1 Double Integral. 1.1 Double Integral over Rectangular Domain

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1 Double Integral. Double Integral over Rectangular Domain As the definite integral of a positive function of one variable represents the area of the region between the graph and the x-asis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function z = f(x, y) and the xy-plane which contains its domain. We start with rectangular domain D = {[x, y] R : x a, b, y c, d } on the xy plane according to Fig. We devide interval a, b, (resp. c, d ) by sequences of points a = x < x < Fig.. Double integral on rectangular domain x <... < x m = b, (resp. c = y < y < y <... < y n = d) to intervals x i, x i, i =,,..., m, (resp. y j, y j, j =,,..., n). We denote sizes of each component x i = x i x i, y j = y j y j. This way is the whole rectangular domain divided into m n small rectangles with area D ij = x i y j. Now we can choose an arbitrary point [ξ i, η j ] in each rectangle D ij and we can evaluate the volume of a prism with basis D ij and and height z = f(ξ i, η j ). m n The sum of the volumes f(ξ i, η j ) x i y j represents the volume of the i= j= body consisted of such prisms over all rectangles D ij if f(x, y) on D. Definition.. If there exists lim m i= n f(ξ i, η j ) x i y j for m, n, x i j=

2 , y j for all i =,,..., m, j =,,..., n, we call it double integral of function f(x, y) over the rectangular domain D and denote it f(x, y) dxdy. Theorem.. (Dirichlet s) Let D = {[x, y] R : x a, b, y c, d }. If function f(x, y) is continuous on rectangle D, then b d d b f(x, y) dxdy = f(x, y) dy dx = f(x, y) dx dy. () D D a c c a In fact there are two ways of computing the double integral. If the inner differential is dy then the limits on the inner integral must have y limits of integration and outer integral must have x limits of integration. We compute the integral d c f(x, y) dy by holding x constant and integrating with respect to y as if this were a single integral (similar approach is used for partial derivatives of function of more than one variable). This will result as a function of a single variable x which we can integrate once again. We use similar approach for the second way of computing of the double integral. We usually write b d f(x, y) dy dx = b d dx f(x, y) dy a c a c and d b f(x, y) dx dy = d b dy f(x, y) dx. c a c a Theorem.. (Properties of the double integral over a rectangular domain). cf(x, y) dxdy = c f(x, y) dxdy,. D (f(x, y) + g(x, y)) dxdy = D f(x, y) dxdy + g(x, y) dxdy, D D D

3 3. D f(x, y) dxdy = f(x, y) dxdy + D D f(x, y) dxdy, where f(x, y), g(x, y) are continuous functions on D, c R and D, D are rectangles that fullfil D = D D. Example.. Compute 6xy dxdy over the domain D = {[x, y] : x 4, y }. D Solution: We will show both ways of the computing a) by integrating the inner integral with respect to variable y D 6xy dxdy = 4 = 4 dx 6xy dy = 4 [ ] 4 xy 3 dx = (6x x) dx = 4x dx = [ 7x ] 4 = 8 = 84. b) by integrating the inner integral with respect to variable x D 6xy dxdy = = 4 dy 6xy dx = [ 3x y ] 4 dy = (48y y ) dy = 36y dy = [ y 3] = 96 = 84. If the integrand f(x, y) can be writen as a multiplication of two functions of one variable f(x, y) = f (x) f (y), it holds: b d f(x, y) dxdy = f (x) dx f (y) dy. () D Example.. Compute the integral from Example.. by using equation (). a c Solution: D 4 6xy dxdy = 6 x dx y dy = [ x ] 4 [y 3] = 7 = 84. 3

4 If the decomposition is not possible, we can always use Dirichlet s Theorem and relation (). Example.3. Compute (x 4y) dxdy over the domain D = {[x, y] : x 3, y }. D Solution: D (x 4y) dxdy = = 3 3 dx (x 4y) dy = 3 4x dx = [ x ] 3 = 6. [ xy y ] dx = Although generally the order of integration doesn t matter, in some cases the integral can be easily solved by using one way of integration while it can be rather complicated using the other way. Everything depends on the integrand f(x, y) itself and the limits of integration. Example.4. Compute x y dxdy over the domain D = {[x, y] : x, y }. D Solution: We will use both ways of the computing to show the situation. a) D x y dxdy = dy x y dx = [ x y+ y + ] dy = ( y + ) dy = y + b) D x y dxdy = = dy y + = [ln y + ] = ln 3 ln = ln 3. dx x y dy = [ x y ln x ] dx = ( x ln x x ) dx =... ln x 4

5 The further computation of the integral is tough. Exercises Compute the following integrals over their domains D.. (x + y ) dxdy, D = { x, y }. [4]. D x x + y dxdy, D = { x, y 3}. [ 5 (3 9 3)] 3. D sin(x + y) dxdy, D = { x π, π 4 y π}. [] 4. D D (x + y + ) dxdy, D = { x, y }. [ln 4 3 ]. Double Integral over General Domain There is no reason to limit our problem to rectangular regions. The integral domain can be of a general shape. We extend the Riemann s definition of the double integral over rectangular domain to a closed connected bounded domain without any problem. The domain is connected if we can connect every two points from it by curve that lies within the domain. We can always find a rectangle D that fullfils D (see Fig..) and we can define function f (x, y) by f(x, y) [x, y], f (x, y) = [x, y] D \. Then it holds f(x, y) dxdy = f (x, y) dxdy. D The properties of the double integral over a general domain must correspond to Theorem.. Theorem.3. (Properties of the double integral over a general domain). cf(x, y) dxdy = c f(x, y) dxdy,. (f(x, y) + g(x, y)) dxdy = f(x, y) dxdy + g(x, y) dxdy, 5

6 Fig.. Extension to closed connected domain. 3. f(x, y) dxdy = f(x, y) dxdy + f(x, y) dxdy, where f(x, y), g(x, y) are continuous functions on, c R and, are domains that fullfil =. There are two types of domains we need to loo at. Definition... Normal domain with respect to the x axis is bounded by lines x = a, x = b, where a < b and continuous curves y = g (x), y = g (x), where g (x) < g (x) for all x a, b.. Normal domain with respect to the y axis is bounded by lines y = c, y = d, where c < d and continuous curves x = h (y), x = h (y), where h (y) < h (y) for all y c, d. Theorem.4. (Fubini s). If the function f(x, y) is continuous on a normal domain with respect to the x axis, then it holds f(x, y) dxdy = b dx g (x) f(x, y) dy. (3) a g (x). If the function f(x, y) is continuous on a normal domain with respect to the 6

7 Fig. 3. a) Normal domain with respect to the x axis, b) Normal domain with respect to the y axis y axis, then it holds f(x, y) dxdy = d dy h (y) f(x, y) dx. (4) c h (y) Example.5. Determine integration limits for f(x, y) dxdy over the domain, which is bounded by curves y = x and x = y. Solution: We need to find intersections of curves y = x and x = y by solving the system of these two equations. We can eliminate variable y, receive one equation x 4 x = and solve it. We obtain real values x =, x = and write the domain as a normal domain with respect to the x axis. It is bounded by lines x =, x = and curves y = x and y = x (see Fig. 4.) We can express inequations for in the form: x, x y x and by Fubini s Theorem x f(x, y) dxdy = dx f(x, y) dy. We can use a similar procedure and express the integral as an integral over normal domain with respect to the y axis with inequations y, y x y. 7 x

