MA7 urface Integrals Parametrically Defined urfaces We discussed earlier the concept of fx,y,zds where is given by z x,y.wehad fds fx,y,x,y1 x y 1 da R where R is the projection of onto the x,y - plane. We shall now develop a generalization of this concept. There are three common ways of defining a surface: I. z x,y 1 as above. Here must be a single-valued, continuous function defined on a region of the plane. II. Often surfaces are represented by equations of the form Fx,y,z If x,y,z is a point on such a surface, we can in many cases represent the portion of the surface near x,y,z in a form analogous to 1 by solving for x,y, orz in terms of the other two variables. III. It is frequently convenient to describe a surface by a parametric representation. Example: x a sin u cosv y a sin u sin v. z a cosu Here u and v are independent parameters. This represents a sphere whose equation is x y z a This equation is gotten by elimination of u and v. Note that u and v are the spherical coordinates and respectively. The set of equations x xu,v y yu,v z zu,v 3 where u and v are parameters represents an arbitrary surface. This can be seen by eliminating u and v from 3, a procedure that leads to an equation of the form Fx,y,z which is case II. In terms of the radius vector r x i y j zk equation (3) for the surface may be written as 1
r r u,v xu,v i yu,v j zu,vk From the parametric equations for a surface it is possible to establish a formula for ds, the element of surface area. In general, ds is obtained by calculating the area between the curves corresponding to: u u, u u du, v v andv v dv. For infinitesimal areas this element will be essentially planar and have area ds AB AC,wherethe vectors are the sides of the differential parallelogram shown in the diagram. A r u,v B r u du,v r u,v r u u,v du C r u,v dv r u,v r v u,v dv Thus AB r u du AC r v dv ds r u r v dudv
Hence, in general, we have for a surface given by that x xu,v y yu,v z zu,v fx,y,zds G fu,v r u r v dudv, where G is the image of the surface in the u,v -plane. uppose the surface is given by the representation z x,y (case I). Let x u, y v z u,v Then also represents the surface. Thus r u,v u i v j u,vk and r u i u k; r v j v k; r u r v k u i v j so ds r u r v dudv 1 u v 1 dudv But since u x, v y we get ds 1 x y 1 dxdy as before. Example We shall find the surface area of a sphere of radius a centered at the origin. The equation of the sphere is x y z a In spherical coordinates the sphere is given by x a sin u cosv y a sinusin v z a cosu Hence r a sin u cosv i a sinusin v j a cosuk 3
r u r v a sin u cosv i sin u sinv j sinucosuk and r u r v a sinu ds a sinududv Thus ds a sin ududv a sindd 4a surface area of a sphere. urface Elements uppose that R is a closed rectangular region in the u,v plane, where a u b, c v d. Then the equations x xu,v, y yu,v, z zu,v, wherex,y,z are continuous, define a set which is part of a surface in x,y,z space. If the functions x,y,z are also 1-1, i.e. distinct points of R are not mapped into the same point of, then the points of in x,y,z space comprise a simple surface element. A simple surface element may be thought of as any configuration which may be obtained from a rectangular plane region by continuous deformation (bending, twisting, stretching, shrinking) without tearing and without bringing together any points which were originally distinct. If is a simple surface element corresponding to a rectangular region R in the u,v plane, the points of which correspond to the boundary of R form what is called the boundary. Other points of are called interior points. All surfaces may be thought of as being built up out of simple surface elements by matching together portions of the edges of the elements. The boundary of