Handbook of mathematics for engineers and scienteists part 51 doc

318 INTEGRALS 7.3.1-2. Classes of integrable functions. Further on, it is assumed that D is a closed bounded domain. 1. If f (x, y) is continuous in D, then the double integral  D f(x, y) dx dy exists. 2. If f (x, y) is bounded and the set of points of discontinuity of f(x, y) has a zero area (e.g., the points of discontinuity lie on finitely many continuous curves in the x, y plane), then the double integral of f (x, y) over the domain D exists. 7.3.1-3. Properties of the double integral. 1. Linearity. If functions f (x, y)andg(x, y) are integrable in D,then  D  af(x, y) bg(x, y)  dx dy = a  D f(x, y) dx dy b  D g(x, y) dx dy, where a and b are any numbers. 2. Additivity. If the domain D is split into two subdomains D 1 and D 2 that do not have common internal points and if the function f (x, y) is integrable in either subdomain, then  D f(x, y) dx dy =  D 1 f(x, y) dx dy +  D 2 f(x, y) dx dy. 3. Estimation theorem.Ifm ≤ f (x, y) ≤ M in D,then mS ≤  D f(x, y) dx dy ≤ MS, where S is the area of the domain D. 4. Mean value theorem.Iff(x, y) is continuous in D, then there exists at least one internal point (¯x, ¯y) D such that  D f(x, y) dx dy = f (¯x, ¯y) S. The number f (¯x, ¯y) is called the mean value of the function f(x, y)inD. 5. Integration of inequalities.Ifϕ(x, y) ≤ f (x, y) ≤ g(x, y)inD,then  D ϕ(x, y) dx dy ≤  D f(x, y) dx dy ≤  D g(x, y) dx dy. In particular, if f(x, y) ≥ 0 in D,then  D f(x, y) dx dy ≥ 0. 6. Absolute value theorem      D f(x, y) dx dy     ≤  D   f(x, y)   dx dy. 7.3. DOUBLE AND TRIPLE INTEGRALS 319 zfxy= (, ) y x D z O Figure 7.4. A double integral of a nonnegative function f(x, y) over a domain D is equal to the volume of a cylindrical body with base D in the plane z = 0 and bounded from above by the surface z = f(x, y). 7.3.1-4. Geometric meaning of the double integral. Let a function f(x, y) be nonnegative in D. Then the double integral  D f(x, y) dx dy is equal to the volume of a cylindrical body with base D in the plane z = 0 and bounded from above by the surface z = f (x, y); see Fig. 7.4. 7.3.2. Computation of the Double Integral 7.3.2-1. Use of iterated integrals. 1 ◦ . If a domain D is definedinthex, y plane by the inequalities a≤x≤ b and y 1 (x)≤ y ≤ y 2 (x) (see Fig. 7.5 a), then*  D f(x, y) dx dy =  b a dx  y 2 (x) y 1 (x) f(x, y) dy.(7.3.2.1) The integral on the right-hand side is called an iterated integral. yyx= 1 () xxy= 2 () xxy= 1 () yyx= 2 () yy xx a c d bOO ()a ()b D D Figure 7.5. Computation of a double integral using iterated integrals: (a) illustration to formula (7.3.2.1), (b) illustration to formula (7.3.2.2). * It is assumed that in (7.3.2.1) and (7.3.2.2) the double integral on the right-hand side and the inner integral on the right-hand side exist. 320 INTEGRALS 2 ◦ .IfD = {c ≤ y ≤ d, x 1 (y) ≤ x ≤ x 2 (y)} (see Fig. 7.5 b), then  D f(x, y) dx dy =  d c dy  x 2 (y) x 1 (y) f(x, y) dx.(7.3.2.2) Example 1. Compute the integral I =  D dx dy (ax + by) 2 , where D = {0 ≤ x ≤ 1, 1 ≤ y ≤ 3} is a rectangle, a > 0,andb > 0. Using formula (7.3.2.2), we get  D dx dy (ax + by) 2 =  3 1 dy  1 0 dx (ax + by) 2 . Compute the inner integral:  1 0 dx (ax + by) 2 =– 1 a(ax + by)     x=1 x=0 = 1 a  1 by – 1 by + a  . It follows that I = 1 a  3 1  1 by – 1 by + a  dy = 1 ab ln 3(a + b) a + 3b . 