Surface integrals, Divergence theorem of Gauss

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1 c jbquig-ucd Februar 13, 3

2 Chapter 6 Surface integrals, Divergence theorem of Gauss introduction A solid D R 3 might be described b an inequalit, see chapter5. A surface R 3 is often described b a single (implicit) equation; a solid is a three, whereas a surface is a two, dimensional object. 6.1 parametriation of a surface See lecture notes See lecture notes Figure 6.1: parametriation of a surface Parametriation of a surface, see figure 6.1 takes the form h : A R R 3 r s h r s r s r s r s r s The domain of parametriation A lies in two-space R and there are two parameters r and s differential matri The differential matri of a surface parametriation mapping h is the 3 matri Dh r s (6.1) 137

3 Dh r s is a linear mapping from R to R 3. Amongst linear mappings Dh r s is the best approimation to the original parameteriation mapping h near 6.1. S a, the spherical surface r s The ball B 3 a is described b the inequalit a The boundar surface of this ball is the sphere A. S a a (6.) described here b an implicit equation. S a can be parameteried, see figure 6., b fiing r a in the spherical polar transform ( 5.6). rectangle missing see lecture notes sphere missing see lecture notes Figure 6.: parametriation of spherical surface h : π π S a R 3 θ φ θ φ h θ φ θ φ θ φ θ φ θ φ asin φ cos θ asin φ sin θ acos φ The differential matri is Dh θ φ θ φ θ φ θ φ asin φ sin θ acos φ cos θ asin φ cos θ acos φ sin θ asin φ T a b, the toroidal surface The boundar surface T a b of the solid torus T 3 a b is called the toroidal surface, (6.3) T a b b a (6.4) described here b an implicit equation, (for derivation of this equation see lecture notes). T a b can be parameteried, see figure 6.3, b fiing r a in the toroidal polar transform 5.7. c jbquig-ucd Februar 13, 3

4 rectangle missing see lecture notes torus surface missing see lecture notes Figure 6.3: parametriation of toroidal surface h : π π T a b R 3 θ φ θ φ h θ φ θ φ θ φ θ φ θ φ b asin φ cos θ b asin φ sin θ acos φ The differential matri is Dh θ φ θ φ θ φ θ φ b asin φ sin θ acos φ cos θ b asin φ cos θ acos φ sin θ asin φ E a b c, ellipsoidal surface The outer surface E a b c of the solid ellipsoid E 3 a b c is called the ellipsoidal surface, (6.5) E a b c a b c 1 (6.6) described here b an implicit equation. E a b c can be parameteried, see figure 6.4, as a variant of the parameteriation of the spherical surface. h : π π E a b c R 3 θ φ θ φ h θ φ θ φ θ φ θ φ θ φ asin φ cos θ bsin φ sin θ ccos φ The differential matri is Dh θ φ θ θ θ φ φ φ asin φ sin θ acos φ cos θ bsin φ cos θ bcos φ sin θ csin φ (6.7) Februar 13, 3 c jbquig-ucd

5 rectangle missing see lecture notes ellipsoidal surface missing see lecture notes Figure 6.4: parametriation of ellipsoidal surface 6. area dilation Parametriation, h : A R n B R n, of a solid is treated in section 5.4. The volume dilation factor or Jacobean plaed an important role. For parametriation h : A R R 3 of a surface, the area dilation factor or area Jacobean plas a similar role. We begin with the easiest case area dilation of a linear mapping theorem 1 The linear mapping or 3 matri dilates area b the factor ADF A A 1 A A î ĵ ˆk a b c d e f a b c d e f : R R 3 (6.8) proof In R the two standard unit basic vectors, î and ĵ, delineate the standard square S whose four vertices are b c î î ĵ and ĵ The area of S is 1, see figure 6.5(i). A carries î and ĵ R to column vectorsa 1 anda Aî A 1 a b c e f A 1 Aĵ A c a 1 f d d e f a b d e R 3 respectivel. (6.9) A (6.1) Thus A carries the square S R to the parallelogram A S R 3 delineated b the vectors A 1 and A, i.e. with vertices A A î A 1 A î ĵ A 1 A and A ĵ A see figure 6.5(ii). The area of the parallelogram A S is sin α α being the angle between vectors  1 and  A  1  µ A S A 1 b definition of the -vector c jbquig-ucd Februar 13, 3