8 Fig. 4. Domain in Example.5. The double integral then taes form f(x, y) dxdy = dy y f(x, y) dx. Example.6. Determine integration limits for y f(x, y) dxdy over the domain, which is a triangle bounded by equations y =, y = x + 4, y = 6 x. Solution: First, we express the domain as normal with respect to the x axis. If we bound the domain by 3 x 5, the upper limit of inner integral can t be written as one curve and we need to divide the domain into two subdomains, by line x = (see Fig. 5.) with inequations : 3 x, : x 5, y x + 4, y 6 x. Using Fubini s Theorem we can express f(x, y) dxdy = dx x+4 5 f(x, y) dy + dx 6 x f(x, y) dy. 3 However, it is much better to express the domain as normal with respect to the y axis. There is no reason to split the domain which is now bounded by inequations y 5, y 4 x 6 y, 8

9 Fig. 5. Domain in Example.6. where we have expressed a variable x from boundary equations. The integral is written in the form f(x, y) dxdy = 5 dy 6 y f(x, y) dx. y 4 Example.7. Compute xy dxdy over the domain bounded by y = x, y = x, x. Solution: By solving the system of equations y = x, y = x we receive intersections of both curves in x =, x = 4 according to Fig. 6. It is better to express the domain as normal with respect to x axis with boundaries and compute the integral x 4, x y x 4 x xy dxdy = dx 4 [ y xy dy = x x/ ] x x/ dx = 4 ( x x3 8 ) dx = [ x 3 = 6 x4 3 9 ] 4 = 6.

10 Fig. 6. Domain in Example.7. The second approach requires splitting the domain into two subdomains. It is a good exercise to compute the example this way. Example.8. Compute (x y ) dxdy over the domain bounded by y = x, y = x +, y = 3. Solution: According to Fig.7. we compute the integral over the normal domain Fig. 7. Domain in Example.8. with respect to y axis. Limits of the integral: y 3, y x y.

11 Computation: (x y ) dxdy = 3 dy y y (x y ) dx = 3 [ x xy ] y y dy = 3 3 = [(y ) (y )y ( y) + ( y)y ] dy = ( y 3 + y ) dy = = [ y4 + ] 3 3 y3 = Exercises Compute the following integrals over their domains.. (5x xy) dxdy, where is triangle ABC, where A = [, ], B =. 3. [, ], C = [, ]. [3] x dxdy, where is bounded by curves y = 6, y = x, x = 8. [576] x 6xy dxdy, where is bounded by curves y =, x =, y = x. [3] xy dxdy, where is given by x + 4y 4, x, y. [ ] (3x y) dxdy, where is given by x + y 4. [].3 Double Integrals in Polar Coordinates At this moment we are able to compute the double integral over an arbitrary domain. In this section we want to loo at some domains that are easier to describe in terms of polar coordinates. We might have a domain that is a disc, ring or part of a disc or ring. Let us consider a double integral of an arbitrary function over the disc with the center in the origin of coordinates and the radius r = (same

12 domain that is used in last Exercise No.5). Using Cartesian coordinates we obtain limits of the integral x, 4 x y 4 x. and by Fubini s Theorem the integral can be written in the form f(x, y) dxdy = dx 4 x f(x, y) dy. 4 x In these cases using Cartesian coordinates can be tedious. However, we are able to replace Cartesian coordinates x, y by polar coordinates ρ, ϕ, where ρ denotes a distance between the point [x, y] and the origin of coordinates and is called a radius, and ϕ denotes the positively oriented angle between positive part of the x axis and the radius vector and is called angular coordinate or azimuth (see Fig. 8). Hence, we obtain transformation equations Fig. 8. Polar Coordinates ρ, ϕ. x = ρ cos ϕ, y = ρ sin ϕ. (5) Transformation to polar coordinates is a special case of mapping region onto that is an image of in polar coordinates in our case. For example a disc with the center in the origin of coordinates and the radius r =, : {[x, y] : x +y 4}, is mapped onto : {[ρ, ϕ] : ρ (,, ϕ, π)}.

13 Then Theorem.5. Let equations x = u(r, s), y = v(r, s) map the region bijectively to the region. Let function f(x, y) be continuous and bounded on and functions x = u(r, s), y = v(r, s) have continuous partial derivatives on ˆ that fulfills ˆ. Let J(u, v) = u r v r Determinant J(u, v) = u s v s f(x, y)dxdy = u r v r in. u s v s f(u(r, s), v(r, s)) J(u, v) drds (6) is called Jacobian or Jacobi determinant. We will use this theorem for transformation of the double integral to polar coordinates as well as the triple integral to cylindrical and spherical coordinates. According to Theorem.5. we replace square element dxdy by J dρ dϕ, where the Jacobian of the transformation to polar coordinates satisfies J(ρ, ϕ) = x ρ y ρ x ϕ y ϕ = cos ϕ sin ϕ ρ sin ϕ ρ cos ϕ = ρ. Transformation of the double integral to polar coordinates can then be written in the form f(x, y) dxdy = f(ρ cos ϕ, ρ sin ϕ)ρ dρ dϕ. (7) Example.9. Compute y dxdy over the domain : x + y 9, y using transformation to polar coordinates. Solution: The domain is an upper half of the disc with the center in the origin 3

14 of coordinates and radius r = 3. We use transformation to polar coordinates and the domain = {[ρ, ϕ] : ρ (, 3, ϕ, π }. Using (7) we obtain y dxdy = ρ sin ϕ ρ dρ dϕ = 3 π ρ dρ = 8. [ ρ 3 sin ϕ dϕ = 3 ] 3 [ cos ϕ] π = Example.. Compute x dxdy over the domain : 4 x + y 9, y x, x. Solution: We can see real advantage of the transformation on this domain (see Fig. 9). While using Cartesian coordinates would be complicated, domain = {[ρ, ϕ] : ρ, 3, ϕ π 4, π } for polar coordinates is rectangular. Fig. 9. Domain in Example.. x dxdy = ρ cos ϕ ρ dρ dϕ = = 9 3 ρ dρ π/ π/4 ( ). [ ρ 3 cos ϕ dϕ = 3 ] 3 [sin ϕ] π/ π/4 = 4

15 Example.. Calculate limits of the integral transformed to polar coordinates for the domain : x + y ax. Solution: First, we find the center and radius of the disc. x + y ax x ax + a + y a (x a) + y a We have found that center S = [a, ] and radius r = a (see Fig ). The azimuth Fig.. Domain in Example.. must fulfill π ϕ π. We can see that the upper limit of coordinate ρ depends on the azimuth ϕ. We obtain the value of the limit by puting transformation equations (5) to boundary equations of x + y = ax ρ cos ϕ + ρ sin ϕ = aρ cos ϕ ρ = aρ cos ϕ ρ(ρ a cos ϕ) = ρ =, ρ = a cos ϕ. We have just obtained the limits of the integral. However, it is necessary to realise the dependancy of coordinate ρ on coordinate ϕ. We can t no more calculate integrals over such domains as an rectangular ones and we need to use Fubini s 5

