Contour and surface integrals

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1 Contour nd surfce integrls Contour integrls of the sclr type These re integrls of the type I = f@rd l C where C is contour nd l is n infinitesiml element of the contour length. If the contour cn be described by {y[x], z[x]), one cn use l = 1+ y x + z x x. For more complicted contours one cn rewrite the contour integrl s n integrl over prmeter t nd use the expression of the contour in the prmetric form {x[t], y[t], z[t]}. In this cse the length element becomes l = x t + y t + z t t. ü Perimeter of n ellipse As n exmple, let us clculte the perimeter of the ellipse x + y b = 1, Here, for the upper prt of the ellipse, one hs y = b 1 x Thus the perimeter is given by, y x = b xë 1 x P = 1+ y x x = 1+ b x ë 4 x = 1 x J1 b Nx x x. In the cse of circle, = b = R, one esily obtins P = pr. However, in the generl cse the integrl cnnot be expressed vi elementry functions.

2 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb IntegrteB +J 1+ b Nx x, 8x,, <, Assumptions 80 < < b<f IntegrteB +J 1+ b Nx x, 8x,, <, Assumptions 80 < b < <F 4EllipticEB1 b F 4EllipticEB1 b F The solution is one of elliptic integrls tht received their nme from the ellipse. For the circle one hs P = 4EllipticE@0D π In the cse of b Ø 0 one hs P = 4EllipticE@1D 4 tht is n expected result s well. A similr result rises in the cse Ø 0 4EllipticEB1 b F P = 4LimitBEllipticEB1 b 4 b 0F F, Let us check tht the solution is symetric in nd b In[31]:= Out[31]= FullSimplifyBEllipticEB1 b bellipticeb1 b EllipticEB1 bf F bellipticeb1 Assumptions 80 < b < <F F bf, This is too difficult for Mthemtic tht seemingly does not know reltions between elliptic integrls. Let us define the function In[34]:= fb@_, b_d := EllipticEB1 b F bellipticeb1 b F This function is zero for ny set of rguments: In[35]:= fb@1, D êê FullSimplify fb@1,.d Out[35]= EllipticE@ 3D EllipticEB 3 4 F Out[36]= tht is zero up to the Mthemtic's stndrd precision. Let us now plot P/(4) tht hs prticulr cses

3 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 3 P 1, b << 4 = : πê, b = bê, b >> s function of rb ª b/. In[1]:= PlotA9rb, EllipticEA1 rb E, πê=, 8rb, 0, 4<, PlotStyle 8Dshed, Thick, Dotted<E 4 3 Out[1]= The rtio of the re of the ellipse S = pb to the squre of its perimeter is S P = π 16 b 1 EllipticEB1 b F This mens tht we hve expressed the re through the perimeter nd the rtio rb ª b/, s two independent vribles: S = π 16 b P EllipticEB1 b F Let us plot this function vs b/ for fixed P = 1 to see tht its mximum corresponds to b/ = 1 In[]:= PlotB π 16 rb P 0.08 ê. P 1, 8rb, 0, 5<F EllipticEA1 rb E 0.06 Out[]= ü Center of mss of prbol segment Let us clculte the position of the center of mss of the prbol segment y = kx, < x <

4 4 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb The verticl coordinte of the center of mss is given by the contour integrl y CM = 1 L y l = 1 C L y 1+ y C x x x = k L 1+k x x where L = l = C 1+k x x is the length of the prbol segment. Clcultion of the integrls yields In[37]:= L = IntegrteB 1+k x, 8x,, <, Assumptions 8 > 0, k > 0<F Out[37]= 1+ k + ArcSinh@kD k nd In[38]:= FullSimplifyByCM = k L IntegrteBx Assumptions k > 0F 1+k x, 8x,, <, Assumptions 8 > 0, k > 0<F, Out[38]= k 1+ k I1+ k M ArcSinh@kD 8k k 1+ k +ArcSinh@kD Prticulr vlue: In[39]:= ycm = ycm ê. 8 1, k 1.< Out[39]= In[49]:= ShowB PlotB x, 8x, 1, 1<, AspectRtio 0.5F, Grphics@Disk@80, ycm<, 0.015DD F Out[49]= ü Length of cycloide Find the length of cycloid x@φd = Sin@φD+ φ; y@φd = Cos@φD; in the intervl œ{0,p}. Here is plot of the cycloid.