a surface consists of the unmatched edges of its surface elements. If there are no unmatched edges, there is no boundary. For example, a hemisphere has a boundary consisting of its equatorial rim. An entire sphere, an ellipsoid, and the surface of a cube are examples of surfaces without boundary. A surface is smooth if the functions which parametrize it are continuously differentiable. If a surface is smooth and has no boundary, it is called a smooth surface without boundary. If a surface is given by Fx,y,z, then the surface is smooth without boundary if F for all x,y,z on the surface. Example: Consider the surface Then Fx,y,z 4x 9y z 8 F 8x i 18y j 4zk and F x y z. But,, is not on this surface. F is smooth without boundary. 4
urface Integrals Example Evaluate fx,y,zds where f x and is the part of the cone between the planes z 1andz. z x y We shall use spherical coordinates x sin cos,y sin sin,z cos. In spherical coordinates the equation of the cone is. Letting u, v we have for x,y, 4 and z on the surface of the cone that xu,v x, cos; yu,v y, sin ; z where and 1 z r x i y j zk cos i sin j k r u r sin i cos j r v r cos i sin j k r r 1 cos i sin j k and r r x ds 1 cos dd 8 3 sin d 15 4 Example Evaluate the integral of 5
fx,y,z x y z over the upper half of the sphere of radius 1 centered at the origin. We shall use spherical coordinates to parametrize the hemisphere. ince 1, we have x, sin cos, y, sin sin, z, cos Thus r, sin cosi sinsinj cosk where and. Then r, sinsini sincosj r, coscosi cossin j sink Hence r r sin Therefore fx,y,zds sin cos sin sin Remark: Very often one is interested in an integral of the form F nds cossin dd where n is a unit normal (perpendicular) vector to the surface pointing in the outward direction. From the discussion above it follows that the vectors r u and r v are both in the plane of the surface. Thus r u r v is to the surface. Hence r u r v r u r v is a unit normal. We choose the appropriate sign (either or-)whichmakesthisunitvectoroutward.onecanselectan appropriate point on the surface and see if r u r v is inward or outward. r u r v If it is inward, then use r u r v. r u r v Note that F nds D F r u r v r u r v dudv r u r v F r u r v dudv D 6
Thus, unless one is asked specifically for the unit vector n, it is not necessary to calculate r u r v. Example Let R be the region bounded by the cylinder x y 1 and the planes z andz x. Let be the entire boundary of R. Find the value of F nds where n is the outward directed unit normal on and F x i 3y j zk. Now is composed of 1,,and 3. On 1 n k F n z. But on z on 1 F n 1 F nds On 3 z x we parametrize as x u y v z u r x i y j zk u i v j u k This is outer so that r u i k r v j r u r v k i F r u r v x z u u u 3 F nds G u dudv Where G is the projection of 3 in the u,v plane. But since u x, v y and the plane z x slices the cylinder x y 1, we see that G is the interior of the circle x y 1. Thus on 3 we have 7
F nds 3 x da x y 1 1 rcosrdrd x y 1 da 1 3 cosd On we shall use cylindrical coordinates x rcos y rsin z z ince our cylinder is x y 1 r 1 r cosi sin j zk where z x cos, and. Taking u v z here, we have r sin i cos j r z k r r z cos i sin j r r z 1 Thus we may use N cos i sin j. This vector is outward, since gives n i. F N cos i 3sin j zk N cos 3sin Hence cos F nds cos 3sin dzd cos 5sin dzd Thus we have finally Remark: tewart uses the notation F nds 1 3 F nds. F d F nds for the surface integral of F over a surface. He also calls the F d the flux of F over. Example Parametrize the surface that is the part of the paraboloid 8