16 Theorem. The integral of an arbritrary function can be written as f(x, y) dxdy = π/ dϕ a cos ϕ f(ρ cos ϕ, ρ sin ϕ) ρ dρ. π/ Example.. Compute 4 x 9 y 4 dxdy over the domain : 4x + 9y 36 using transformation to polar coordinates. Solution: The boundary of the domain can be written in the form x 9 + y 4 =. Therefore, the domain is ellipse with center in the origin of coordinates and semi-axis a = 3, b = (see Fig. ). In such case we use generalized polar Fig.. Domain in Example.. coordinates in the form x = aρ cos ϕ For Jacobian of the transformation we obtain x x J(ρ, ϕ) = ρ ϕ = a cos ϕ b sin ϕ y ρ y ϕ y = bρ sin ϕ. (8) aρ sin ϕ bρ cos ϕ = abρ. Using generalized polar coordinates we obtained transformated domain = {[ρ, ϕ] : ρ (,, ϕ, π)} and we can solve the integral now. 4 x 9 y 4 dxdy = (3ρ cos ϕ) (ρ sin ϕ) 4 6ρ dρ dϕ = 9 4 6

17 = 6 4 ρ ρ dρ dϕ = 6 π dϕ = 4π(8 3 3). Integral over coordinate ρ was calculated using substitution 4 ρ = t ρ dρ = dt 4 ρ ρ dρ = 6 π 3 (8 3 3) = Exercises Compute the following integrals over their domains.. ( x 3y) dxdy, : x + y. [π] x y dxdy, : x + y, x, y. [ π 6 ] sin x + y dxdy, : π x + y 4π. [ 6π ] xy dxdy, : x + y 4y, y x. [ 8 3 ] (x + y) dxdy, : 4x + y 6, y, x. [ 3].4 Practical Applications of the Double Integral.4. Volume of a Body According to section., we now that D f(x, y) dxdy of a positive function f(x, y) > over a rectangular domain D has a meaning of the volume of the prism with a rectangular base D bounded from above by the function f(x, y). If we replace rectangle D by general domain, we obtain f(x, y) dxdy (see Fig. ) and we calculate volume of the cylindrical body with basis and bounded from above by the function f(x, y). Therefore, we are able to define the volume of 7

18 the cylindrical body with basis and bounded by an arbitrary function f(x, y) as V = f(x, y) dxdy. (9) Example.3. Calculate the volume of the body bounded by surfaces x + 3y =, z = y, x =, y =, z =. Solution: The basis of the body lies in the plane z =. Planes x + 3y =, x =, y = are perpendicular to the basis, thus they define the triangular domain (see Fig. ). Surface z = y for all [x, y], therefore it bounds the Fig.. Domain of the body in Example.3. body from above. We write the domain as a normal with respect to the x axis with inequations for in the form: x 6, y 4 3 x. Using (9) we compute the volume of the body V = y dxdy = 6 dx 4 3 x y dy = 6 [ y 3 3 ] 4 3 x dx = = 6 6 ( 4 ) 3 3 x dx = ( 6 3 ) [ (4 3 x)4 4 ] 6 = 6. 8

19 Exercises Compute the volumes of the bodies bounded by surfaces:. z =, z = xy, y =, y = x, x + y =. [ 3 8 ]. z =, y = x, z = 4 y. [ 56 ] 3. z =, z = x y. [ π ].4. Area of a Region Let us consider a cylindrical body with a domain bounded from above by a constant function f(x, y) =. The volume of such a body as well as the area of a region can be expressed as A = dxdy. () Example.4. Calculate the area of a region bounded by curves y = x, y = 4 x. Solution: We need to find intersections of both parabolas y = x, y = 4 x that are x = ±. We write the domain as a normal with respect to the x axis with Fig. 3. Domain of the area in Example.4. inequations in the form (see Fig. 3): x, x y 4 x. 9

20 Using () we compute the area of our region using symmetry of the domain with respect to the y axis A = dxdy = 4 x dx dy = (4 x ) dx = [4x 3 ] x3 x = 6. 3 Exercises Compute the areas of the regions bounded by surfaces:. y = x, y = 5x, x =. []. y = x, y = x, y = 4. [ 9 ln 4 ] 3. x + y = 4, x + y = 4y. [ 8 3 π 3].4.3 Surface Area We are able to compute the area of the surface z = f(x, y) where [x, y] is a point in the region. Function z = f(x, y) must have continuous partial derivatives on. In this case surface area is given by S = + ( ) z + x ( ) z dxdy. () y Example.5. Calculate the area of a surface x + y + z = 4 bounded by planes x =, y =, x =, y =. Solution: We express the surface in the form z = 4 x y and hence both partial derivatives z z =, =. According to Fig. 4 the domain is x y a rectangle given by inequations x, y and we calculate the surface area ( ) ( ) z z S = + + dxdy = + ( ) + ( ) x y dxdy = = 3 dx dy = 3[x] [y] = 4 3.

21 Fig. 4. Domain of the area in Example.5. Exercises Compute the areas of the surfaces given by:. z = xy bounded by planes x =, y =, x =, y = 4. [ 6 ] 3. y + z = 9 bounded by planes x =, x =, y = 3, y = 3. [6π] 3. z = xy bounded by cylinder x + y = 4. [ 3 π(5 5 )].4.4 Center of Mass Let σ(x, y) be a surface density defined for each [x, y]. The mass of a domain is defined by m = σ(x, y) dxdy. Static moment of the domain with respect to the x axis (resp. y-axis) is given by S x = y σ(x, y) dxdy, resp. S y = x σ(x, y) dxdy. The coordinations of the center of mass C = [ξ, η] can be expressed as ξ = S y m, η = S x m. ()

22 Example.6. Calculate the coordinates of the center of mass of homogeneous disc bounded by equation (x ) + y =. Solution: The disc is homogeneous with constant surface density σ. All integrals for the calculation of the mass of the disc and its static moments will be transformed to the polar coordinates with the same limits π ϕ π, ρ cos ϕ. We now that the center of mass of the homogeneous disc must be in the center of the disc. We will prove it by computing the mass of the disc m = σ(x, y) dxdy = σ ρ dρ dϕ = σ π/ dϕ cos ϕ ρ dρ = π/ = σ π/ π/ [ ρ ] cos ϕ and both static moments. dϕ = σ π/ π/ cos ϕ dϕ = σ = σ [ϕ + ] π/ sin ϕ π/ π/ π/ = σπ ( + cos ϕ) dϕ = S x = y σ(x, y) dxdy = σ ρ sin ϕ ρ dρ dϕ = σ π/ sin ϕ dϕ cos ϕ ρ dρ = π/ = σ π/ π/ [ ρ 3 sin ϕ 3 ] cos ϕ dϕ = 8 π/ 3 σ π/ sin ϕ cos 3 ϕ dϕ =. The last integral is an integral of an odd function on a symetric interval, therefore must be equal to. S y = x σ(x, y) dxdy = σ ρ cos ϕ ρ dρ dϕ = σ π/ cos ϕ dϕ cos ϕ ρ dρ = π/