6 standard square missing see lecture notes parallelogram missing see lecture notes Figure 6.5: (i) standard square S (ii) parallelogram A S 6.. area dilation under surface parametriation remark The area dilation factor under the surface parameteriation mapping h : A varies from point to point in the domain A of paramertiation. But at a particular point, r s rs r s, the area dilation factor of h is equal to the area dilation factor of the linear approimate mapping Dh r s. We have proven theorem 11. theorem 11 The area dilation factor under h : A R R 3 r s h r s r s r s r s r s a C 1 parametriation of the surface R 3 is ADF h Dh 1 Dh î ĵ ˆk (6.11) computational trick Direct computation of the ADF of h is onerous but, if the columns of the differential Dh, then Dh 1 Dh Dh 1 Dh sin α Dh 1 Dh (6.1) since the angle between vectors Dh 1 and DH is α π. eercise Find Dh 1 Dh, where matri are orthogonal, i.e. Dh 1 If h is the parameteriation in 6.1. of the spherical surface S a then, Dh 1 Dh a sin φ. If h is the parameteriation in of the toroidal surface T a b then, Dh 1 Dh a b a sin φ If h is the parameteriation in of the ellipsoidal surface E a b c, the columns of Dh are not orthogonal and computation of Dh 1 Dh is long and mess. Februar 13, 3 c jbquig-ucd

7 6.3 integration of a scalar field over a surface remark The actual definition of the integral, f dσ, of a vector field f over a surface R3, was covered in lectures (and will be tped up when time permits). There is a formula for evaluation of this integral, in terms of a parametriation of the surface, see theorem 1. theorem 1 Let f : R 3 R be a C scalar vector field over the surface R 3. Let h : A R R 3 f f r s h r s r s r s r s r s be a C 1 parameteriation, over the domain A R, of the surface, see figure 6.6, then domain of parametriation missing, see lecture notes surface missing see lecture notes target R missing see lecture notes Figure 6.6: scalar field over a parametried surface f dσ 3 R A R f h r s Dh 1 Dh A R f h r s A R f h r s i j k d r s d r s proof Reread the proof of the Change of Variable theorem 9. Make appropriate alterations, in particular replace the volume dilation factor (volume Jacobean) with the area dilation factor (area Jacobean). d r s eample Find the surface area c jbquig-ucd Februar 13, 3

8 dsµ S a S a 1 dσ of the spherical surface S a. dsµ T a b T a b 1 dσ of the toroidal surface T a b. 6.4 the normal field of a surface remark There are three was to think of a surface R 3 and consequentl three different forms of the equation of the surface. You ma have met the first two methods in an earlier course: the are not so important in the theor of integration. The third and final method, which uses parametriation of the surface, plas an important role in integration. At each point there is a unique (up to 1) unit normal vector n. This leads to the concept of the normal vector field, n : σ R 3, to the surface. Each of the three equations for the surface leads to a different epression for the vector field n. The epression for n in terms of parametriation of plas an important role in the theor of integration. We have a scalar function f : R 3 R, the set on which f has constant value d R is known as the level surface,, or contour on which f has constant value d. Thus f d Here we see the implicit equation of the surface. From earlier courses in advanced calculus, a normal field to in terms of the implicit equation is n grad f f i f j f k eample Let f and d a S a a n i j k and here we have the implicit equation of the spherical surface, S a R 3, center, radius a and the normal field derived from the implicit equation. Given a subset A R and a scalar function g : A R R the graph of g is a surface Gr g g n g i g j Here we see the eplicit equation of the surface. and the corresponding formula for the normal vector field. eample Let g a S a B g a for B n a 1 a R Here we have the eplicit equation of the spherical surface, S a (rather, onl get the upper hemisphere, this method has its limitations) and the normal vector field derived from the eplicit equation. i a j k Februar 13, 3 c jbquig-ucd

9 Now we come to the third and most important wa to describe a surface R 3. We use a parametriation h mapping from a domain of parametriation, A R h : A R R 3 r s h r s r s r s r s r s Here r and s are called parameters. The differential matri Dh r s of the parameteriation mapping h is the 3 matri Dh r s and is that linear mapping Dh r s : R mapping, h : A R R 3 which most closel approimates the parametriation R 3 r, close to A. See lecture notes for the derivation of derived s the formula for the unit normal vector field n to, in terms of the parametriation mapping Dh n 1 Dh Dh 1 Dh (6.13) eample polar transform b fiing r a h : The parametriation mapping for the spherical surface S a is derived from the spherical Dh θ φ π π π S a R 3 θ φ h r s r s θ θ θ φ φ φ We leave the reader to derive the normal vector field. 6.5 vector flu integral across a surface Let v be a vector field defined over the surface R 3. v : Consider also the unit normal vector field n : r s r s r s asin φ cos θ asin φ sin θ acos φ asin φ sin θ acos φ cos θ asin φ cos θ acos φ sin θ v M N P asin φ R 3 over. Taking the inner product of these two we obtain v ndσ a scalar field v n v n over the surface. This scalar field can be integrated over, we have called the total flu of the vector field v across the surface. Here dσ can be considered to be an infinitesimal piece of surface area; v n is the (scalar) resolution of the vector field v in the normal direction at each point of the surface. The integral is sometimes written as v dσ Here dσ ndσ is a piece of directed area, i.e. with magnitude dσ and direction n. c jbquig-ucd Februar 13, 3

