Multiple Integration

Transcription

1 Contents 7 Multiple Integration 7. Introduction to Surface Integrals 7. Multiple Integrals over Non-rectangular Regions 7. Volume Integrals Changing Coordinates 66 Learning outcomes In this Workbook ou will learn to integrate a function of two variables over various rectangular and non-rectangular areas. You will learn how to do this for various other coordinate sstems. You will learn to integrate a function of three variables over a volume.

2 Introduction to Surface Integrals 7. Introduction Often in Engineering it is necessar to find the sum of a quantit over an area or surface. This can be achieved b means of a surface integral also known as a double integral i.e. a function is integrated twice, once with respect to one variable and subsequentl with respect to another variable. This Section looks at the concept of the double integral and how to evaluate a double integral over a rectangular area. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... thoroughl understand the various techniques of integration be familiar with the concept of a function of two variables understand the concept of a surface integral integrate a function over a rectangular region HELM (8): Workbook 7: Multiple Integration

3 . n eample of a surface integral n engineer involved with the construction of a dam to hold back the water in a reservoir needs to be able to calculate the total force the water eerts on the dam so that the dam is built with sufficient strength. In order to calculate this force, two results are required: (a) The pressure p of the water is proportional to the depth. That is p = kd () where k is a constant. (b) The force on an area subjected to constant pressure is given b force = pressure area () The diagram shows the face of the dam. The depth of water is h and δ is a small area in the face of the dam with coordinates (, ). h δ h Figure Using (), the pressure at δ k(h ). Using (), the force on an area δ k(h )δ. Both of these epressions are approimate as is slightl different at the top of δ to the bottom. Now Total force on dam = sum of forces on all areas δ making up the face of the dam k(h )δ all δ For a better approimation let δ become smaller, and for the eact result find the limit as δ. Then Total force on the dam = lim k(h )δ δ = k(h ) d where k(h ) d stands for the surface integral of k(h ) over the area. Surface integrals are evaluated using double integrals. The following Section shows a double integral being developed in the case of the volume under a surface. HELM (8): Section 7.: Introduction to Surface Integrals

4 . Single and double integrals s has been seen in given b b a 4., the area under the curve = f () between = a and = b is f() d (assuming that the curve lies above the ais for all in the range a b). This is illustrated b the figure below. area required = f() a b Figure In a similar manner, the volume under a surface (given b a function of two variables z = f(, )) and above the plane can be found b integrating the function z = f(, ) twice, once with respect to and once with respect to. f(, ) = c = d = a = b Figure The above figure shows the part of a surface given b f (, ) which lies above the rectangle a b, c d. This rectangle is shaded and the volume above this rectangle but below the surface can be seen. 4 HELM (8): Workbook 7: Multiple Integration

5 f(, ) δ = c = d = a = b Figure 4 Imagine a vertical slice taken through this volume at right angles to the -ais (figure above). This slice has thickness δ and lies at position. ssuming that δ is small enough that the areas of both sides (left and right) of this slice are virtuall the same, the area of each face of the slice is given b the integral =d =c f (, ) d and the volume of the slice will be given b δ =d =c f (, ) d (where measures the position of the slice) To find the total volume between the surface and the plane, this quantit should be summed over all possible such slices, each for a different value of. Thus V i =d =c f ( i, ) d δ When δ becomes infinitesmall small, it can be considered to be d and the summation will change into an integral. Hence V = =b =d =a =c f (, ) dd Thus the volume is given b integrating the function twice, once with respect to and once with respect to. The procedure shown here considers the volume above a rectangular area and below the surface. The volume beneath the surface over a non-rectangular area can also be found b integrating twice (see Section 7.). HELM (8): Section 7.: Introduction to Surface Integrals 5

6 Ke Point Volume Integral The volume under the surface z = f(, ) and above a rectangular region in the plane (that is the rectangle a b, c d) is given b the integral: V = b =a d =c f(, ) dd. Inner and Outer integrals tpical double integral ma be epressed as =b [ =d ] f (, ) d d =a =c where the part in the centre i.e. =d =c f (, ) d (known as the inner integral) is the integral of a function of and with respect to. s the integration takes place with respect to, the variable ma be regarded as a fied quantit (a constant) but for ever different value of, the inner integral will take a different value. Thus, the inner integral will be a function of e.g. g () = =d =c f (, ) d. This inner integral, being a function of, once evaluated, can take its place within the outer integral i.e. =b =a double integral. g () d which can then be integrated with respect to to give the value of the The limits on the outer integral will be constants; the limits on the inner integral ma be constants (in which case the integration takes place over a rectangular area) or ma be functions of the variable used for the outer integral (in this case ). In this latter case, the integration takes place over a non-rectangular area (see Section 7.). In the Eamples quoted in this Section or in the earl parts of the net Section, the limits include the name of the relevant variable; this can be omitted once more familiarit has been gained with the concept. It will be assumed that the limits on the inner integral appl to the variable used to integrate the inner integral and the limits on the outer integral appl to the variable used to integrate this outer integral. 6 HELM (8): Workbook 7: Multiple Integration