23 = 8 π/ 3 σ π/ = σ π/ π/ [ ρ 3 cos ϕ 3 ] cos ϕ [ ] ( + cos ϕ) dϕ = 3 σ = π/ 3 σ π/ dϕ = 8 π/ 3 σ π/ π/ π/ cos 4 ϕ dϕ = ( + cos ϕ + cos ϕ) dϕ = ( + cos ϕ + + cos 4ϕ) dϕ = = 3 σ [ 3 ϕ + sin ϕ + 8 sin 4ϕ ] π/ π/ = 3 σ 3 π = σπ. Now, we can use () to show that ξ = σπ σπ =, η = σπ hypothesis. C = [, ]. = to prove our Exercises Compute the coordinates of the center of mass of the homogeneous regions bounded by curves:. y = x, y = x [C = [, 5 ]]. y = x, x = 4, y = [C = [3, 4 5 ]] Triple Integral. Triple Integral over a Rectangular Hexahedron Now, that we now how to integrate over a two-dimensional region, we need to move on to the integration over a three-dimensional regions. While we used a double integral to integrate over a two-dimensional regions, we use triple integral to integrate over three-dimensional regions. The definition of the three-dimensional integral over a rectangular hexahedron is similar to Definition.. we used for double integral over a rectangle. Let u = f(x, y, z) is a function of three variables that is continuous and bounded on the rectangular hexahedron G = {[x, y, x] R 3 : x a, b, y c, d, z e, h }. We devide intervals a, b, c, d, e, h by three sequences of points a = x < x < x <... < x m = b, c = y < y < y <... < 3

24 y n = d and e = z < z < z <... < z p = h to intervals x i, x i, i =,,..., m, y j, y j, j =,,..., n, z, z, =,,..., p. We denote x i = x i x i, y j = y j y j, z = z z. The planes that lead through points x i (resp. y j or z ) parallel to coordinate planes devide the hexahedron G to m n p small hexahedrons G ij (see. Fig. 5) with volume of each G ij = x i y j z. We choose an arbitrary point [ξ i, η j, ζ ] in each hex- Fig. 5. Triple integral on a rectangular hexahedron ahedron G ij and we create products f(ξ i, η j, ζ ) G ij = f(ξ i, η j, ζ ) x i y j z that for positive function f(x, y, z) has a physical meaning of the mass of the hexahedron G ij with density f(ξ i, η j, ζ ). The sum of these products m n p f(ξ i, η j, ζ ) x i y j z represents the mass of the body consisted i= j= = of such hexahedrons. Definition.. If there exists lim m i= n j= = p f(ξ i, η j, ζ ) x i y j z for m, n, p, x i, y j, z for all i =,,..., m, j =,,..., n, z =,,...,, we call it a triple integral of function f(x, y, z) over the rectangular hexahedron G and denote it by f(x, y, z) dxdydz. The triple integral over a hexahedron G of a positive function f(x, y, z) has a meaning of the mass of a hexahedron G with density f(x, y, z). Theorem.. (Dirichlet s) Let G = {[x, y, z] R 3 : x a, b, y c, d, z e, h }. If a function G 4

25 f(x, y, z) is continuous on the hexahedron G, then b d h f(x, y, z) dxdydz = f(x, y, z) dz dy dx. (3) G a c e The Theorem is similar to two-dimensional Dirichlet s Theorem.. We can rewrite (3) by using a different order of integration in five more ways. The triple integral is then converted to three one-dimensional integrals. Similarly to the double integral, we can write b d h b d h f(x, y, z)dz dy dx = dx dy f(x, y, z)dz. (4) a c e a c e Theorem.. (Properties of the triple integral over a rectangular hexahedron). cf(x, y, z) dxdydz = c f(x, y, z) dxdydz,. G (f(x, y, z) + g(x, y, z)) dxdydz = G = f(x, y, z) dxdydz + g(x, y, z) dxdydz, G 3. G f(x, y, z) dxdydz = G = f(x, y, z) dxdydz + G G G f(x, y, z) dxdydz, where f(x, y, z), g(x, y, z) are continuous functions on G, c R and G, G are hexahedrons that fullfil G = G G. If the integrand f(x, y, z) can be writen as a product of three functions of one variable f(x, y, z) = f (x) f (y) f 3 (z), it holds: G f(x, y, z)dxdydz = b a f (x)dx 5 d c f (y)dy h e f 3 (z)dz. (5)

26 Example.. Compute xy zdxdydz over the rectangular hexahedron G = {[x, y, z] : G x,, y, 3, z, }. Solution: G xy zdxdydz = = x dx ( 9 ) 3 3 y dy ( ) [ x z dz = ] [ y 3 3 = = 6. ] 3 [ z ] = If it is not possible to decompose the integrand using (5), we can always use Dirichlet s Theorem and formula (3). Example.. Compute (x + y) dxdydz over the rectangular hexahedron G = {[x, y, z] : G x,, y,, z, 3 }. Solution: (x + y) dxdydz = dx dy 3 (x + y) dz = dx (x + y) [z] 3 dy = G = 3 dx (x+y) dy = 3 ] [xy + y dx = 3 = 6 3 = 9. [ ] x (x+) dx = 6 + x = Exercises Compute the following integrals over their domains G.. ( x + y + 3 ) dxdydz, G = {x,, y,, z, }. z G [6 ln ] 6

27 . 3. G G x y dxdydz, G = {x,, y, 5, z, 4 }. [ ln 4 5 ] xy z dxdydz, G = {x,, y, 3, z, 4 }. [ 5 3 ( 4)]. Triple Integral over a General Domain Similarly lie in two-dimensional case, we are able to generalize our problem of solving triple integrals over any three-dimensional region that is bounded by a closed surface. We consider only such surfaces that don t intersect themselves and lines parallel with z axis leading through an arbitrary inner point of the surface intersect with the surface in two points. Such domain will be called normal domain with respect to the coordinate plane x, y (see Fig. 6). We create an orthogonal projection of the domain into coordinate plane Fig. 6. Normal domain with respect to coordinate plane x, y x, y. A variable z must fulfill f (x, y) z f (x, y). The domain is either normal with respect to x axis or y axis and we describe it using approach from section. by inequalities x x x, g (x) y g (x) (resp. y y y, h (y) x h (y)). If the function f(x, y, z) is continuous on we use a method similar to Fubini s Theorem.4 and express f(x, y, z) dxdydz = x dx g (x) dy f (x,y) f(x, y, z) dz (6) x g (x) f (x,y) 7