10 Given a parametriation h : A R v ndσ R 3 we can compute Use the epression 6.13 for n in terms of h to obtain v ndσ A R v n h Dh 1 Dh v ndσ using theorem 13. dσ Dh v 1 Dh A R Dh 1 Dh Dh 1 Dh dσ Cancelling the area dilation factor, above and below, we obtain v ndσ A R v Dh 1 Dh d s t On the right hand side we see a scalar triple product of vectors which is a erminant. We have arrived at the evaluation theorem, using surface parametriation, of a vector flu integral, across a surface. The above discourse is the proof. The integral v ndσ can be epressed in two form notation. This notation is suggested b writing the scalar vector triple as a erminant and epanding along the first row A A A v Dh 1 Dh d s t M s t s t s t N s t s t s t P s t s t s t M N N P P N Now the two form notations suggests itself and is v nσ P P M M P M M M d d N d d Pd d We have arrived at the evaluation theorem (b surface parametriation) for a vector flu integral across a surface. theorem 13 Let v : R 3 σ be a C vector field on the surface R 3. Let h : R be a C 1 parameteriation of the surface. Then v ndσ v dσ A M d d N d d Pd d s t h s t s t M N P s t s t s t N N R 3 R 3 d s t d s t Februar 13, 3 c jbquig-ucd

11 A A v Dh 1 Dh d s t M s t s t s t N s t s t s t P s t s t s t d s t 6.6 eamples of the surface vector flu integral flu of field across spherical surface We will compute I S a i j k 3 ndσ d d d d d d S a 3 This is the most basic flu integral of all, being the flu of the inverse square law field across a spherical surface of radius a, S S a a. Use spherical polar parameteriation of the surface S S a, see 6.3, and the formula for evaluation of surface flu integrals b parameteriation, see theorem13. S d d d d d d 3 π π π π π π π π π π π π π π π M N P θ θ θ φ φ φ 3 d θ φ 3 3 asinφsinθ asinφcosθ acosφcosθ acosφsinθ asinφ a 3 a 3 a 3 d θ φ asinφsinθ asinφcosθ acosφcosθ acosφsinθ asinφ asinφcosθ a 3 asinφsinθ a 3 acosφ a 3 asinφsinθ asinφcosθ acosφcosθ acosφsinθ asinφ a a 3 a a sinφcosθ sinφsinθ cosφ sinφsinθ sinφcosθ cosφcosθ cosφsinθ sinφ d θ φ d θ φ d θ φ But we recognie this erminant as essentiall the Jacobian erminant of the well known spherical polar transform with well known value sinφ. I a 3 π π π π π π a 3 π π π sinφd θ φ π sinφdθdφ sinφdφ π cosφ cosπ cos c jbquig-ucd Februar 13, 3

12 π π 4π flu integral across ellipsoidal surface We will attempt to compute, the flu of the inverse square law field but this time across the ellipsoidal surface Thus we seek I E E E a b c R 3 ndσ i j k 3 a b c 1 E d d d d d d 3 Use elliptical polar parameteriation of the surface E E a b c, see 6.7, and the formula for evaluation of surface flu integrals b parameteriation, see theorem 13. E d d d d d d 3 π π π π π π π π π π π π π π π M N P θ θ θ φ φ φ 3 d θ φ 3 3 asinφsinθ bsinφcosθ acosφcosθ bcosφsinθ csinφ a sin φcos θ b sin φsin θ c cos φ 3 asinφsinθ acosφcosθ d θ φ a sin φcos θ b sin φsin θ c cos φ bsinφcosθ bcosφsinθ 1 a sin φcos θ b sin φsin θ c cos 3 φ asinφcosθ bsinφsinθ ccosφ asinφsinθ bsinφcosθ acosφcosθ bcosφsinθ csinφ abc a sin φcos θ b sin φsin θ c cos φ 3 sinφcosθ sinφsinθ cosφ sinφsinθ sinφcosθ cosφcosθ cosφsinθ sinφ But we recognie this erminant as essentiall the Jacobian erminant of the well known spherical polar transform with well known value sin φ. I π π π abcsinφ a sin φcos θ b sin φsin θ d θ φ c cos 3 φ It is too too dificult to finish this computation, we give up: but see 6.8., we will succeed with a different approach using the divergence theorem of Gauss flu integral across toroidal surface Let a b and be the toroidal surface obtained b rotating the circle C : b a which lies in the -plane, about the -ais. Evaluate b d d b d d d d d θ φ d θ φ Februar 13, 3 c jbquig-ucd