7 4. Integration over rectangular areas Consider the double integral 5 = = ( + ) dd This represents an integral over the rectangle shown below. 5 Here, the inner integral is g () = ( + ) d and the outer integral is 5 = g () d Looking in more detail at the inner integral g () = ( + ) d Figure 5 the function ( + ) can be integrated with respect to (keeping constant) to give + +C (where C is a constant and can be omitted as the integral is a definite integral) i.e. [ g () = + ] ( = + ) ( + ) = + + = 4. This is a function of as epected. This inner integral can be placed into the outer integral to get 5 = 4 d which becomes [ ] 5 = 5 = 5 = 5 Hence the double integral 5 = = ( + ) dd = 5 HELM (8): Section 7.: Introduction to Surface Integrals 7

8 Ke Point Double Integral When evaluating a double integral, evaluate the inner integral first and substitute the result into the outer integral. Eample Evaluate the double integral = = This integral is evaluated over the area shown below. dd Figure 6 Solution Here, the inner integral is g () = = d = and hence the outer integral is [ 5 5 d = = ] [ = 9 4 = 5 ] = () = 5 8 HELM (8): Workbook 7: Multiple Integration

9 Eample Use the above approach to evaluate the double integral 5 = = cos π dd Note that the limits are the same as in a previous case but that the function itself has changed. Solution The inner integral is = cos π d = so the outer integral becomes 5 = 4 π d = [ π sin π ] = π π () = 4 π [ 4 π ] 5 = 4 π 5 4 π = 5 π 5. Clearl, variables other than and ma be used. Eample Evaluate the double integral 4 π s= t= s sin t dtds Solution This integral becomes (dispensing with the step of formall writing the inner integral), = 4 s= 4 [ ] π s cos t ds = 4 [ ] 4 s ds = s = 6 = 5 [ s cos π + s cos ] ds = 4 [ s () + s ()] ds HELM (8): Section 7.: Introduction to Surface Integrals 9

10 Clearl, evaluating the integrals can involve further tools of integration, e.g. integration b parts or b substitution. Eample 4 Evaluate the double integral e + dd Here, the limits have not formall been linked with a variable name but the limits on the outer integral appl to and the limits on the inner integral appl to. s the integrations are more complicated, the inner integral will be evaluated eplicitl. Solution Inner integral = e + d which can be evaluated b means of the substitution U = +. If U = + then du = d so d = du. lso if = then U = 5 and if = then U =. So the inner integral becomes (remembering that ma be treated as a constant) e [ ] e du du = U U = e ln U = e ln (ln ln 5) = e 5 and so the double integral becomes ln e 5 d = ln e d which can be evaluated b integration b parts. [ [ ln ] e ( ] e ) d = ln [ ] e + () e + e d [ = ln [ ] ] e e + e = ln [ e e e + e ] = ln [ ] e.4 5 HELM (8): Workbook 7: Multiple Integration

11 Task Evaluate the following double integral. ( + ) dd Your solution nswer The inner integral = ( + ) d = [ ] + This can be put in the outer integral to give ( + 8 ) [ ] d = + 8 = + 8 ( 8) = 4 = ( + ) = = 5 Eercises Evaluate the following double integrals over rectangular areas.. = = dd. 4 ( + ) dd. π sin dd 4. st dsdt 5. 5z w ( w ) 4 dwdz (Requires integration b substitution.) 6. nswers π t sin t ddt (Requires integration b parts.).,. 46/,., 4. 6, 5. 9/, 6. π HELM (8): Section 7.: Introduction to Surface Integrals

12 5. Special cases If the integrand can be written as f (, ) = g () h () then the double integral b d a can be written as b a c g () h () g () d d c dd h () d i.e. the product of the two individual integrals. For eample, the integral = = = = dd which was evaluated earlier can be written as [ ] [ d d = = 5 = 5 the same result as before. ] = [ 8 () ] [ 9 4 ] The integral Ke Point Double Integral as a Product b d a c g()h() dd can be written as b a g() d d c h() d Imagine the integral e dd pproached directl, this would involve evaluating the integral b algebraic means (i.e. it can onl be determined numericall). e d which cannot be done HELM (8): Workbook 7: Multiple Integration

13 However, the integral can be re-written as [ d e d = ] e d = e d = and the result can be found without the need to evaluate the difficult integral. If the integrand is independent of one of the variables and is simpl a function of the other variable, then onl one integration need be carried out. The integral I = I = b d a c b d a c h () dd ma be written as I = (b a) g () dd ma be written as I = (d c) b a d c h () d and the integral g () d i.e. the integral in the variable upon which the integrand depends multiplied b the length of the range of integration for the other variable. Eample 5 Evaluate the double integral dd Solution s the integral in can be multiplied b the range of integration in, the double integral will equal [ ] [ ] ( ) d = = () = 6 Note that the two integrations can be carried out in either order as long as the limits are associated with the correct variable. For eample [ ] 4 4 dd = d = [ 4 ] 4 d = = = = and = = = [ ] 4 d = 5 = = = 5 d = 4 dd = [ ] = [ 5 5 = 4 = ] d = [ 5 ] d HELM (8): Section 7.: Introduction to Surface Integrals