28 or f(x, y, z) dxdydz = y dy h (y) dx f (x,y) f(x, y, z) dz (7) y h (y) f (x,y) We start with integration over variable z with bounds that are functions of two variables x, y and after that we calculate a double integral over a regular domain. We are able to use analogical approach and create an orthogonal projection of the domain into coordinate planes either x, z or y, z. That way we can use six different orders of integration for an arbitrary domain. The triple integral over a general closed domain has analogical properties as the triple integral over a rectangular hexahedron. Theorem.3. (Properties of the triple integral over a general domain). cf(x, y, z) dxdydz = c f(x, y, z) dxdydz,. (f(x, y, z) + g(x, y, z)) dxdydz = f(x, y, z) dxdydz + g(x, y, z) dxdydz, 3. f(x, y, z) dxdydz = = f(x, y, z) dxdydz + f(x, y, z) dxdydz, where f(x, y, z), g(x, y, z) are continuous functions on G, c R and, are domains that fullfil =. Example.3. Determine integration limits for f(x, y, z) dxdydz over the domain that is bounded by surfaces z = (x + y ), z = 4 (x + y ). Solution: Both surfaces are paraboloids and it is obvious that (x + y ) z 4 (x + y ). We can see the projection of the domain to coordinate plane y, z on Fig. 7. We obtain an equation of the intersection of both surfaces from 8

29 Fig. 7. Projection of the Domain to coordinate plane y, z in Example.3. the equation (x + y ) = 4 (x + y ) which leads to x + y = 4. Therefore, the orthogonal projection of the domain to coordinate plane x, y is a ring with the center in the origin of coordinates and radius r =. That domain can be treated as a normal domain with respect to the x axis with inequations x, 4 x y 4 x as well as a normal domain with respect to the y axis with inequations y, 4 y x 4 y. Example.4. Compute x dxdydz over the domain bounded by surfaces x =, y =, z =, x + y + z 6 =. Solution: The domain must fulfill z 6 x y (see Fig. 8). The Fig. 8. Domain in Example.4. orthogonal projection of the domain to coordinate plane x, y is the triangle 9

30 in plane z = bounded by lines x =, y =, y = 3 x. The last equation is obtained from x + y + z 6 = by letting z =. We describe as normal with respect to x axis by inequations x 3, y 3 x and we calculate the integral using (6). 3 x dxdydz = dx 3 x dy 6 x y x dz = 3 x dx 3 x [z] 6 x y dy = = 3 = 3 x dx 3 x (6 x y) dy = 3 x [ 6(3 x) x(3 x) (3 x) ] dx = x [ 6y xy y ] 3 x dx = 3 (x 3 6x + 9x) dx = [ x 4 = 4 x3 + 9 x ] 3 = 7 4. Exercises Compute the following integrals over their domains.. x 3 y z dxdydz, = {y, y x, x, z, z xy}. [ ]. 3. dxdydz, = {x, y, z, x + y + z }. + x + y [ 3 ln ] y cos(x + z) dxdydz, = {y x, y, z, x + z π } [ π 6 ] dxdydz, = {x + y, y, y x, z, z x }. [ 4 6 ] dxdydz, = {z 3x + 3y, z x y }. [ π 8 ] 3

31 .3 Transformation of the Triple Integral Similarly to the double integral, using Cartesian coordinates for some domains can be rather complicated. Especially in case of cylinders, cones or spheres. We formulate theorem analogical to Theorem.5. that describes general transformation of the Triple Integral. Theorem.4. Let equations x = u(r, s, t), y = v(r, s, t), z = w(r, s, t) map the region bijectively to the region. Let function f(x, y, z) be continuous and bounded on and functions x = u(r, s, t), y = v(r, s, t), z = w(r, s, t) have continuous partial derivatives on ˆ that fulfills ˆ. Let J(u, v, w) = u r v r w r u s v s w s u t v t w t in. Then f(x, y, z) dxdydz = f(u(r, s, t), v(r, s, t), w(r, s, t)) J drdsdt (8) is again called Jacobian or Jacobi de- Determinant J(u, v, w) = terminant. u r v r w r u s v s w s u t v t w t.3. Transformation to Cylindrical Coordinates Trasformation to cylindrical coordinates is suitable for integration domains such as cylinders, cones, their parts and it is used in cases when orthogonal projection of the domain to coordinate plane x, y is a disc or a part of a disc. We replace Cartesian coordinates x, y, z by cylindrical coordinates ρ, ϕ, z (see Fig. 9). The meaning of coordinates ρ, ϕ is the same as we have already used for polar coordinates and the third coordinate z doesn t change. Therefore, the transformation to 3

32 Fig. 9. Transformation to cylindrical coordinates cylindrical coordinates is given by transformation equations x = ρ cos ϕ y = ρ sin ϕ (9) z = z According to Theorem.4. we replace volume element dxdydz by J dρdϕdz, where the Jacobian of the trasformation to cylindrical coordinates satisfies x x x cos ϕ ρ sin ϕ ρ ϕ z J(ρ, ϕ, z) = y y y = sin ϕ ρ cos ϕ = ρ. ρ ϕ z z ρ z ϕ z z The transformation of the triple integral to cylindrical coordinates can then be written in the form f(x, y, z)dxdydz = f(ρ cos ϕ, ρ sin ϕ, z)ρ dρ dϕ dz. () Example.5. Compute dxdydz over the domain bounded by surfaces x +y =, z =, z =. Solution: The domain is the cylinder symetrical with respect to the z axis, with radius of the base ρ = and height z = (see Fig. ). We need to determine the bounds of the transformated domain. It is obvious that z. Inequations 3

33 Fig.. Domain in Example.5. for coordinates ρ, ϕ are the same as for transformation to polar coordinates, i.e. ρ, ϕ π. Therefore dxdydz = ρ dρ dϕ dz = π dϕ ρ dρ z dz = = [ϕ] π [ ρ ] [z] = π. Example.6. Compute dxdydz over the domain bounded by surfaces z 3x + 3y, z x y. Solution: The domain is similar to the domain in Example.3. Both surfaces are paraboloids and the orthogonal projection of the domain to coordinate plane x, y is a ring. We obtain its equation from the intersection of both paraboloids 3x + 3y = x y x + y = 4. Therefore, inequations for coordinates ρ, ϕ must fulfill ρ, ϕ π. We obtain limits of the variable z from equations of both paraboloids by substituting of variables z = 3x + 3y = 3ρ cos ϕ + 3ρ sin ϕ = 3ρ, 33

34 z = x y = ρ cos ϕ ρ sin ϕ = ρ. Hence, 3ρ z ρ and we compute the integral by using () dxdydz = ρ dρ dϕ dz = π dϕ / ρ dρ ρ 3ρ dz = = π / ρ [z] ρ 3ρ dρ = π / [ ρ = π ρ4 / ρ( 4ρ ) dρ = π (ρ 4ρ 3 ) dρ = ] / = π 6 = π 8. Exercises Compute the following integrals over their domains.. z dxdydz, = {x + y 4, z, y }. [4π]. dxdydz, = {x + y, x, z 6}. [3π] z x + y dxdydz, = {x + y x, z }. [ 6 9 ] z dxdydz = {z x + y, z }. [ π 4 ].3. Transformation to Spherical Coordinates Transformation to spherical coordinates is suitable for integrals where integration domains are spheres, ellipsoids or their parts or in cases when orthogonal projection of the domain to coordinate plane x, y is a disc or a part of a disc. We replace Cartesian coordinates x, y, z by spherical coordinates ρ, ϕ, ϑ (see Fig. ). The coordinate ρ denotes a distance between the point [x, y, z] and the origin of the coordinates, ϕ denotes positively oriented angle in coordinate plane x, y between positive part of the x axis and the projection ρ of the radius vector ρ to coordinate plane x, y and ϑ denotes positively oriented angle between positive 34