13 both directl b using a parametriation of. This integral is again computed in b a different method, the famous divergence theorem of Gauss, see 6.7. In the problem we are asked to compute the flu I across the surface of the vector field v v ndσ Mi Nj Pk M d d N d d Pd d b i b j k Recall the parameteriation of the toroidal surface, 6.5. and, using this, the formula for evaluation of a vector flu integral across a surface, theorem 13. Thus v ndσ M d d N d d Pd d Now we must slug out all nine matri entries. π π M 1 b b asinφ cosθ 1 b b asinφ b asinφ cosθ asinφ b acosφ asinφcosθ M N P θ θ θ φ φ φ B similar methods we obtain N asinφsinθ and of course P acosφ. Net hack out si partial derivatives I π π π π asinφcosθ asinφsinθ acosφ b asinφ sinθ b sinφ cosθ acosφcosθ π π π π a b asinφ a acosφsinθ asinφ sinφcosθ sinφsinθ cosφ sinθ cosθ cosφcosθ cosφsinθ sinφ d θ dφ d θ φ where we have divided constants out of all three rows. The remaining erminant has mutuall perpendicular rows and so is row row row 3 Thus I π π π π a b asinφ a 1 1 1d θ φ π π π π a b asinφ d θ φ a π a π a b π π π π π π π π π 4π a b π b asinφ dθdφ b Fubini s Theorem b dθ dφ 1dθdφ since π π sin We remind ou that the above computation can also be performed as a scalar integral over the solid torus, b use of the divergence theorem of Gauss, see c jbquig-ucd Februar 13, 3

14 6.7 divergence theorem of Gauss remark If is a surface which bounds a solid region D and if v is a vector field defined over D the divergence theorem of Gauss epresses the vector flu integral of v across as the integral of the divergence scalar field divv over the solid region D. The statement of the theorem follows. theorem 14 Let be a surface bounding the solid region D R 3. Let v be a vector field defined over all of the region D. Then v ndσ Using two form notation and epanding the divergence M d d N d d Pd d D divvd D M N 6.8 eamples of the divergence theorem of Gauss flu integral across the spherical surface P d We tr to appl the divergence theorem of Gauss to compute the flu integra, I, of the inverse square law across the spherical surface S S a. From chapter we know that div v. S is the boundar surface of the solid ball B B 3 a R 3 a. From we have field v 3 prior knowledge that I 4π. Thus 4π I S S B B ndσ i j k 3 d d d d d d 3 div v d d We have proven that 4π! What has gone wrong? This error has been included so that the reader ma be warned to take care when using the divergence theorem of Gauss, which requires that v be defined over ALL of the solid region B. However v is undefined ( ). The mistake lies with us, not in the Gauss theorem. The net section 6.8. shows a valid and important use of this theorem in a similar contet flu integral across the ellipsoid surface Let E E a b c be the ellipsoidal surface E E a b c a b c 1 Recall failure in 6.6. of an attempt to compute the flu integral of the inverse square law field across the surface E. We tr again using the divergence theorem of Gauss to compute I ndσ S i j k 3 S d d d d d d 3 a b c. Let S S r be tha small spherical surface enclosed b the ellipsoidal surface E, see Let r figure 6.7. Let D denote the solid region between the spherical surface S and the elliptical surface E, see Februar 13, 3 c jbquig-ucd