x y z that lies between the planes x 4andx, and give an expression for xds ketch the surface. olution: Let y u sinv, z u cosv, x u where v, and x 4 implies u. ru,v u i u sin vj u cosvk so r u ui sin vj cosvk r v u cosvj u sin vk r u r v i j k u sinv cosv u cosv u sin v : Thus Thus j u sin v k u cosv i sin v u ui u sin vj u cosvk r u r v 4u 4 cos v 4u 4 sin v u u 4u 1 xds y z 4 x r u r v dudv cos v u Example 5 page 956 in tewart u u 4u 1 dudv 391 6 17 1 6 9
Evaluate F d where F x,y,z yi xj zk and if the boundary of the solid region E enclosed by the paraboloid z 1 x y and the plane z. olution: The graph of the surface is shown below. The closed surface consists of a parabolic top surface 1 and a circular bottom surface : x y 1, z. We may parametrize the surface 1 as x u, y v, z 1 u v Then or ru,v ui vj 1 u v k ru,v u,v,1 u v Thus ru,v 1,, u u ru,v,1, v v Hence r u r v 1,, u,1, v u,v,1 ui vj k The projection of 1 onto the u,v plane, which is this case is the x,y plane, since x u and y v, is the circle D : x y 1. Thus using x and y instead of u and v we have 1
1 F d D yi xj zk xi yj k da D yi xj 1 x y k xi yj k da x y 1 4xy 1 x y da ince we are integrating over a circle of radius 1 centered at the origin, we switch to polar coordinates and have 1 F d 1 4r cossin 1 r rdrd 1 Now on Finally n k, z, and F x,y,z yi xj so that F n and F d F d 1 F d F d 1 Example Let be the surface of the solid cylinder T bounded by z andz 3andx y 4. Evaluate F nd, where F x y z x i y j zk and n is the outward unit normal. ketch the surface. OLUTION is composed of 1,,and 3. On 1 n k F n zx y z. 11
But z on 1 F n 1 F nds. On 3 z 3, n k F n zx y z 3x y 9 3x 3y 7. ince 3 is a disk of radius we introduce polar coordinates: x rcos, y rsin, ds rdrd and r x y so 3 F nds 3 3x 3y 7dxdy 3 r 9rdrd 13 On we shall use cylindrical coordinates x rcos y rsin z z. ince our cylinder is x y 4 r r cos i sin j zk where z 3. Taking u v z here, we have r sin i cos j r z k r r z cos i sin j Thus we may use N cos i sin j for a normal, since this is outward. F x y z x i y j zk so F 4cos 4sin z cos i sin j zk Then F N 4 z cos i sin j zk cos i sin j 4 z cos sin 4 4 z Hence Thus we have finally F nds 4 3 4 z dzd 168 F nds 1 3 F nds 13 168 3. 1
tokes Theorem and the Divergence Theorem tokes Theorem: Let be a regular surface bounded by a closed curve denoted by (boundary of ). Let F and curl F be continuous over. Then curl F nds F nds F d r Here the direction of integration around is positive with respect to the side of on which the normal n is drawn. Remark: Example Verify tokes Theorem when F y i 3z j 3xk and is the hemispheric surface z 1 x y. x y z 1 13
We shall use the outward normal n. We calculate F d r first. Now is the circle x y 1, z. We parametrize this as x cost, y sin t, z t F sin t i j 3costk r t x i y j zk cost i sint j k r t sin t i cost j Thus Now consider F d r sin tdt. curl F nds. curlf i j k x y z y 3z 3x 3 i 3 j k is the surface x y z 1 z. In spherical coordinates 1 x sin cos, y sinsin, z cos 14
Let u so that v and therefore r u,v sin u cosv i sinusin v j cosuk r u r v sin u cosv i sin u sinv j sin u cosuk At,, i.e. u v r u r v i which is outward. Let N r u r v is outward. Then curl F N 3sin u cosv 3sin u sin v sin u cosu curl F Nds 3sin u cosv 3sin u sinv sin u cosududv 3 cosv sinvsin ududv cosusinududv 3 cosv sin vu sinu dv 1 dv 3 cosv sinvdv 3 4 sinv cosv as before. Example Verify tokes Theorem is true for the vector field F x,y,z x i y j z k and is the part of the paraboloid z 1 x y that lies above the x,y plane and has upward orientation. ketch. 1 x y z 15
We must show curlf nds F dr curlf i j k x y z x y z x,y,z,, Thus curlf nds For the line integral we parametrize the boundary of, namely the circle x y 1inthex,y plane, as x cost, y sint, z t so rt costi sintj k r t sinti costj F t cos ti sin tj k F dr cos tsint sin tcost dt cos 3 t 3 t 3 Example Evaluate the surface integral F nd, where F x,y,z 3zi 5xj yk and is the part of the parabolic surface z x y that lies below the plane z 4 and whose orientation is given by the upward unit normal vector. olution: The surface is shown below: x y 16