5 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 5 In[56]:= x@φ_d := Sin@φD + φ; y@φ_d := Cos@φD; PrmetricPlot@8x@φD ê. 1ê, y@φd<, 8φ, 0, 4 π<d H Cycloide L Out[57]= One cn see tht for < 1 the dependence y[x] is non-monotonic. In this cse prmetric representtion is the only convenient wy to clculte the contour integrl. In[4]:= Out[4]= SimplifyAH φ x@φdl +H φ y@φdl E 1+ +Cos@φD For < 1 the length of the cycloid is given by In[6]:= IntegrteB H φ x@φdl +H φ y@φdl, 8φ, 0, π<, Assumptions 8 Rels, 1, 0<F Out[6]= 4 4 Abs@ 1 + D EllipticEB H 1+L F In[58]:= Out[58]= L1@_D = IntegrteB H φ x@φdl +H φ y@φdl, 8φ, 0, π<, Assumptions 0 < < 1F 4 4 H 1+LEllipticEB H 1+L F nd for > 1 it is given by In[4]:= Out[4]= L@_D = IntegrteB H φ x@φdl +H φ y@φdl, 8φ, 0, π<, Assumptions 1 < F 4 4 H 1+LEllipticEB H 1+L F The full dependence for > 0 cn be represented by In[5]:= L@_D := If@0 < < 1, L1@D, L@DD PlotB L@D, 8, 0, <, PlotRnge 80, <F π Out[6]= L[] hs singulrity t = 1. tht becomes better visible if we plot the derivtive of L[]

6 6 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb In[75]:= = H Immedite set is mndtory! L LDer@_D = L@D; LDer@_D := If@0 < < 1, L1Der@D, LDer@DD Plot@LDer@D, 8, 0, <D Out[77]= 3 1 Note tht LDer cnnot be determined s LDer@_D := If@0 < < 1, L1@D, L@DD LDer@D H with delyed set, Mthemtic tries to clculte derivtives when they re needed, i.e., when is set to. But it cnnot differentite with respect to L Generl::ivr:isnotvlidvrible. à Out[18]= H4EllipticE@ 8DL or s LDer@_D = If@0 < < 1, L1@D, L@DD H This remins delyed set becuse it depends on condition, nd delyed set does not work L LDer@D Out[]= If@0 < < 1, L1@D, L@DD Generl::ivr:isnotvlidvrible. à Out[3]= H4EllipticE@ 8DL Contour integrls of the work / circultion type An exmple of contour integrl is work long the trjectory C tht is line in 3d spce W = F r C Such integrls cn be introduced in usul wy by splitting of the trjectory into sum of mny smll finite displcements r n nd tking the limit of the sum N W = lim N n=1 F n r n Prcticlly contour integrls cn be clculted in the prmetric form s ordinry integrls over prmeter. The nturl prmeter for the mechnicl integrl bove is time t. Chnging the vrible to, one obtins

7 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 7 t r t W = F t 1 t t = F v t t 1 tht is, the work is the integrl of the power P = F v over time. As n exmple, consider the work of the grvity force F = mg = mge z long contour being prt of circle z = RCos@φD, x = RSin@φD, φ 1 φ φ to compute the work, one could consider motion of the prticle long the circulr trjectory with f = f[t] nd write t IFx W = v x +F y v y +F z v z M t = t 1 t mg z t mg t 1 t t = z φ t mgrsin@φd t 1 φ t t = φ t 1 t t. The result does not depend on the ngulr velocity f/ t with which the prticle moves long the trjectory, becuse one cn chnge the integrtion vrible to f φ mgrsin@φd W = φ. φ 1 In fct, f could be used s the prmeter for the contour integrtion insted of t from the very beginning. Moreover, one cn firther chnge the integrtion vrible to z nd finlly clculte the integrl φ mg z W = = mghz z 1 L = mgrhcos@φ D Cos@φ 1 DL. φ 1 Of course, this ws toy exmple becuse the grvity force is potentil force nd its work cn be relted to the difference of potentil energies, W = HU U 1 L This reltion cn be proven in generl form. If the force is potentil, F = U(r), then the infinitesiml work is simply the negtive differentil of the potentil energy: W = F r = - U HrL r = U x e x+ U y e y+ U z e z r = U x U U x+ y+ z = U y z Thus W = - U HrL r = - U = HU U 1 L. C C Work of ny potentil force over closed contour is zero. If insted of the force one hs vector function tht is not grdient of sclr, the integrl of this type over closed contour cn be non-zero. In this cse it cn be clled circultion. ü Circultion of the mgnetic field As n illustrtion of the circultion, consider the mgnetic field produced by n infinite thin stright wire directed long the z xis nd crrying current I in the positive z direction. The mgnetic field is directed in the x-y plne nd in the polr coordinte system is given by B = e φ r, = µ 0I π,

