure aABb of Fig. 6.2, which is called a curvilinear trapezoid. Thisfigure is bounded by the vertical line segments aA and bB, the closed interval [a, b]on the x-axis, and the curve AB, which is the graph of an integrable function y =f (x) on [a, b].Let [α, β] be a closed interval contained in [a, b].
We denote by S(α, β) the areaof the curvilinear trapezoid αf (α)f (β)β corresponding to it.3846IntegrationOur ideas about area are as follows: if a ≤ α < β < γ ≤ b, thenS(α, γ ) = S(α, β) + S(β, γ )(additivity of areas) andinf f (x)(β − α) ≤ S(α, β) ≤ sup f (x)(β − α).x∈[α,β]x∈[α,β](The area of an enclosing figure is not less than the area of the figure enclosed.)Hence by Proposition 1, the area of this figure must be computed from the formula bS(a, b) =f (x) dx.(6.57)aExample 5 Let us use formula (6.57) to compute the area of the ellipse given by thecanonical equation (6.56).By the symmetry of the figure and the assumed additivity of areas, it suffices tofind the area of just the part of the ellipse in the first quadrant, then quadruple theresult.
Here are the computations: a' π/2 #x2S=4b2 1 − 2 dx = 4b1 − sin2 ta cos t dt =a00 π/2 π/22= 4abcos t dt = 2ab(1 − cos 2t) dt = πab.00Along the way we have made the change of variable x = a sin t, 0 ≤ t ≤ π/2.Thus S = πab. In particular, when a = b = R, we obtain the formula πR 2 forthe area of a disk of radius R.Remark It should be noted that formula (6.57) gives the area of the curvilinear trapezoid under the condition that f (x) ≥ 0 on [a, b]. If f is an arbitrary integrablefunction, then the integral (6.57) obviously gives the algebraic sum of the areas ofcorresponding curvilinear trapezoids lying above and below the x-axis.
When thisis done, the areas of trapezoids lying above the x-axis are summed with a positivesign and those below with a negative sign.6.4.4 Volume of a Solid of RevolutionNow suppose the curvilinear trapezoid shown in Fig. 6.2 is revolved about the closedinterval [a, b]. Let us determine the volume of the solid that results.We denote by V (α, β) the volume of the solid obtained by revolving the curvilinear trapezoid αf (α)f (β)β (see Fig. 6.2) corresponding to the closed interval[α, β] ⊂ [a, b].6.4 Some Applications of Integration385According to our ideas about volume the following relations must hold: if a ≤α < β < γ ≤ b, thenV (α, γ ) = V (α, β) + V (β, γ )andπ22inf f (x) (β − α) ≤ V (α, β) ≤ π sup f (x) (β − α).x∈[α,β]x∈[α,β]In this last relation we have estimated the volume V (α, β) by the volumes ofinscribed and circumscribed cylinders and used the formula for the volume of acylinder (which is not difficult to obtain, once the area of a disk has been found).Then by Proposition 1 bV (a, b) = πf 2 (x) dx.(6.58)aExample 6 By revolving about the x-axis the semicircle√ bounded by the closed interval [−R, R] of the axis and the arc of the circle y = R 2 − x 2 , −R ≤ x ≤ R, onecan obtain a three-dimensional ball of radius R whose volume is easily computedfrom (6.58): R 24V =πR − x 2 dx = πR 3 .3−RMore details on the measurement of lengths, areas, and volumes will be given inPart 2 of this course.
At that time we shall solve the problem of the invariance of thedefinitions we have given.6.4.5 Work and EnergyThe energy expenditure connected with the movement of a body under the influenceof a constant force in the direction in which the force acts is measured by the productF · S of the magnitude of the force and the magnitude of the displacement.
Thisquantity is called the work done by the force in the displacement. In general thedirections of the force and displacement may be noncollinear (for example, whenwe pull a sled by a rope), and then the work is defined as the inner product &F, S' ofthe force vector and the displacement vector.Let us consider some examples of the computation of work and the use of therelated concept of energy.Example 7 The work that must be performed against the force of gravity to lift abody of mass m vertically from height h1 above the surface of the Earth to heighth2 is, by the definition just given, mg(h2 −h1 ). It is assumed that the entire operationoccurs near the surface of the Earth, so that the variation of the gravitational forcemg can be neglected. The general case is studied in Example 10.3866IntegrationExample 8 Suppose we have a perfectly elastic spring, one end of which is attachedat the point 0 of the real line, while the other is at the point x.