14 Task Evaluate the following integral: z (w + ) dwdz. Your solution nswer Eercises. Evaluate the following integrals: (a) (b) (c) π/ 8 5. Evaluate the integrals ( cos ) dd dd (s + ) 4 dtds dd and dd and show that the are equal. s eplained in the tet, the order in which these integrations are carried out does not matter for integrations over rectangular areas. nswers. (a) /, (b) /, (c). 4 4 HELM (8): Workbook 7: Multiple Integration

15 6. pplications of surface integration over rectangular areas Force on a dam t the beginning of this Section, the total force on a dam was given b the surface integral k(h ) d Imagine that the dam is rectangular in profile with a width of m and a height h of 4 m. The epression d is replaced b dd and the limits on the variables and are to m and to 4 m respectivel. The constant k ma be assumed to be 4 kg m s. The surface integral becomes the double integral 4 k(h ) dd that is 4 4 (4 ) dd s the integral in this double integral does not contain, the integral ma be written 4 4 (4 ) dd = ( ) that is the total force is 8 meganewtons. Centre of pressure 4 4 (4 ) d [ ] 4 = 4 4 = 6 [(4 4 4 /) ] = 6 8 = 8 8 N We wish to find the centre of pressure ( p, p ) of a plane area immersed verticall in a fluid. Take the ais to be in the surface of the fluid and the ais to be verticall down, so that the plane O contains the area. O surface δ Figure 7 We require the following results: (a) The pressure p is proportional to the depth h, so that p = ωh where ω is a constant. (b) The force F on an area δ subjected to constant pressure p is given b F = pδ HELM (8): Section 7.: Introduction to Surface Integrals 5

16 Consider a small element of area δ at the position shown. The pressure at δ is ω. Then the force acting on δ is ωδ. Hence the total force acting on the area is ω d = ω d. Moment of force on δ about O = ωδ Total moment of force on δ about O = ω d Moment of force on δ about O = ω δ Total moment of force on δ about O = ω Taking moments about O: total force p = total moment ( ) ω d p = ω d p d = d Taking moments about O: Hence p = total force p = total moment ( ) ω d p = ω d p d = d d d and p = d. d d 6 HELM (8): Workbook 7: Multiple Integration

17 Eample 6 rectangle of sides a and b is immersed verticall in a fluid with one of its edges in the surface as shown in Figure 8. Where is the centre of pressure? O surface a b Figure 8 Solution To epress the surface integral as double integrals we will use cartesian coordinates and vertical slices. We need the following integrals. Hence d = d = p = d = d d b a b a b a = dd = dd = dd = b [ b b ] a ] a [ [ a b a b = a and p = ] a d = b d = d = d d The centre of pressure is ( b, a), so is at a depth of a. b b [ ] b a d = a = a b [ ] b a d = 4 a = 4 a b [ ] b a d = a = a b = 4 a b a b = b HELM (8): Section 7.: Introduction to Surface Integrals 7

18 reas and moments The surface integral f(, ) d can represent a number of phsical quantities, depending on the function f(, ) that is used. Properties: (a) If f(, ) = then the integral represents the area of. (b) If f(, ) = then the integral represents the first moment of about the ais. (c) If f(, ) = then the integral represents the first moment of about the ais. (d) If f(, ) = then the integral represents the second moment of about the ais. (e) If f(, ) = then the integral represents the second moment of about the ais. (f) If f(, ) = + then the integral represents the second moment of about the z ais. Eample 7 Given a rectangular lamina of length l, width b, thickness t (small) and densit ρ (see Figure 9), find the second moment of area of this lamina (moment of inertia) about the -ais. b O densit: ρ thickness: t (small) l Figure 9 Solution B propert (e) above, the moment of inertia is given b b l ρt dd = ρt(l ) [ = lρt = lρt b b ] b d s the mass of the lamina is M = lbtρ, the moment of inertia simplifies to Mb. The t and ρ are included in the integral to make it a moment of inertia rather than simpl a second moment. 8 HELM (8): Workbook 7: Multiple Integration

19 Task B a similar method to that in Eample 7, find the moment of inertia of the same lamina about the -ais. Your solution nswer From propert (d) above, the moment of inertia (or second moment of area) is given b the integral l b l ρt dd = ρt(b ) [ = bρt ] l d = bρt l s the mass of the lamina is M = lbρt, the moment of inertia simplifies to Ml. gain, the t and ρ are included in the integral to make it a moment of inertia rather than simple a second moment. Eercises B making use of the form of the integrand, evaluate the following double integrals:. π cos dd. 8 dd. 5 (s + ) 4 dtds nswers. π 4,.. HELM (8): Section 7.: Introduction to Surface Integrals 9