f dr. (6.1) f(x i, y i, z i ) r i. (6.2) N i=1

Transcription

1 hapter 6 Integrals In this chapter we will look at integrals in more detail. We will look at integrals along a curve, and multi-dimensional integrals in 2 or more dimensions. In physics we use these integrals to calculate moments of inertia, or the mass of an object from the density function. 6.1 Line integrals Useful line integrals in physics are often of a vector quantity f along a curve denoted by f dr. (6.1) We can express this integral as the limit of a sum over sections of the curve f dr = lim N i=1 N f(x i, y i, z i ) r i. (6.2) In practice when we come to evaluate a line integral it usually reduces to a sum over scalar integrals. For example, in artesian coordinates we have f = f x i + f y j + f z k and dr = idx + jdy + kdz. In this case f dr = (f x dx + f y dy + f z dz). (6.3) The curve that we use for this particular integral could be expressed parametrically e.g. (x(t), y(t), z(t)), or may be that y and z are parameterised in terms of x. onsequently our line integral will reduce to an integral such as Q(x, y)dx (6.4) where the curve is defined by y = y(x). We will consider integrals such as this in what follows. As an example, consider the line integral from (, a) to (a, ) over the upper part of the circle defined by x 2 + y 2 = a 2 of the following xdy. (6.5) 6

2 HAPTER 6. INTEGRALS 61 On our path x = a 2 y 2, so xdy = = a π/2 a2 y 2 dy (6.6) a 2 cos 2 udu = a2 π 4. (6.7) Path dependence In general the line integral beetween two points A and B will depend on the route chosen for the integral. However, for some special types of integral we may find that the line integral is independent of the path chosen. If our line integral is along a path defined parametrically by x = f(t) and y = g(t), then for a function F(x, y) the line integral of the total differential of F is df = ( df F dt dt = dx x dt + F y dy dt ) dt = F F dx + x y dy = F(t 2) F(t 1 ) (6.8) where t 1 is the starting point of the integral, and t 2 is the end point. onsequently if we have a line integral P(x, y)dx + Q(x, y)dy (6.9) where P(x, y) = F x, F Q(x, y) = y (6.1) then the line integral is path independent. It can be shown that under suitable conditions of continuity of Q and P, then the condition P y = Q x (6.11) is enough to ensure path independence of our integral. This is the condition that we have a total differential in our integrand that we discussed in the last chapter. 6.2 Double integrals A double integral of a function f(x, y) over a region a < x < b and c < y < d can be defined using a Riemann integral. We divide the domain of integration into parts by dividing the x domain into m parts and the y domain into n parts as follows a < x 1 < x 2 <... < x m 1 < b (6.12) c < y 1 < y 2 <... < y n 1 < d (6.13) Thus we have mn rectangular areas. We choose a point lying in each rectangle (ξ r, η s ) x r 1 ξ r x r (6.14) y s 1 η s y s (6.15)

3 HAPTER 6. INTEGRALS 62 The integral is then defined as the limit of the double sum I = lim m n,m r=1 s=1 n f(ξ r, η s )δr rs (6.16) where δr rs = (x r x r 1 )(y s y s 1 ) is the area of the corresponding rectangle. This defines the double integral f(x, y)dxdy (6.17) R over the region we have defined, R. Under suitable technical conditions we can change the order in which we do the integrals over x and y, corresponding to either summing over strips in the x direction or strips in the y direction. Transforming coordinate system In one dimensional integrals it is often convenient to change variables to simplify the integration. This is also true in higher dimensions, but we must be careful to correct for any change in the element of infinitesimal area over which we are summing. Suppose we change from our artesian coordinate system (x, y), to a new coordinate system related as follows x = x(u, v) (6.18) y = y(u, v). (6.19) We must then include in our integration the change in area through the Jacobian, J f(x, y)dxdy = f(x(u, v), y(u, v)) (x, y) (u, v) dudv (6.2) }{{} J where J = (x, y) (u, v) = x y u v x y v u. (6.21) Polar oordinates Polar coordinates are related to artesian coordinates by x = r cos θ and y = r sin θ. An element of area in polar coordinates should be rδθδr on geometrical grounds. If we evaluate our Jacobian then we find J = cosθr cosθ ( r sinθ)(sin θ) = r (6.22) so f(x, y)dxdy = g(r, θ)rdθdr (6.23)