15 Figure 6.7: solid between spheroidal and ellipsoidal surfaces figure 6.7. The boundar surface, S theorem of Gauss E, of D consists of two separate pieces. Appling the divergence i j k E S ndσ 3 i j k ndσ E S D D 3 i j k div d 3 d In passing we mention that the inverse square law vector field is defined everwhere on D since D and that the divergence of this field ero, see chapter The minus in the first line is because the normal out of D on S is the inward normal on S. Thus E d d d d d d 3 S 4π d d d d d d 3 remark See 6.6.; we have computed a ver difficult integral π π π abc sinφ a sin φcos θ b sin φsin θ d θ φ c cos 3 φ 4π flu integral across the toroidal surface Let a b and be the toroidal surface obtained b rotating the circle, C which lies in the -plane, about the -ais. Denote b T the solid torus bounded b the surface of. Thus R 3 Evaluate b a T R 3 b d d b d d d d b a, b a c jbquig-ucd Februar 13, 3

16 using the divergence theorem of Gauss (this computation has alread been carried out b direct evaluation of the surface integral I, see 6.8.3). or in more ail We must slug out divv M v ndσ M d d N d d Pd d N P T divvd T M N P d b 1 b 1 1 b 1 b 3 1 b 1 b b 1 b 3 3 b 1 We return to the main computation I T 3 b 1 d To evaluate this scalar integral over the solid region T we of course appeal to the COV theorem 9 and use the Toroidal Polar transform chapter h : a π π π π with its Jacobian Dh θ φ r b r sinφ. I θ φ b r sinφ cosθ b r sinφ sinθ r cosφ 3 b 1 d T a π π π 3 b 1 Dhd r θ φ π a π π π π a π π π 3 b b r sinφ 1 r b r sinφ d r θ φ π a π π π 3r b r sinφ br d r θ φ π a π π π π a π π π a π π π 4π a rbdr 4π a b 3 b b r sinφ 1 r b r sinφ d r θ φ rb r sinφ dθdφdr b Fubini s th rb r sinφ dφdr rb dφdr since π π sin Do check back, that we have obtained the same result in section T Februar 13, 3 c jbquig-ucd

17 6.9 problem set Surface integration and the divergence theorem of Gauss This problem set has been gleaned from an earlier document without much editing. There is repetition and no diagrams but still much useful material. Will not be further edited this academic ear, jbquig Let B, a solid bod with surface, be totall immersed in a liquid of densit ρ, whose surface is the plane. Let g denote the acceleration of gravit. The pressure at a point p is ρgn, thus the magnitude of pressure is proportional to depth and the direction is that of n the unit normal to at p. The bouanc force per unit area at p is ρgn k (the pressure resolved in the vertical direction). B taking a surface integral of the bouanc force over the surface and using the divergence theorem prove Archimedes Principle, The total buoanc force on the bod B equals the weight of the displaced liquid.. Let a b and let A be that region in the -plane where b b and b a. Let B R 3 be that volume obtained b revolution of A about the ais. Let be the surface bounding B and I b d d (i) Sketch A R. Sketch B R 3. (ii) Compute I, directl as a surface integral. (iii) Use the Divergence theorem to compute I as an integral over B. b d d d d 3. Let a b and T be the solid torus obtained b rotating the disc D : b a, which lies in the -plane, about the -ais. Denote b the surface of T. Evaluate b d d b d d d d both directl using a parametriation of and b using the divergence theorem. 4. Let D R 3 be the region given b and 9. Let be the surface 3 being one of the two surfaces which together form the boundar of D. Let (i) Sketch D showing. (ii) Use the divergence theorem to prove, I volume integral. I d d d d d d (iii) Compute I again b parametriation of the surface. 5. Let a b and T be the solid torus obtained b rotating the disc D D d. Compute I b evaluation of this : b a about the -ais and let be the toroidal surface which bounds T. Consider the vector field b v i b j k c jbquig-ucd Februar 13, 3

18 (i) Parametrie. (ii) Compute v Show that the integral I does not depend on a. v (iii) Prove that, (even though div v div v I v ndσ v is infinite on the circle b ), the integral T d 6. eists and is ero. You might consider the region between T and a smaller torus where a is replaced b ε which tends to ero. E v i j k (i) Compute divv. : a b c 1 R 3 F I E : a b 1 R v n dσ (ii) B manipulating I in two was, first using a parametriation of the elliptic surface E and second using the divergence theorem prove π π F abc sinφ a sin φcos θ b sin φsin θ c cos φ dθdφ arctan c 1 a b d 7. Let λ be a fied positive real number. Define a vector function F on R 3 b (i) Show that div F. F λ i j k 3 (ii) Suppose that S is a piecewise smooth orientable surface in three dimensional Euclidean space, and that the origin is an interior point of S. Let n be the outward unit normal to S. Use the divergence theorem to show that regardless of the shape of S. S F n dσ 4πλ Februar 13, 3 c jbquig-ucd