8 8 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb where r is the distnce from the wire tht is put into the origin of the coordinte system, nd e φ is the unit vector perpendiculr to r (in the x-y plne) nd directed in the direction of incresing of f, in ccordnce with the screw rule e φ = Sin@φDe x +Cos@φDe y. Let us first clculte the contour integrl B r C with the contour C being circle of rdius R centered t the wire nd going in the positive f direction. Since for this contour r =e φ r = e φ R f, this integrl simplifies to independently of R. π π B r = C 0 R He φ e φ LR φ = 0 R R φ = π = µ 0I, In fct, the contour integrl (circultion of the mgnetic field) depends only on the current tht crosses the surfce bounded by the contour nd not on the shpe of the contour (Ampere's lw). To illustrte this, let us use the contour tht consists of n infinite horizontl line y = -y 1 going to the right nd nother horizontl line y = y going to the left. With B x = B e x = Sin@φD the contour integrl is given by r = y x +y B r = B x y= y1 x+ B x y=y x = C y 1 x +y x+ y 1 x +y = u+ u +1 u +1 u = π+π = π = µ 0 I, the sme vlue s tht long the circulr contour. Surfce integrls of the sclr type These re the integrls of the type f@rd S S where S is surfce in 3d spce. Surfce cn be prmetrized by two prmeters, r = r(u,v). The surfce element S creted by chnging u nd v by u nd v is given by the bsolute vlue of the vector product S = r u r v u v For instnce, the surfce of sphere cn be prmetrized by the two ngles J nd f, nd the surfce element is elementrily given by S = R Sin@ϑD ϑ φ, where R is the rdius of the sphere. With nd r = 8RSin@ϑDCos@φD, RSin@ϑDSin@φD, RCos@ϑD<

9 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 9 r = 8RCos@ϑDCos@φD, RCos@ϑDSin@φD, RSin@ϑD< ϑ r = 8 RSin@ϑDSin@φD, RSin@ϑDCos@φD, 0< φ one obtins r ϑ r φ = 9 R Cos@ϑDCos@φDSin@ϑD, R Cos@ϑDSin@ϑDSin@φD, R Sin@ϑD = tht cn be computed s Cross@8R Sin@ϑD Cos@φD, R Sin@ϑD Sin@φD, R Cos@ϑD<, 8 R Sin@ϑDSin@φD, R Sin@ϑDCos@φD, 0<D êê Simplify 9 R Cos@ϑDCos@φDSin@ϑD, R Cos@ϑDSin@ϑDSin@φD, R Sin@ϑD = The bsolute vlue of this vector is SimplifyB %.%, Assumptions 8R > 0, 0 < ϑ < πê<f R Sin@ϑD the sme result. Let us now compute r ϑ r φ in one step: r = 8R Sin@ϑDCos@φD, R Sin@ϑDSin@φD, RCos@ϑD<; SimplifyB Cross@ ϑ r, φ rd.cross@ ϑ r, φ rd, Assumptions 8R > 0, 0 < ϑ < πê<f R Sin@ϑD In most cses, like the sphere, one cn write down the expression for S without using the generl formul. For instnt, for the prboloid z = k Ix +y M = k ρ

10 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb In[81]:= PrmetricPlot3DB:ρCos@φD, ρsin@φd, k ρ ê.

10 10 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb In[81]:= k ρ ê. k 1>, 8φ, 0, π<, 8ρ, 0, 3<F Out[81]= in the cylindricl coordinte system one hs S = 1+ z ρ ρ ρ φ = 1+Hk ρl ρ ρ φ Excersice: derive this from the generl formul. The re of the prboloid in the region r < is In[4]:= S = πintegrteb 1+Hk ρl ρ, 8ρ, 0, <, Assumptions 8 > 0, k > 0<F k + k 1+ k π Out[4]= 3k This result cn be simplified by the Simplify commnd pplied loclly 1+SimplifyB 1+ k + k 1+ k F π 3k I 1+I1+ k M 3ê M π 3k wheres Simplify pplied to the whole expression does not work. In the limit k <<1 the re of the prboloid reduces to the re of the circle p,s it should be. In fct, this integrl is very simple becuse r r = 1 r. Similr integrl describing the length of prbol bove is much more complicted. Surfce integrls of the flux type These re integrls of the type F@rD S, S = n S S where n is unit vector perpendiculr to the surfce t the point r. If the surfce is prmetrized by two prmeters, r = r(u,v), then S generted by chnging u nd v by u nd v is given by