It is known that theforce necessary to hold this end of the spring is kx, where k is the modulus of thespring.Let us compute the work that must be done to move the free end of the springfrom position x = a to x = b.Regarding the work A(α, β) as an additive function of the interval [α, β] andassuming valid the estimatesinf (kx)(β − α) ≤ A(α, β) ≤ sup (kx)(β − α),x∈[α,β]x∈[α,β]we arrive via Proposition 1 at the conclusion thatA(a, b) =abkx 2 bkx dx =.2 aThis work is done against the force. The work done by the spring during the samedisplacement differs only in sign.2The function U (x) = kx2 that we have found enables us to compute the workwe do in changing the state of the spring, and hence the work that the spring mustdo in returning to its initial state.
Such a function U (x), which depends only on theconfiguration of the system, is called the potential energy of the system. It is clearfrom the construction that the derivative of the potential energy gives the force ofthe spring with the opposite sign.If a point of mass m moves along the axis subject to this elastic force F = −kx,its coordinate x(t) as a function of time satisfies the equationmẍ = −kx.(6.59)We have already verified once (see Sect.
5.6.6) that the quantitymv 2 kx 2+= K(t) + U x(t) = E,22(6.60)which is the sum of the kinetic and (as we now understand) potential energies of thesystem, remains constant during the motion.Example 9 We now consider another example. In this example we shall encounter anumber of concepts that we have introduced and become familiar with in differentialand integral calculus.We begin by remarking that by analogy with the function (6.60), which was written for a particular mechanical system satisfying Eq. (6.59), one can verify that foran arbitrary equation of the forms̈(t) = f s(t) ,(6.61)6.4 Some Applications of Integration387Fig. 6.3where f (s) is a given function, the sumṡ 2+ U (s) = E2(6.62)does not vary over time if U (s) = −f (s).Indeed,dU dsdE 1 dṡ 2 dU (s)=+= ṡ s̈ +·= ṡ s̈ − f (s) = 0.dt2 dtdtds dtThus by (6.62), regarding E as a constant, we obtain successively, first ṡ = ± 2 E − U (s)(where the sign must correspond to the sign of the derivativedsdt ),then1dt= ±√,ds2(E − U (s))and finallyt = c1 ±ds.√2(E − U (s))Consequently, using the law of conservation of the “energy” (6.62) in Eq.
(6.61),we have succeeded theoretically in solving this equation by finding not the functions(t), but its inverse t (s).Equation (6.61) arises, for example, in describing the motion of a point along agiven curve. Suppose a particle moves under the influence of the force of gravityalong a narrow ideally smooth track (Fig. 6.3).Let s(t) be the distance along the track (that is, the length of the path) from afixed point O – the origin of the measurement – to the point where the particle is attime t. It is clear that then ṡ(t) is the magnitude of the velocity of the particle ands̈(t) is the magnitude of the tangential component of its acceleration, which mustequal the magnitude of the tangential component of the force of gravity at a givenpoint of the track.
It is also clear that the tangential component of the force of gravitydepends only on the point of the track, that is, it depends only on s, since s can be3886IntegrationFig. 6.4regarded as a parameter that parametrizes the curve9 with which we are identifyingthe track. If we denote this component of the force of gravity by f (s), we find thatms̈ = f (s).For this equation the following quantity will be preserved:1 2mṡ + U (s) = E,2where U (s) = −f (s).Since the term 12 mṡ 2 is the kinetic energy of the point and the motion along thetrack is frictionless, we can guess, avoiding calculations, that the function U (s),up to a constant term, must have the form mgh(s), where mgh(s) is the potentialenergy of a point of height h(s) in the gravitational field.If the relations ṡ(0) = 0, s(0) = s0 , and h(s0 ) = h0 held at the initial time t = 0,then by the relations2E= ṡ 2 + 2gh(s) = Cmwe find that C = 2gh(s0 ), and therefore ṡ 2 = 2g(h0 − h(s)) and sdst=.√2g(h0 − h(s))s0(6.63)In particular if, as in the case of a pendulum, the point moves along a circle ofradius R, the length s is measured from the lowest point O of the circle, and theinitial conditions amount to the equality ṡ(0) = 0 at t = 0 and a given initial angleof displacement −ϕ0 (see Fig.
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