3 ◦ . Consider a domain D inscribed in a rectangle {a ≤ x ≤ b, c ≤ y ≤ d}. Let the boundary of D, within the rectangle, be intersected by straight lines parallel to the coordinate axes at two points only, as shown in Fig. 7.6 a. Then, by comparing formulas (7.3.2.1) and (7.3.2.2), we arrive at the relation  b a dx  y 2 (x) y 1 (x) f(x, y) dy =  d c dy  x 2 (y) x 1 (y) f(x, y) dx, which shows how the order of integration can be changed. yy xx aa c D D DD D 1 22 3 d bOO ()a ()b Figure 7.6. Illustrations to the computation of a double integral in a simple (a) and a complex (b) domain. 4 ◦ . In the general case, the domain D is first split into subdomains considered in Items 1 ◦ and 2 ◦ , and then the property of additivity of the double integral is used. For example, the domain D showninFig.7.6b is divided by the straight line x = a into three subdomains D 1 , D 2 ,andD 3 . Then the integral over D is represented as the sum of three integrals over the resulting subdomains. 7.3. DOUBLE AND TRIPLE INTEGRALS 321 7.3.2-2. Change of variables in the double integral. 1 ◦ .Letx = x(u, v)andy = y(u, v) be continuously differentiable functions that map one-to-one a domain D 1 in the u, v plane onto a domain D in the x, y plane, and let f(x, y) be a continuous function in D.Then  D f(x, y) dx dy =  D 1 f  x(u, v), y(u, v)  |J(u, v)|du dv, where J(u, v)istheJacobian (or Jacobian determinant) of the mapping of D 1 onto D: J(u, v)= ∂(x, y) ∂(u, v) =      ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v      = ∂x ∂u ∂y ∂v – ∂x ∂v ∂y ∂u . The fraction before the determinant is a common notation for a Jacobian. The absolute value of the Jacobian characterizes the extension (contraction) of an infinitesimal area element when passing from x, y to u, v. 2 ◦ . The Jacobian of the mapping defining the change from the Cartesian coordinates x, y to the polar coordinates ρ, ϕ, x = ρ cos ϕ, y = ρ sin ϕ,(7.3.2.3) is equal to J(ρ, ϕ)=ρ.(7.3.2.4) Example 2. Given a sphere of radius R and a right circular cylinder of radius a < R whose axis passes through the sphere center, find the volume of the figure the cylinder cuts out of the sphere. The volume of this figure is calculated as V =  x 2 +y 2 ≤a 2  R 2 – x 2 – y 2 dx dy. Passing in the integral from x, y to the polar coordinates (7.3.2.3) and taking into account (7.3.2.4), we obtain V =  2π 0  a 0  R 2 – ρ 2 ρdρdϕ= 4π 3  R 3 –(R 2 – a 2 ) 3/2  . 3 ◦ . The Jacobians of some common transformations in the plane are listed in Table 7.1. TABLE 7.1 Some common curvilinear coordinates in the plane and the respective Jacobians Name of coordinates Transformation Jacobian, J Polar coordinates ρ, ϕ x = ρ cos ϕ, y = ρ sin ϕ ρ Generalized polar coordinates ρ, ϕ x = aρ cos ϕ, y = bρ sin ϕ abρ Elliptic coordinates u, v (special system; u ≥ 0, 0 ≤ v ≤ π) x = a cosh u cos v, y = a sinh u sin v a 2 (sinh 2 u +sin 2 v) Parabolic coordinates σ, τ x = aστ, y = 1 2 a(τ 2 – σ 2 ) a 2 (σ 2 + τ 2 ) Bipolar coordinates σ, τ x = a sinh τ cosh τ –cosσ , y = a sin σ cosh τ –cosσ a 2 (cosh τ –cosσ) 2 322 INTEGRALS 7.3.2-3. Differentiation of the double integral with respect to a parameter. 