11 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 11 S = r u r v nd the flux integrl becomes F@rD r S u r v u v u v, S = n S As in the cse of surfce integrls of the sclr type, in most cses the expression for S cn be written without clcultion. For instnce, for the sphere of rdius R one hs S = r r R Sin@ϑD ϑ φ ü Flux of the electric field As n illustrtion we clculte the flux of the Coulomb field produced by point chrge Q E = r r 3, = Q 4 πε 0 through the sphere of rdius R: π π Φ = E@rD S = r r π S 0 0 R 3 R R Sin@ϑD ϑ φ = π Sin@ϑD ϑ = 4 π = Q. 0 ε 0 Guss theorem sttes tht the flux of the electric field depends only on the totl chrge inside the surfce but not on the position of the chrge (chrges) nd the surfce shpe. To illustrte this, let us clculte the flux of the sme Coulomb field through the prboloid In the integrl z = k Ix +y M z 0 = k ρ z 0 Φ = E@rD S S S is perpendiculr to the prboloid t given point of its surfce. It is convenient to mke generl clcultion of S in the cylindricl coordinte system The vector product I r r = 8ρCos@φD, ρsin@φd, z< = :ρcos@φd, ρsin@φd, k ρ r ρ φ r = :ρcos@φd, ρsin@φd, k ρ Simplify@ Cross@ φ r, ρ rd D 9k ρ Cos@φD, k ρ Sin@φD, ρ= Mcn be computed s follows z 0>; z 0>. This vector is directed out of the prboloid. Then, the electric field on the prboloid becomes E = r :ρcos@φd, ρsin@φd, k ρ = r 3 Kρ +J k ρ z 0 N 3ê O z 0 >

12 1 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb Thus E I r r ρ φ M is given by :ρcos@φd, ρsin@φd, k ρ Kρ +J k ρ z 0 N 3ê O 4ρIk ρ +z 0 M I4 ρ +k ρ 4 4kρ z 0 +4z 0 M 3ê Now the flux through the prboloid is given by tht is Φ = π 0 z 0 >.9k ρ Cos@φD, k ρ Sin@φD, ρ= êê Simplify 4ρIk ρ +z 0 M I4 ρ +k ρ 4 4kρ z 0 +4z 0 M 3ê ρ = π 0 Hku+z 0 L u I4u+k u 4kuz 0 +4z 3ê 0 M In[7]:= Out[7]= π Hku+z 0 L Φ = IntegrteB I4u+k u 4kuz 0 +4z 0 M 3ê, 8u, 0, <, Assumptions 8z 0 > 0, k > 0<F 4π This result is correct nd in ccordnce with the Guss theorem sying tht the flux must be the sme s the flux through the sphere. In the cse z 0 < 0 the chrge is outside the prboloid nd the flux is zero: π Hku+z 0 L Φ = IntegrteB I4u+k u 4kuz 0 +4z 0 M 3ê, 8u, 0, <, Assumptions 8z 0 < 0, k > 0<F 0 In the cse z 0 < 0 the chrge is exctly on the surfce of the prboloid nd the flux is π Hku+z 0 L Φ = IntegrteB I4u+k u 4kuz 0 +4z 0 M 3ê, 8u, 0, <, Assumptions 8z 0 == 0, k > 0<F π The sme result rises if the chrge is plced on the surfce of sphere. Vector integrl reltions: Stokes nd Guss theorems There re importnt vector integrl reltions between integrls over region nd over the border of this region. The simplest exmple of these reltions is the formul b x F@xD x = F@bD F@D nd slightly more complicted formul for the integrl of grdient over contour. In both cses, the region is onedimensionl nd its border is zero-dimensionl, thus there is no integrl over the border, just vlues of the functions t the border points. Next of these reltions is the Stokes' theorem tht reltes the flux through n open surfce ( d integrl) with work-type integrl over contour bounding this surfce. It hs the form ı F S = F r. S C

13 Mthemticl_physics-08-Contour_nd_surfce_integrls.nb 13 There is one-dimensionl integrl on the right but two-dimensionl integrl on the left, however, of differentited function, the curl. The physicl impliction of the Stokes theorem is the Ampere's lw, where F is the mgnetic field B nd õ F is proportionl to the density of current flowing through the surfce. Next there is the Guss theorem relting volume integrl with flux through the closed surfce bounding this volume ı F V = F S. V S Here there is two-dimensionl integrl on the right but three-dimensionl integrl on the left, however, of differentited function, the divergence. The physicl impliction of the Guss theorem is the reltion between the electric-field flux out of the volume V through its surfce S, F being the electric field E, nd the totl electric chrge in the volume, the chrge density being proportionl to õ E. There is vector vrint of the Guss theorem ı F V = F S V S rising in the theory of mgnetism.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space. Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)