4 HAPTER 6. INTEGRALS 63 Example: Gaussian integrals I = e x2 dx (6.24) can be evaluated by using a double integral. The square of this integral is given by I 2 = e x2 dy This double integral can be calculated if we transform to polar coordinates. I 2 = = π/2 = e r2 rdr e y2 dx. (6.25) e r2 rdrdθ (6.26) π/2 dθ (6.27) 2 π (6.28) I = 1 2 π. (6.29) We were quick to make this coordinate transformation here, but we should take more care with the shape of the infinite region. We started with a square shaped region, but finished the calculation with a sector of a circle. It can be shown by using a circumscribed and an inscribed circle that the two calculations produce the same result. 6.3 Triple integrals Triple integrals can be used to calculate volumes, and are defined in much the same way as double integrals, as the limit of a sum. Their application for the calculation of volumes will be discussed in lectures. Spherical Polar oordinates Here we have x = r cosφsin θ, y = r sinφsin θ and z = r cosθ. The Jacobian can be calculated in the same way as for the polar coordinates example above J = The volume element is thus given by dv = r 2 sin θdθdφdr. = = (x, y, z) (r, θ, φ) (6.3) x x x r θ φ y y y r θ φ (6.31) z z z r θ φ cosφsin θ r cosφcos θ r sin φsin θ sin φsin θ r sin φcosθ r cosφsin θ (6.32) cosθ r sin θ = r 2 sin θ (6.33)

5 HAPTER 6. INTEGRALS 64 ylindrical Polar oordinates Here we have x = r cos φ, y = r sin φ and z remains unchanged, so dv = rdrdφdz. 6.4 Applications of line and multiple integrals Maxwell s Equations You will meet Maxwell s equations of electromagnetism next year. These equations can be written in integral forms, for example Ampére s law B dl = I enc µ. (6.34) This equation can be used to calculate the magnetic flux density, for example around a straight wire. Another example is calculating the electromotive force (i.e. voltage) around a loop due to a changing magnetic flux density E = E dl = B ds (6.35) t Moments of Inertia To start an object rotating we have to apply a torque to that object. The angular acceleration of the object depends on its moment of inertia, that is the distribution of mass around the axis of rotation. The moment of inertia is defined as m i ri 2 = ρri 2 dv (6.36) I = i where we have written the discrete version in terms of the mass m i and the distance from the axis r i of an element in the object, and the continuous version in terms of the density distribution ρ. As an example let s calculate the moment of inertia of a square laminate, of uniform density ρ, rotated about an axis through its centre. This is given by I = where the total mass M = a 2 ρ. a/2 a/2 a/2 a/2 ρ(x 2 + y 2 )dxdy = a4 6 ρ = Ma2 6 (6.37) As a second example, let s calculate the moment of inertia of a sphere of radius R about an axis r through its centre, with a non-uniform mass distribution ρ = ρ R. First let s calculate the total mass of the sphere, which is most easily expressed in spherical polar coordinates M = R π 2π ρ r R r2 sin θdrdθdφ (6.38) R 3 = ρ 4 4π = ρ R 3 π. (6.39) Now the moment of inertia can be calculated in a similar way I = R π 2π = ρ R 5 6 ρ r R (r sin θ)2 r 2 sin θdrdθdφ (6.4) 4 3 2π = 4 9 MR2 (6.41)