1 ◦ . In various applications, situations may arise where the integrand function and the integration domain depend on a parameter, t. The derivative of such a double integral with respect to t is expressed as d dt  D(t) f(x, y, t) dx dy =  D(t) ∂ ∂t f(x, y, t) dx dy +  L(t) (n ⋅ v)f(x, y, t) dl,(7.3.2.5) where L(t) is the boundary of the domain D(t), n is outer unit normal to L(t), and v is the velocity of motion of the points of L(t). 2 ◦ . If the boundary L(t) is specified by equations in parametric form x = X(λ, t), y = Y (λ, t), α ≤ λ ≤ β,(7.3.2.6) then n =  Y λ  X 2 λ + Y 2 λ ,– X λ  X 2 λ + Y 2 λ  , v =  X t , Y t  ,(n ⋅ v)= Y λ X t – X λ Y t  X 2 λ + Y 2 λ , dl =  X 2 λ + Y 2 λ ; the subscripts λ and t denote the respective partial derivatives. The last integral in (7.3.2.5) becomes  L(t) (n ⋅ v)f(x, y, t) dl =  β α f  X(λ, t), Y (λ, t), t  (Y λ X t – X λ Y t ) dλ.(7.3.2.7) Example 3. Let D(t) be a deformable plane domain bounded by an ellipse x 2 a 2 (t) + y 2 b 2 (t) = 1.(7.3.2.8) Let us rewrite the equation of the ellipse (7.3.2.8) in parametric form as x = a(t)cosλ, y = b(t)sinλ, 0 ≤ λ ≤ 2π, which corresponds to the special case of (7.3.2.6) with X(λ, t)=a(t)cosλ and Y (λ, t)=b(t)sinλ.Taking into account the aforesaid and using formula (7.3.2.7), we obtain  L(t) (n ⋅ v)f(x, y, t) dl =  2π 0 f  a(t)cosλ, b(t)sinλ, t  (a  t b cos 2 λ + ab  t sin 2 λ) dλ.(7.3.2.9) Let us dwell on the simple special case of f(x, y, t)=1, when the first integral on the right-hand side of (7.3.2.5) vanishes. Evaluate the second integral by formula (7.3.2.9) to obtain  L(t) (n ⋅ v) dl =  2π 0 (a  t b cos 2 λ + ab  t sin 2 λ) dλ = a  t b  2π 0 cos 2 λdλ+ ab  t  2π 0 sin 2 λdλ = π[a  (t)b(t)+a(t)b  (t)]. (7.3.2.10) Remark. Formula (7.3.2.10) is easy to derive directly from (7.3.2.5) noting that, for f(x, y, t)=1,the integral on the left-hand side of (7.3.2.5) gives the area of the ellipse, S = πa(t)b(t). Differentiating this formula yields (7.3.2.10). 3 ◦ . If the boundary L(t) is specified by an equation in implicit form, F (x, y, t)=0, then one should take into account in (7.3.2.5) that n =  F x  F 2 x + F 2 y , F y  F 2 x + F 2 y  , v =  – F t F x ,– F t F y  ,(n ⋅ v)=– 2F t  F 2 x + F 2 y . For an elliptic domain, specified by equation (7.3.2.8), we have F (x, y, t)= x 2 a 2 (t) + y 2 b 2 (t) – 1. 7.3. DOUBLE AND TRIPLE INTEGRALS 323 7.3.3. Geometric and Physical Applications of the Double Integral 7.3.3-1. Geometric applications of the double integral. 1. Area of a domain D in the x, y plane: S =  D dx dy. 2. Area of a surface defined by an equation z = f(x, y) with (x, y) D (the surface is projected onto a domain D in the x, y plane): S =  D   ∂f ∂x  2 +  ∂f ∂y  2 + 1 dx dy. 3. Area of a surface defined parametrically by equations x = x(u, v), y = y(u, v), z = z(u, v), with (u, v) D 1 : S =  D 1 √ EG – F 2 du dv. Notation used: E =  ∂x ∂u  2 +  ∂y ∂u  2 +  ∂z ∂u  2 , G =  ∂x ∂v  2 +  ∂y ∂v  2 +  ∂z ∂v  2 , F = ∂x ∂u ∂x ∂v + ∂y ∂u ∂y ∂v + ∂z ∂u ∂z ∂v . 