35 Fig.. Transformation to spherical coordinates part of the z axis and the radius vector ρ. We obtain transformation equations from Fig. x = ρ cos ϕ sin ϑ y = ρ sin ϕ sin ϑ () z = ρ cos ϑ Jacobian of the transformation to spherical coordinates satisfies x x x cos ϕ sin ϑ ρ sin ϕ sin ϑ ρ cos ϕ cos ϑ ρ ϕ ϑ J = y y y = sin ϕ sin ϑ ρ cos ϕ sin ϑ ρ sin ϕ cos ϑ ρ ϕ ϑ cos ϑ ρ sin ϑ z ρ z ϕ z ϑ = ρ sin ϑ. Transformation of the triple integral to spherical coordinates can be written in the form f(x, y, z)dxdydz = = f(ρ cos ϕ sin ϑ, ρ sin ϕ sin ϑ, ρ cos ϑ) ρ sin ϑ dρ dϕ dϑ. () The sphere with the center in the origin of coordinates and radius a transformed to spherical coordinates is mapped to domain given by inequations ρ a, ϕ π, ϑ π. Example.7. Compute (x + y + z ) dxdydz over the domain bounded by surfaces 35

36 x + y + z, x + y + z 4. Solution: The domain is bounded by two spherical surfaces with the center in the origin of coordinates and radiuses a =, b =. We use transformation to spherical coordinates with inequations ρ, ϕ π, ϑ π and calculate the integral by using () (x + y + z ) dxdydz = = (ρ cos ϕ sin ϑ + ρ sin ϕ sin ϑ + ρ cos ϑ)ρ sin ϑ dρ dϕ dϑ = = Example.8. Compute dρ π [ ρ 5 = 5 π dϕ ] 4, x, y, z. ρ 4 sin ϑ dϑ = ρ 4 dρ π π dϕ sin ϑ dϑ = [ϕ] π [ cos ϑ]π = 3 4 π = 5 5 π. z dxdydz over the domain bounded by surfaces x + y + z Solution: The domain is one eighth of the sphere in the first octant with the center in the origin of coordinates and radius a =. Therefore, ρ, ϕ π, ϑ π and we can calculate the integral z dxdydz = ρ cos ϑ ρ sin ϑ dρ dϕ dϑ = = dρ π/ dϕ [ ρ 4 4 π/ ] ρ 3 sin ϑ cos ϑ dϑ = ρ 3 dρ π/ dϕ π/ [ϕ] π/ [ 4 ] π/ cos ϑ = 4 π = π. Exercises Compute the following integrals over their domains. sin ϑ dϑ = 36

37 . z(x + y ) dxdydz, = {x + y + z, x, y, z }.. 3. }. [ π 48 ] x + y + z dxdydz, = {x + y + z, x, y, z (x + y ) dxdydz, = {4 x + y + z 9, z }. [ π] [ π 8 ].4 Practical Applications of the Triple Integral.4. Volume of a Body As we have already mentioned in section.., the physical meaning of the triple integral of the function f(x, y, z) over the domain is the mass of the domain with the density σ = f(x, y, z). If σ =, the triple integral has a meaning of volume of the body V = dxdydz (3) Example.9. Compute the volume of the body bounded by cylindrical surfaces z = 5 y, z = y + 3 and planes x =, x =. Solution: While limits of variables x, z are obvious, we need to calculate intersections of cylindrical surfaces to determine the limits of variable y. 5 y = y + 3 y = y = ± Therefore, inequations for the domain are in the form x y y + 3 z 5 y 37

38 We calculate volume of the body using (3) V = dxdydz = dx dy 5 y y +3 dz = [x] [z] 5 y y +3 dy = = ( y ) dy = 4 ] [y y3 = = 6 3. Exercises Compute the volumes of bodies bounded by their surfaces.. z = x + y, z = y. [ π 3 ]. x y + z = 6, x + y =, x = y, y =, z =. [ 6 3 ] 3. y = x, z =, y + z =. [ 3 5 ].4. Mass of a Body Mass of the domain is given by m = σ(x, y, z) dxdydz, (4) where σ(x, y, z) denotes volume density in each point of the domain. Example.. Compute the mass of the body bounded by spherical surfaces x + y + z =, x + y + z = 4 with the density defined by σ = x + y + z. Solution: The problem was solved in Example.7. Exercises Compute the mass of bodies.. x + y + z 4, σ = x + y + z. [6π]. x + y + z, σ = x + y + z. [8π] 3. x = y, y + z =, y + z =, σ = y. [ 8 35 ] 38

39 3 Line Integral 3. The Curve and Its Orientation To study the line integral we must be first able to express analytically its domain (the curve in two- or three-dimensional space) and implement orientation of the curve. Definition 3.. Let x = x(t), y = y(t), z = z(t) be continuous functions for t a, b.. The curve with parametrical equations x = x(t), y = y(t), z = z(t), t a, b is called positively oriented with respect to the parameter t, if and only if its points are ordered so that for arbitrary values t, t a, b, t < t, the point M = [x(t ), y(t ), z(t )] lies before the point M = [x(t ), y(t ), z(t )], i.e. Reversely, t, t a, b : t < t M M. t, t a, b : t < t M M, the curve is called negatively oriented with respect to the parameter t.. If the curve is positively oriented with respect to the parameter t a, b, then the point A = [x(a), y(a), z(a)] is called the starting point of the curve and the point B = [x(b), y(b), z(b)] is called the end point of the curve. 3. The curve is called closed, if A = B. 4. The curve is called smooth in a, b, if there exists continuous derivatives of parametrical equations x = x (t), y = y (t), z = z (t) and t a, b : (x (t), y (t), z (t)) (,, ). 5. The curve is called piecewise smooth in a, b, if it is smooth in a, b except for a finite number of points t i a, b, i =,..., n. 6. The curve is called simple in a, b, if it doesn t intersect itselfs, i.e. t, t (a, b) : t t M M. The symbol means "precedes" or "lies before". Example 3.. Write parametrizations of 39

40 . the line segment between points A = [a, a, a 3 ], B = [b, b, b 3 ],. the circle x + y = r, r >, 3. the circle (x m) + (y n) = r, r >, 4. the ellipse x a + y =, a, b >. b Solution:. We should now from the analytical geometry that the parametrical equations of the line segment between points A = [a, a, a 3 ], B = [b, b, b 3 ] are in the form x = a + u t, y = a + u t, t,. z = a 3 + u 3 t, where u = AB = (u, u, u 3 ) is the vector parallel to the line segment AB.. Parametrical equations of the circle with the center in the origin of coordinates and radius r are in the form x = r cos t, y = r sin t, t, π). 3. Analogically for the circle with the center in the point C = [m, n] and radius r, the parametrical equations are in the form x = m + r cos t, y = n + r sin t, t, π). 4. Parametrical equations of the ellipse with the center in the origin of coordinates and semi-axis a, b are in the form x = a cos t, y = b sin t, t, π). 4