6 HAPTER 6. INTEGRALS Problems 1B. Evaluate the following line integrals (a) (x2 + 2y)dx from (, 1) to (2, 3), where is the line y = x + 1 (b) xdy from (, ) to (π, ), where is the curve y = sin x 2B. Evaluate (x 2 ydx + (x 2 y 2 )dy) from (, ) to (1, 4), where (i) is the curve y = 4x 2, (ii) is the line y = 4x. 3B. Evaluate (xdy ydx) around the curve x = a cos 3 t, y = a sin 3 t where a is a constant. 4B. Show that (2xsin ydx + x 2 cosydy) is independent of the path, and evaluate it from (, to (1, π/2). (Hint: onsider the total differential of f(x, y) = x 2 sin y) 5B. Integrate both sin(x + y) and xe xy (a) over the square < x < π/2, < y < π/2, (b) over the square < x < π, < y < π, (c) over the triangle formed by the lines y =, x = π/2, x = y. (Leave your answer in the form of a single integral in the one case in which it cannot be evaluated by standard means.) 6A. Show that x 2 ydxdy = ( )/42 D where D is the smaller of the two areas bounded by the curves xy = 1, y = x 2 and y = 2. 7B. Evaluate the integral 2 x= 1 y=x/2 2xy 2 dydx hange the order of integration (integrate with respect to x first) and thence evaluate the integral again. 8. In each of the following, (a) evaluate the integral, (b) sketch the region of integration, (c) write down the integral with the order of integration reversed, (d) evaluate again and compare with (a).

7 HAPTER 6. INTEGRALS 66 (i) a dx a x dy (ii) a dx x (x2 + y 2 )dy (iii) 1 dx x x xy 2 dy (iv) π/2 π/2 dθ 2acos θ r 2 cosθdr 9A. alculate the Jacobians for the following coordinate transformations, and sketch a diagram illustrating the change in the infinitesimal volume in each case (a) ζ = 3x, η = y/2; (b) x = r cosθ, y = r sin θ; (c) x = r cosφsin θ, y = r sin φsin θ, z = r cosθ. 1. The variables x, y can be expressed as functions of ξ, η, which in turn can be expressed as functions of λ, µ. Show that the Jacobian (x, y) x y (λ, µ) = λ λ satisfies the equation and deduce that x µ y µ (x, y) (x, y) (ξ, η) = (λ, µ) (ξ, η) (λ, µ) (x, y) η) = 1/ (ξ, (ξ, η) (x, y). Find (ξ,η) (x,y) in the case where x = c coshξ cosη, y = c sinhξ sin η. 11B. (i) Transform to polars and evaluate a/ 2 a 2 x 2 dx x2 + y 2 dy x (ii) Reverse the order of integration and hence evaluate 1 2 x x dx x y dy. 12B. By suitably transforming the variables x and y to another pair of variables, calculate the integral x 2 dxdy (1 + x 2 + y 2 ) Evaluate the volume of the integral e (x2 +y 2 +z 2 )/a 2 over the whole of three-dimensional space 14B. By evaluating the appropriate triple integrals find

8 HAPTER 6. INTEGRALS 67 (a) the volume of a sphere of radius a; (b) the volume of a flat-topped cone, described in cylindrical polar coordinates by {(r, θ, z) z a, θ 2π, r z}; (c) the volume of a round-topped cone, described in spherical polar coordinates by {(r, θ, φ) r a, φ 2π, θ π/4};. 15. Prove that the volume enclosed by the sphere x 2 + y 2 + z 2 = a 2, and the cylinder x 2 + y 2 = ay is given by V = 2 a2 x 2 y 2 dxdy the integral being taken over the circle x 2 + y 2 = ay. Transform the integral into polar coordinates, and hence show that V = 4 ( π 3 a3 2 2 ). 3 16A. A thin uniform circular disc has mass M and radius a. (a) Prove that its moment of inertia about an axis perpendicular to its plane and passing through its centre is 1 2 Ma2 (b) Prove that the moment of inertia of the same disc about a diameter is 1 4 Ma2. This is an example of a general result for planar bodies that the moment of inertia of the body about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two perpendicular axes lying in the plane: i.e. I z = I x + I y. Additional questions: 17B. Evaluate (x 2 xy)ds from (, 4) to (4, ), where is the circle x 2 + y 2 = B. (i) Evaluate the line integral ( ) xdy ydx x 2 + y 2, + 1 where is the boundary of the minor segment of the circle x 2 +y 2 = 1 cut off by the chord x + y = 1. (ii) Show that the line integral PQ (yzdx + zxdy + xydz) is independent of path of integration, and evaluate it when P and Q are the points (1, 1, 2) and (3, 2, 1) respectively.

