V9. Surface Integrals - MIT Mathematics

1 V9. Surface IntegralsSurface Integrals are a natural generalization of line Integrals : instead of integrating overa curve, we integrate over a Surface in 3-space. Such Integrals are important in any of thesubjects that deal with continuous media (solids, fluids, gases), as well as subjects that dealwith force fields, like electromagnetic or gravitational most of our work will be spent seeing how Surface Integrals can be calculated andwhat they are used for, we first want to indicate briefly how they are defined. The surfaceintegral of the (continuous) functionf(x, y, z) over the surfaceSis denoted by(1) Sf(x, y, z)dS .You can think ofdSas the area of an infinitesimal piece of the surfaces . To define theintegral (1), we subdivide the surfaceSinto small pieces having area Si, pick a point(xi, yi, zi) in thei-th piece, and form the Riemann sum(2) f(xi, yi, zi) the subdivision ofSgets finer and finer, the corresponding sums (2) approach a limitwhich does not depend on the choice of the points or how the Surface was subdivided.

2 Thesurface integral (1) is defined to be this limit. (The surfacehas to be smooth and not infinitein extent, and the subdivisions have to be made reasonably, otherwise the limit may notexist, or it may not be unique.)1. The Surface integral for most important type of Surface integral is the one which calculates the flux of avector field acrossS. Earlier, we calculated the flux of a plane vector fieldF(x, y) across adirected curve in thexy-plane. What we are doing now is the analog of this in assume thatSisoriented: this means thatShas two sides and one of them has beendesignated to be thepositive side. At each point ofSthere are two unit normal vectors,pointing in opposite directions; thepositively directedunit normal vector, denoted byn, isthe one standing with its base ( , tail) on the positive side.

3 IfSis a closed Surface , likea sphere or cube that is, a Surface with no boundaries, so that it completely encloses aportion of 3-space then by convention it is oriented so thatthe outer side is the positiveone, , so thatnalways points towards the outside (x, y, z) be a continuous vector field in space, andSan oriented Surface . We define(3)flux ofFthroughS= S(F n)dS= SF dS;the two Integrals are the same, but the second is written using the commonand suggestive abbreviationdS= the velocity field for the flow of an incompressible fluid of density 1, thenF nrepresents the component of the velocity in the positive perpendicular direction to the12V. VECTOR INTEGRAL CALCLUS Surface , andF ndSrepresents the flow rate across the little infinitesimal piece of surfacehaving areadS.

4 The integral in (3) adds up these flows across the pieces of Surface , so thatwe may interpret (3) as saying(4)flux ofFthroughS= net flow rate acrossS,where we count flow in the direction ofnas positive, flow in the opposite direction asnegative. More generally, if the fluid has varying density, then the right side of (4) is thenet mass transport rate of fluid acrossS(per unit area, per time unit).IfFis a force field, then nothing is physically flowing, and one just uses the term flux to denote the Surface integral, as in (3).2. Flux through a cylinder and now show how to calculate the flux integral, beginning withtwo surfaces wherenanddSare easy to calculate the cylinder and the the flux ofF=zi+xj+ykoutward through the portion of the cylinderx2+y2=a2in the first octant and below the planez= piece of cylinder is pictured.

5 The word outward suggeststhat we orient the cylinder so thatnpoints outward, , away from thez-axis. Since by inspectionnis radially outward and horizontal,(5)n=xi+yja.(This is the outward normal to the circlex2+y2=a2in thexy-plane;nhasnoz-component since it is horizontal. We divide byato make its length 1.)dzad nad aahTo getdS, the infinitesimal element of Surface area, we use cylindrical coordinates toparametrize the cylinder:(6)x=acos ,y=asin z=z .As the parameters andzvary, the whole cylinder is traced out ; the piece we want satisfies0 /2,0 z h. The natural way to subdivide the cylinder is to use little piecesof curved rectangle like the one shown, bounded by two horizontal circles and two verticallines on the Surface . Its areadSis the product of its height and width:(7)dS=dz a d.