41 Fig.. Dividing of the curve - the domain of the line integral 3. Line Integral of a Scalar Field Analogically to the other types of integral, we need to divide our domain - the curve - into small elements. Let us consider the simple smooth curve with parametrization x = x(t), y = y(t), z = z(t), t a, b positively oriented with respect to the parameter t. We divide interval a, b by sequence of points a = t < t < < t n = b into n partial curves,,..., n (see Fig. ). For each i =,..., n, we denote by s i the length of each element i and we choose an arbitrary point M i = [x(t i ), y(t i ), z(t i )] in each element i. Moreover, we are able to construct positively oriented unitary tangential vector τ i (M i ) at each point M i that we will use in section 3.3. to define the line integral of a vector field. The curve lies within the domain and we consider a bounded continuous scalar function u(x) = u(x, y, z) defined for each X. Now we can create the sum of products n u(m i ) s i = i= n u(x(t i ), y(t i ), z(t i )) s i i= and define the line integral of a scalar field. Definition 3.. If there exists lim n u(x(t i ), y(t i ), z(t i )) s i for n and s i we call i= it the line integral of a scalar field u(x, y, z) along the curve and denote it u(x, y, z) ds. The line integral of the scalar field doesn t depend on the orientation of the 4

42 curve, because the lengths of all components s i are always positive. The element of the curve ds in the three-dimensional space is the body diagonal of the rectangular hexahedron with sides dx, dy, dz. Therefore we obtain ds = (dx) + (dy) + (dz) = (ẋ(t)dt) + (ẏ(t)dt) + (ż(t)dt) = = (ẋ(t)) + (ẏ(t)) + (ż(t)) dt and the line integral of the scalar field can then be written in the form b u(x, y, z) ds = u(x(t), y(t), z(t)) (ẋ(t)) + (ẏ(t)) + (ż(t)) dt. (5) a That way we transform the line integral of the scalar field into the one-dimensional definite integral. Theorem 3.. (Properties of the line integral of a scalar field). cu(x)ds = c u(x)ds,. (u(x) + v(x))ds = u(x)ds + v(x)ds, 3. u(x)ds = u(x)ds + u(x)ds, where c R,, are curves such that curve fulfills = and u(x), v(x) are bounded continuous scalar functions for all X that contains the curve. Example 3.. Calculate the line integral (x + z) ds along the line segment between points A = [,, 3], B = [3,, ]. Solution: We create the parametrization of the line segment x = + t y = t, z = 3 t 4

43 and its derivatives: ẋ =, ẏ =, ż =. We express the element ds ds = (ẋ(t)) + (ẏ(t)) + (ż(t)) dt = 8 dt = dt. We use (5) to calculate the line integral (x + z) ds = ( + t + 3 t) dt = 8 dt = 8. If we consider just the two-dimensional problem, i.e. the curve is in x, y plane, the element ds = (ẋ(t)) + (ẏ(t)) dt and the line integral of the scalar field is then in the form b u(x, y) ds = u(x(t), y(t)) (ẋ(t)) + (ẏ(t)) dt. (6) a Example 3.3. Calculate the line integral ds, where is the first arc of cycloid x = a(t sin t), y = a( cos t), a >, t, π. Solution: We express derivatives ẋ = a( cos t), ẏ = a sin t and the element of the curve ds = (ẋ(t)) + (ẏ(t)) dt = a ( cos t) + a sin t dt = = a cos t + cos t + sin t dt = a cos t dt = a cos t dt = = a Now, we calculate the integral using (6) sin t dt = a sin t dt. ds = π a sin t [ dt = a cos t ] π = 4a ( ) = 8a. 43

44 Suppose the curve in x, y plane can be expressed explicitly by y = y(x), x a, b. The element of the curve is then given by ds = (dx) + (dy) = (dx) + (y (x)dx) = + (y (x)) dx and the line integral is then in the form b u(x, y) ds = u(x, y(x)) + (y (x)) dx. (7) a Example 3.4. Calculate the line integral y ds, where is a part of the function y = x 3 between points A = [, ], B = [, ]. Solution: We calculate a derivative of the function an element of the curve and finally the integral by using (7) y = 3x, ds = + (3x ) dx = + 9x 4 dx y ds = x 3 + 9x 4 dx = 36 t dt = 36 [ t 3] = 3 54 ( ), where we used substitution + 9x 4 = t, 36x 3 dx = dt. Exercises Compute the line integrals of scalar fields along given curves. z. x + y ds, is one thread of the spiral x = cos t, y = sin t, z = t. x ds, is a line segment between points A = [, ], B = [, ] [ 44 [ 8 3 π 3 ] 5 ]

45 3. x ds, is an upper half of the circle x + y = a, a > [ πa3 ] 4. x ds, : y = ln x, x, 3 [ 3 ( )] 3.3 Line Integral of a Vector Field In the previous section we looed at line integral of a scalar field u(x, y, z). In this section we are going to evaluate line integral of vector fields. We consider simple smooth curve with the same parametrization and division to small elements as we have already discribed in section 3.. Furthermore, we consider bounded continuous vector field F(X) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z) defined for each X. We create scalar products of the vector field and the positively oriented tangential vector of each element of the curve F(M i ) s i, where s i = s i τ i. Now we can create sum of such products n F(M i ) s i = i= n F(x(t i ), y(t i ), z(t i )) s i τ i i= and define the line integral of a vector field. Definition 3.. n If there exists F(x(t i ), y(t i ), z(t i )) s i τ i for n and s i, we call i= it the line integral of a vector field F(x, y, z) along the curve + and denote it F(x, y, z) ds, where + denotes positively oriented curve. While would + denote negatively oriented curve. The line integral of the vector field depends on the orientation of the curve, because coordinates of the unitary tagential vectors τ i depend on the orientation of the curve. To derive the form of the line integral of the vector field, we need to express the scalar product F(x, y, z) ds = (P (x, y, z), Q(x, y, z), R(x, y, z)) (dx, dy, dz) = = P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz and the differentials dx = ẋ(t) dt, dy = ẏ(t) dt, dz = ż(t) dt by using parametrization of the curve. The line integral of the vector field then can be written in the 45

46 form b F(x, y, z) ds = ε [P (x(t), y(t), z(t))ẋ(t) + Q(x(t), y(t), z(t))ẏ(t)+ a +R(x(t), y(t), z(t))ż(t)] dt, (8) where ε = in case of positively oriented curve with respect to the parameter t, while ε = in case of negatively oriented curve. This way we transform the line integral of the vector field into the one-dimensional definite integral, similarly to the case of the line integral of a scalar field. Theorem 3.. (Properties of the line integral of a vector field). cf(x)ds = c F(X)ds,. (F(X) + G(X))ds = F(X)ds + G(X)ds, F(X)ds = F(X)ds + F(X)ds = F(X)ds. F(X)ds, + where c R,, are curves such that curve fulfills = (considering the same orientation of these curves) and F(X), G(X) are bounded continuous vector functions for all X that contains the curve. Example 3.5. Calculate the line integral of the vector field F = (x, y, z) along the curve, that is one thread of the spiral x = cos t, y = sin t, z = 3t, t, π. Solution: We express derivatives of parametrization ẋ = sin t ẏ = cos t ż = 3 and we can calculate the integral by using (8) F(x, y, z)ds = x dx + y dy + z dz = 46