4. Area of a surface defined by a vector equation r = r(u, v)=x(u, v)  i + y(u, v)  j + z(u, v)  k, with (u, v) D 1 : S =  D 1   n(u, v)   du dv. Here, the unit normal is calculated as n(u, v)=r  u × r  v . Remark. The formulas from Items 3 and 4 are equivalent—they define one and the same surface in two forms, scalar and vector, respectively. 5. Calculation of volumes. If a domain U of the three-dimensional space is defined by  (x, y) D, f (x, y) ≤ z ≤ g(x, y)  ,whereD is a domain in the x, y plane, the volume of U is calculated as V =  D  g(x, y)–f (x, y)  dx dy. The three-dimensional domain U is a cylinder with base D bounded bythe surface z =f(x, y) from below and the surface z = g(x, y) from above. The lateral surface of this body consists of segments of straight lines parallel to the z-axis. 324 INTEGRALS 7.3.3-2. Physical applications of the double integral. Consider a flat plate that occupies a domain D in the x, y plane. Let γ(x, y) be the surface density of the plate material (the case γ = const corresponds to a homogeneous plate). 1. Mass of a flat plate: m =  D γ(x, y) dx dy. 2. Coordinates of the center of mass of a flat plate: x c = 1 m  D xγ(x, y) dx dy, y c = 1 m  D yγ(x, y) dx dy, where m is the mass of the plate. 3. Moments of inertia of a flat plate about the coordinate axes: I x =  D y 2 γ(x, y) dx dy, I y =  D x 2 γ(x, y) dx dy. The moment of inertia of the plate about the origin of coordinates is calculated as I 0 =I x +I y . 7.3.4. Definition and Properties of the Triple Integral 7.3.4-1. Definition of the triple integral. Let a function f(x, y, z)bedefined in a domain U of the three-dimensional space. Let us break up U into n subdomains (cells) that do not have common internal points. Denote by λ = λ(U n )thediameter of the resulting partition U n , i.e., the maximum of the cell diameters (the diameter of a domain in space is the diameter of the minimal sphere enclosing the domain). Select an arbitrary point, (x i , y i , z i ), i = 1, 2, , n, in each cell and make up an integral sum s n = n  i=1 f(x i , y i , z i ) ΔV i , where ΔV i is the volume of the ith cell. If there exists a finite limit of the sums s n as λ(U n ) → 0 that depends on neither the partition U n nor the selection of the points (x i , y i , z i ), then it is called the triple integral of the function f(x, y, z) over the domain U and is denoted  U f(x, y, z) dx dy dz = lim λ→0 s n . 7.3.4-2. Properties of the triple integral. The properties of triple integrals are similar to those of double integrals. 1. Linearity. If functions f (x, y, z)andg(x, y, z) are integrable in a domain U ,then  U  af(x, y, z) bg(x, y, z)  dx dy dz = a  U f(x, y, z) dx dy dz b  U g(x, y, z) dx dy dz, where a and b are any numbers. . r  v . Remark. The formulas from Items 3 and 4 are equivalent—they define one and the same surface in two forms, scalar and vector, respectively. 5. Calculation of volumes. If a domain U of the three-dimensional. sphere of radius R and a right circular cylinder of radius a < R whose axis passes through the sphere center, find the volume of the figure the cylinder cuts out of the sphere. The volume of this. boundary of the domain D(t), n is outer unit normal to L(t), and v is the velocity of motion of the points of L(t). 2 ◦ . If the boundary L(t) is specified by equations in parametric form x =