6 Having obtainednanddS, the rest of the work is routine. We express the integrand ofour Surface integral (3) in terms ofzand :F ndS=zx+xya a dz d ,by (5) and (7);= (azcos +a2sin cos )dz d ,using (6).V9. Surface INTEGRALS3 This last step is essential, since thedzandd tell us the Surface integral will be calculatedin terms ofzand , and therefore the integrand must use these variables can nowcalculate the flux throughS: SF ndS= /20 h0(azcos +a2sin cos )dz d inner integral =ah22cos +a2hsin cos outer integral =[ah22sin +a2hsin2 2] /20=ah2(a+h).Example the flux ofF=xzi+yzj+z2koutward through that part of thespherex2+y2+z2=a2lying in the first octant (x, y, z, 0). again, we begin by findingnanddSfor the sphere. We take theoutside of the sphere as the positive side, sonpoints radially outward from the origin; wesee by inspection therefore that(8)n=xi+yj+zka,where we have divided byato makena unit do the integration, we use spherical coordinates.

7 On the Surface of the sphere, =a, so the coordinates are just the two angles and . The area elementdSis mosteasily found using the volume element:dV= 2sin d d d =dS d = area thicknessso that dividing by the thicknessd and setting =a, we get(9)dS=a2sin d d .asin dSaaaad a sin d d Finally since the area elementdSis expressed in terms of and , the integration willbe done using these variables, which means we need to expressx, y, zin terms of and .We use the formulas expressing Cartesian in terms of spherical coordinates (setting =asince (x, y, z) is on the sphere):(10)x=asin cos ,y=asin sin ,z=acos .We can now calculate the flux integral (3). By (8) and (9), the integrand isF ndS=1a(x2z+y2z+z2z) a2sin d d .Using (10), and noting thatx2+y2+z2=a2, the integral becomes SF ndS=a4 /20 /20cos sin d d =a4 212sin2 ] /20= VECTOR INTEGRAL CALCLUS3.

8 Flux through general a general Surface , we will usexyz-coordinates. It turns out that here it is simplerto calculate the infinitesimal vectordS=ndSdirectly, rather than calculatenanddSseparately and multiply them, as we did in the previous section. Below are the two standardforms for the equation of a Surface , and the corresponding expressions fordS. In the firstwe usezboth for the dependent variable and the function which givesits dependence onxandy; you can usef(x, y) for the function if you prefer, but that s one more letter tokeeptrack (x, y),dS= ( zxi zyj+k)dx dy(npoints up )(11a)F(x, y, z) =c,dS= FFzdx dy(choose the right sign);(11b)Derivation of formulas (x,y)RSdxdSdyRefer to the pictures at the right. The surfaceSlies over its projectionR,a region in thexy-plane. We divide upRinto infinitesimal rectangles havingareadx dyand sides parallel to thexy-axes one of these is shown.

9 Over itlies a piecedSof the Surface , which is approximately a parallelogram, since itssides are approximately infinitesimal vectordS=ndSwe are looking for hasdirection: perpendicular to the Surface , in the up direction;magnitude: the areadSof the infinitesimal shows our infinitesimal vector is the cross-productdS=A BwhereAandBare the two infinitesimal vectors forming adjacent sides ofthe parallelogram. To calculate these vectors, from the definition of thepartial derivative, we haveABdxdyndSABdydxf dyf dxyxAlies over the vectordxiand has slopefxin theidirection, soA=dxi+fxdxk;Blies over the vectordyjand has slopefyin thejdirection, soB=dyj+ B= i j kdx0fxdx0dy fydy = ( fxi fyj+k)dx dy ,which is (11a).To get (11b) from (11a), , our Surface is given by(12)F(x, y, z) =c,z=z(x, y)where the right-hand equation is the result of solvingF(x, y, z) =cforzin terms of theindependent variablesxandy.

10 We differentiate the left-hand equation in (12) with respectto the independent variablesxandy, using the chain rule and remembering thatz=z(x, y):F(x, y, z) =c Fx x x+Fy y x+Fz z x= 0 Fx+Fz z x= 0V9. Surface INTEGRALS5from which we get z x= FxFz,and similarly, z y= by (11a),dS=( z xi z yj+ 1)dx dy=(FxFzi+FyFzj+ 1)dx dy= FFzdx dy ,which is (11b).Example portion of the plane 2x 2y+z= 1 lying in the first octant forms atriangleS. Find the flux ofF=xi+yj+zkthroughS; take the positive side ofSas theone where the normal points up . the plane in the formz= 1 2x+ 2y, we get using (11a),dS= (2i 2j+k)dx dy ,so SF dS= S(2x 2y+z)dy dx= R(2x 2y+ (1 2x+ 2y))dy dx ,Ry1/2x1/2whereRis the region in thexy-plane over whichSlies. (Note that since the integrationis to be in terms ofxandy, we had to expresszin terms ofxandyfor this last step.)