47 = π = π [ cos t ( sin t) + sin t cos t + 3t 3] dt = [ 4 sin t cos t + 4 sin t cos t + 9t] dt = π 9t dt = 9 [ t ] π = 8π. If we consider the two-dimensional problem, i.e. the curve is in x, y plane, the vector field F = (P (x, y), Q(x, y)) and the tangential vector of the element ds = (dx, dy) = (ẋ(t)dt, ẏ(t)dt). The line integral of the vector field is then in the form b F(x, y) ds = ε [P (x(t), y(t))ẋ(t) + Q(x(t), y(t))ẏ(t)]dt (9) a Suppose the curve in x, y plane is expressed explicitly by y = y(x), x a, b. The tagential vector of the curve is given by ds = (dx, dy) = (dx, y (x) dx) and the line integral of a vector field is then in the form b F(x, y) ds = ε [P (x, y(x)) + Q(x, y(x))y (x)] dx (3) a Example 3.6. Calculate the line integral (x+y) dx+(x y) dy, where : y =, x, 3. x Solution: The curve is given explicitly y = x, dy = x dx. We calculate the integral by using (3) 3 (x + y) dx + (x y) dy = [(x + x ) + (x x ) ( x ] ) dx = 47

48 3 = [x + x x + x ] dx = 3 3 (x + [ x x ) dx = 3 ] 3 = x = = Exercises Compute the line integrals of vector fields along given curves.. x dx y dy+z dz, is an oriented line segment AB : A = [,, ], B = [4, 3, ] [5] y dx + x dy, is one quarter of the circle x = a cos t, y = a sin t, a >, t, π [] (xy ) dx + x y dy, is an arc of the ellipse x = cos t, y = sin t from the starting point A = [, ] to the end point B = [, ]. [ 4 3 ] xy dx + (y x) dy, is a part of the parabola y = x from the starting point A = [, ] to the end point B = [, ]. [ 7 3 ] 3.4 Green s Theorem Green s Theorem expresses the relation between a line integral of a vector field along a plane (two-dimensional) closed curve and a double integral. Definition 3.3. Let be a domain in a plane bounded by a simple closed curve. The curve is orientated positively with respect to the domain if traveling on the curve we always have the domain on the left side (see Fig. 3). Positive orientation of the curve means traveling in a counterclocwise direction, while negative orientation means traveling in a clocwise direction. We denote the line integral along a closed curve by the symbol. 48

49 Fig. 3. Positively orientated curve with respect to the domain Theorem 3.3. (Green s) Let two-dimensional vector field F(x, y) = P (x, y)i + Q(x, y)j have continuous derivatives on the plane domain, which is bounded by simple piecewise smooth closed curve orientated positively with respect to the domain. Then P (x, y)dx + Q(x, y)dy = [ Q(x, y) x P (x, y) y ] dxdy. (3) The Green s Theorem transforms the line integral of a vector field along a plane closed curve to a double integral over the domain, that is bounded by the curve. It is especially useful in situations when is a polygon and we have to calculate as many line integrals as the lines the polygon consists of. Example 3.7. Calculate the integral (xy 5y) dx + (x + y) dy, where is the circle with the center in the origin of coordinates and radius r. Solution: The disc x + y r is two-dimensional domain that is normal with respect to both coordinate axis. The curve is simple and closed and we can orientate it positively with respect to. Also both functions P (x, y) = xy 5y and Q(x, y) = x + y fulfill assumptions of the Green s Theorem. We calculate derivatives P y = x 5 Q x = x and evaluate the integral by using (3) (xy 5y) dx+(x +y) dy = (x (x 5)) dxdy = 5 dxdy = 5πr. 49

50 Fig. 4. Curve and domain in Example 3.8. Example 3.8. Calculate the integral (x + y ) dx + (x + y) dy, where consists of the sides of the triangle ABC: A = [, ], B = [, 3], C = [3, 3]. Solution: All assumptions of the Green s Theorem are fulfilled. P = x + y, Q = (x + y), P y = y, Q x = (x + y). The domain is shown on the Fig. 4. We calculate the double integral as a normal one with respect to the x axis with inequations for in the form By using (3) we obtain (x + y ) dx + (x + y) dy = 3 = x[y] 3 x dx = 3 x 3, x y 3. 3 (x + y y) dxdy = [ ] 3 3 (3x x ) dx = x x3 3 3 x dx = 3. x dy = Exercises Compute the line integrals of vector fields along given curves by using the Green s Theorem. 5

51 . (x + y ) dy, are sides of the rectangle x =, y =, x =, y = [6] y dx (x + y) dy, are sides of the triangle x =, y =, x + y = 4. [ ] (x + y) dx (x y) dy, is the ellipse 4x + 9y = 36. [ π] (e x sin y 6y) dx + (e x cos y + 6) dy, is the circle x + y = x. [6π] 3.5 Path Independence of the Line Integral We can use another method of calculation of the line integral of a vector field in situations when the value of the integral doesn t depend on an integration curve and depends only on its starting and end point. Definition 3.4. Let points A, B. Let the vector function F(X) = P (X)i + Q(X)j + R(X) be continuous over the domain. If the value of the line integral of the vector field F(X) ds = P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz doesn t depend on the integration curve that lies within the domain, we say the integral is path independent between points A, B. If this property is fulfilled for arbitrary points A, B, we say integral is path independent over. Theorem 3.4. Let F(X) = P (X)i + Q(X)j + R(X) have continuous partial derivatives in a domain. Let a curve lie within, A be the starting point of the curve, while B be its end point. Then:. The line integral F(X)ds = P (X)dx + Q(X)dy + R(X)dz is path independent over if and only if there exists some scalar function Φ(x, y, z) 5

52 over such that F (x, y, z) = grad Φ(x, y, z), i.e. P = Φ x, Q = Φ y, R = Φ z. The vector field F(x, y, z) is called a conservative field, function Φ(x, y, z) is called a scalar potential.. The line integral F(X)ds = P (X)dx + Q(X)dy + R(X)dz is path independent over if and only if i j curl F = x y z P Q R = o. 3. In such case, the line integral of the vector field F(X) along a curve from the starting point A to the end point B is given by the difference of the values of scalar potential in the end point B and starting point A : F(X)ds = P (X)dx + Q(X)dy + R(X)dz = Φ(B) Φ(A). 4. Hence, if the line integral is path independent and the curve is closed F(X)ds =. If the problem is considered only in xy plane, the path independent line integral of the vector field is given by F(X)ds = P (x, y)dx + Q(x, y)dz = Φ(B) Φ(A). The test for determining if a vector field is conservative can be written in the form P y = Q x. We will show the way of calculation of the scalar potential at following examples. 5