We use tokes Theorem to evaluate this integral where C is the circle x y 4, z 4, t Then F nds F dr x y 4 C may be parametrized as x cost,y sint, z 4, so r costi sintj 4k and F nds x y 4 F dr F t r tdt 34i 1costj 4sintk sinti costj k dt 4sint cos t dt 4sint 1 1costdt Alternatively, we can directly compute the surface integral. First we calculate the integrand. F i j k x y z 3z 5x y i 3j 5k For the surface, we use x and y as parameters and have r x,y,x y, x y 4 r x 1,,x r y,1,y r x r y x, y,1 Then 17
F nd,3,5 x, y,1da xy D 4x 6y 5dA xy D 4rcos 6rsin 5rdrd. Here, the integral is changed into polar coordinates, since the region of integration is the disc x y 4. The Divergence Theorem (Gauss s Theorem) Remark: We shall call a surface positively oriented if the normal N is an outer normal; otherwise, is negatively oriented. Theorem: uppose is a regular, positively oriented, closed surface, and that F and div F are continuous over and the region V is enclosed by. Then F d F nds V div FdV V FdV where n is the outward unit normal to. Note: nmustbe outward. Example: Check the validity of the divergence theorem if F x i y j zk, wherev is the volume of the cube x,y,z l. Hence divf 1 1 1 3. V divfdv 3 V dv 3V 3l 3 18
Now we must calculate F nds over all six faces of the cube. On x lwe use n i F n l i y j zk i l Face xl F nds l Face xl ds l area of face l 3 On x F y j zk we may take n i. Thus F n Thus the contribution from this face is. We get similarly for y l, Face yl F nds l 3, whereas for y, Face y F nds. And for the face z l, Face zl F nds l 3 and on z, Face z F nds,. Finally we have where is the entire surface of the cube. Example Verify Gauss s Divergence theorem, namely F nds l 3 l 3 l 3 3l 3 F nds div FdV V where F x y z i x j k and is the closed parabolic bowl consisting of the two pieces and 1 :thecircle x y 1, z 1, : z x y ; x y 1 19
Thus is the bowl proper and 1 is the circular cap on top. ince F 1 V 1 1 x Fdv 1dv 1 1 1 x dzdydx x y V 1 1 1 x 1 x 1 x y dydx 1 1 r rdrd r r4 4 1 d We now evaluate On we use cylindrical coordinates F nds 1 F nds x rcos, y rsin z z x rcos, y rsin z x y r, Let r u, v x u cosv, y u sin v, z u u 1 v r u,v u cosv i u sin v j u k r u cosv i sinv j uk r v u sin vi u cosv j
r u r v i j k cosv sin v u u sin v ucosv u sinv j u cos vk u sin vk u cosv i u cosv i u sin v j uk Note that for v, r u 1, and we have which is inner. Therefore we use r u r v i k r u r v u cosv i u sinv j uk Therefore F u cosv u sin v u i u cosv j k F n u 3 cos v u 3 sinvcosv u 4 cosv 4u 3 sinvcosv u F nds 1 u 3 cos v u 3 sin vcosv u 4 cosv ududv 1 cos v 1 sin vcosv 5 cosv 1 dv 1 4 1 cosvdv 1 4 sin v 5 sin v 1 v Thus v sinv 4 8 F nds On 1 :thisisthecirclex y 1, z 1. We use the parametrization x rcos, y rsin, z 1 Therefore r u,v u cosv i u sin v j k u 1, v r u cos v sin v j r v u sinv i u cosv j 1
r u r v i j k cosv sin v u sin v ucosv u cos vk u sin vk uk As expected this is outward since u 1. F u cosv u sin v 1 i u cosv j k F r u r v u o that 1 F nds ududv 1 1 Example Evaluate F nds, where F x,y,z x 3 i y 3 j z 3 k and is the positively oriented surface of the solid bounded by the cylinder x y 1andz and z andn. Use the Divergence Theorem. Then V F nds divf dv V divf 3 x y z divf dv 3 x y z 1 dv 3 r z rdzdrd V 3 1 r 3 8 3 r drd 3 1 4 3 d 3 11 6 11 Example Let be the closed surface of the solid cylinder T bounded by the planes z andz 3 and the cylinder x y 4. Calculate the surface integral F nd where F x y z xi yj zk olution: The surface is x y 4