9 HAPTER 6. INTEGRALS 68 19B. Show that, if P(x, y, z), Q(x, y, z) and R(x, y, z) are continuous and single-valued functions with continuous and single-valued partial derivatives then the condition that the line integral [Pdx + Qdy + Rdz] shall be independent of path in some (simply connected) region R is that for every point in R. Hence show that P y = Q x, Q z = R y, R x = P z. ( y z dx + x xy ) dy z z 2 dz is independent of path, and evaluate it from (,, 1) to (1, 1, 1). 2B. Sketch the curve given by r = 2a(1 + cosθ) and evaluate (x 2 + y 2 ) 1/2 dxdy where D is the interior of the curve. 21B. Show, by a diagram, the region over which the following integral extends a dy D a y dx a2 x 2 + b 2 y 2. Transform to polar coordinates and hence show that the value of the integral is a b b sinh 1 a. 22. By means of the substitutions x = a sinθ cosφ, y = b sinθ sin φ evaluate x 2 ydxdy 1 x2 /a 2 y 2 /b 2 over the area inside the ellipse x 2 /a 2 + y 2 /b 2 = 1 and in the positive quadrant. 23. By putting x = r cosθ, y = r sin θ prove that e (x2 +2xy cos α+y 2) dxdy = α 2 sin α, α < π 24B. Evaluate the integral x 2 (1 x 2 y 2 )dxdy over the interior of the circle of radius 1 centred at x =, y =. 25B. (a) (x 2 4xy y 2 ) 2 dxdy over the interior of the circle x 2 + y 2 = a 2. (b) xdxdy over the first quadrant of the ellipse x 2 /a 2 + y 2 /b 2 = 1

10 HAPTER 6. INTEGRALS 69 Answers to hapter 6 Problem Note: you must attempt the problems before referring to the answers 1. (a) 32 3 (b) 2 2. (i) (ii) dx = 3a cos 2 t sin tdt dy = 3a cost sin 2 tdt The integral simplifies to 2π 3a 2 cos 2 t sin 2 tdt = 3a 2 π/4 (use double angle formulae to calculate) 4. The differential in the question is exact (2xsin ydx + x2 cosydy) = d(x2 sin y) = [ x 2 sin y ] (1,π/2) (,) 5. For sin(x + y) (a) 2 (b) (c) 1 For xe xy ) 1 (a) 2π (4e π2 /4 4 π 2 ) (b) 1 π (e π2 1 π 2 (c) π/2 π/2 y xe xy dxdy (cannot be done analytically in terms of simple functions) 6. Answer given in the question. 2 y 1 1/y x2 ydxdy 7. 4/5 1 y= 2y x= 2x2 ydxdy 8. (i) a 2 /2 (ii) a 4 /3 (iii) 1/35 (iv) 8a 2 /3 9. (a) J = 1/3 2 = 2/3

11 HAPTER 6. INTEGRALS 7 (b) J = cosθ r sinθ sin θ r cosθ = r cosφsin θ r sinφsin θ r cosφcos θ (c) J = sin φsin θ r cosφsin θ r sin φcosθ cosθ sinθ = r2 sin θ 1. J = (ξ,η) (x,y) (x,y) = 1/ (ξ,η) = c sinhξ cosη c coshξ sinη 11. (i) π/2 a π/4 rdrdθ = a2 π 8 (ii) 1 dy y x y dx dy 2 y x ydx = ln Transform to polar coordinates 2π c coshξ sin η c sinhξ cosη = c2 cosh 2 ξ sin 2 η + c 2 cos 2 η sinh 2 ξ r 2 cos 2 θ dθ rdr = π/4 (1 + r) (a) (b) (c) e x2 +y 2 +z 2 )/a 2 dxdydz = 2π π a a 2π z π/4 2π a r 2 sin θdrdθdφ = 4πa3 3 rdrdθdz = a3 π 3 ( ) 3 π a r 2 sinθdrdφdθ = 1 3 a3 π(2 2) 15. Answers given in the question. 16. Answers given in the question.