We use the divergence Theorem Then F nds F dv T F x y z xi x y z yj x y z zk divf 3x y z x 3y z x y 3z 5 x y z F nds 5 x y z dv T Using cylindrical coordinates to evaluate the integral we have F nds 5 x y z dv T 3 5 r z r dzdrd 3 As an alternative, we can calculate the surface integral directly. ince the surface,, is made up of three components, top (disc), side (cylinder) and bottom (disc), we deal with each components separately and then add the results. F nds 5 x y z 3 dv 5 r z r dzdrd 3 T As an alternative, we can calculate the surface integral directly. ince the surface,, is made up of three components, top (disc), side (cylinder) and bottom (disc), we deal with each components separately and then add the results. (i) On the top, z 3andn k. Thus F x y 9 xi yj 3k and 3
F nds top x y 4 3 x y 9 da 3 r 9 rdrd 3 13. r 3 9r drd (ii) On the bottom, z andn k. (Remember, we must use the outward normal.) Thus F x y xi yj k and F n. Hence, F nds. bottom (iii) On the side, we use cylindrical coordinates to parametrize with r. o, we have r cosi sinj zk r sini cosj r z k r r z sini k cosj k sinj cosi Before proceding, we check that we have the correct (outward) normal. This is OK, so we move to the integrand. On the surface, we have F 4 z cosi sinj zk Thus, Finally, F r r z 4 z 4cos 4sin 4 4 z F 3 nds 4 4 z ddz side 168 F nds F nds F nds top side 13 168 3 F nds bottom 4
Ma7 Additional Problems Green s, tokes and the Divergence Theorems Example Use tokes Theorem to evaluate F nd where F z i 3xj x 3 y 3 k and is the part of the surface z 5 x y above the plane z 1. Assume that oriented upwards. ketch. olution: r,,5 r tokes Theorem is F nd F dr C Now the boundary C of will be where the surface intersects z 1, that is, when 1 5 x y or x y 4. Thus C : x cost,y sint; t ; z 1 and F z i 3xj x 3 y 3 k rt costi sintj k F t 1 i 3costj cost 3 sint 3 k Then r t sinti costj and F t r t sint 1cos t Thus 5
F dr F t r tdt C sint 1cos t dt cost 6costsin t t 1 Example Verify Green s theorem for the line integral x ydx x ydy C where C is the positively oriented unit circle centered at the origin. olution: A parametrization of C is x cost,y sin t t. Thus x ydx x ydy cost sint sin t cost sin tcostdt C sintcost sin t cos t dt sin t 1 costsin t 1 t 1 costsint 1 t Here P x y and Q x y so Example where x y 1 Evaluate the surface integral Q x P y da F x y 1 F nd xyi 1 y j zk 1 1dA and the closed surface consists of the two surfaces z 4 3x 3y, z 4onthetoponthetop with normal upward, and z on the bottom with normal downward. olution: r,,4 3r 6
We use the divergence theorem, namely F nd F dv E where E is the volume enclosed by. F y y 1 1 Note that z x y 4 3. Using cylindrical coordinates we have z 4 3r, r,and 3 F 4 3r dv 3 1rdzdrd 3 4r 3r 3 drd r 3 4 r4 3 d E 4 3 3 4 16 9 d 4 3 d 8 3 Alternatively, we will calculate the suface integral directly. Let 1 denote the portion of the paraboloid on top and denote the disc on the bottom. For 1 we parametrize the surface using x and y as the parameters. Thus rx,y x,y,4 3 x y r x x,y 1,, 6x r y x,y,1, 6y The domain of rx,y is the disc D x,y x y 4 3 Then,. 7
r x r y i j k 1 6x 1 6y 6xi 6yj k We observe that the z component is positive, so we have the correct orientaion of the normal. F nd 1 D D xy, 1 y,4 3 x y 6x y 3y 3 4 3 x y 6x,6y,1dA xy da xy 3 6r 3 cos sin 3r 3 sin 3 4 3r rdrd 3 6r 4 cos sin 3r 4 sin 3 4r 3r 3 drd 6 r 5 cos sin 3 5 4 r5 sin 3 r 3 r 3 4 r4 r d 6 5 3 6 5 8 3 3 5 5 cos sin 3 4 3 cos 3 3 3 4 3 5 sin 3 4 3 3 4 16 9 5 d sin cos cos 4 3 3 For, we also parametrize using x and y, but now z soit smuchsimpler. rx,y x,y, r x x,y 1,, i r y x,y,1, j The domain is the same disc,d, asfor 1. r x r y i j k We observe that this vector points upward and we need the downward normal, so we use the negative and have Finally F nd D xy, 1 y,,, 1dA xy 8
F nd F nd 1 8 3 F nd 9