Stokes’ Theorem

1 Stokes ~F(x,y,z) = y,x,xyz and~G= curl~F. LetSbe the part of the spherex2+y2+z2= 25that liesbelow the planez= 4, oriented so that the unit normal vector at(0,0, 5)is 0,0, 1 . Use Stokes Theorem to find S~G d~ s a picture of the use Stokes Theorem , we need to first find the boundaryCofSand figure out how it should beoriented. The boundary is wherex2+y2+z2= 25 andz= 4. Substitutingz= 4 into the firstequation, we can also describe the boundary as wherex2+y2= 9 andz= figure out howCshould be oriented, we first need to understand the orientation ofS. We are toldthatSis oriented so that the unit normal vector at (0,0, 5) (which is the lowest point of the sphere)is 0,0, 1 (which points down). This tells us that the blue side must be the positive want to orient the boundary so that, if a penguin walks near the boundary ofSon the positive side (which we ve already decided is the blue side), he keeps the surface on his left.

2 If we imaginelooking down on the surface from a really high point like (0,0,100), then the penguin should walkclockwise (from our vantage point).So, using Stokes Theorem , we have changed the original problem into a new one:Evaluate the line integral C~F d~r, whereCis the curve described byx2+y2= 9 andz= 4,oriented clockwise when viewed from , we just need to evaluate the line integral, using the definition of the line integral. (This is like#4(a) on the worksheet Vector Fields and Line Integrals .) We start by parameterizingC. Onepossible parameterization is~r(t) = 3 cost,3 sint,4 , 0 t <2 .(1)If we look at this from above, it isoriented counterclockwise, which is the wrong orientation. Therefore, C~F d~r= 2 0~F(~r(t)) ~r (t)dt= 2 0 3 sint,3 cost,36 costsint 3 sint,3 cost,0 dt= 2 09dt= 18 (1)To come up with this, remember that we can parameterize a circlex2+y2= 1 inR2byx= cost,y= sint(and, astincreases, this goes around the circle counterclockwise).

3 Here, we re looking atx2+y2= 9; if we rewrite this as(x3)2+(y3)2= 1,then we can writex3= cost,y3= ~F(x,y,z) = y,x,z . LetSbe the part of the paraboloidz= 7 x2 4y2that lies above the planez= 3, oriented with upward pointing normals. Use Stokes Theorem to find Scurl~F d~ is a picture of the strategy is exactly the same as in #1. The boundary is wherez= 7 x2 4y2andz= 3, whichis the same asx2+ 4y2= 4 andz= oriented with normals pointing upward, the top side of the paraboloid (the yellow side in thepicture) is the positive side. If we imagine looking down on the surface from above, then a penguinwalking around on the positive (yellow, in this case) side keeps the surface on his left by , by Stokes Theorem , the original problem can be rewritten asEvaluate the line integral C~F d~r, whereCis the curve described byx2+ 4y2= 4 andz= 3,oriented counterclockwise when viewed from parameterization of this curve is~r(t) = 2 cost,sint,3.

4 (2)This goes counterclockwise when viewedfrom above (as we want), so C~F d~r= 2 0~F(~r(t)) ~r (t)dt= 2 0 sint,2 cost,3 2 sint,cost,0 dt= 2 0(2 sin2t+ 2 cos2t)dt= 2 02dt=4 (2)To come up with this parameterization, rewritex2+ 4y2= 4 as(x2)2+y2= 1 and then usex2= cost,y= sint. It s easyto check that it s reasonable: if we plug inx= 2 cost,y= sint, andz= 3, then the equationsx2+ 4y2= 4 andz= 3 areindeed planez=x+ 4and the cylinderx2+y2= 4intersect in a curveC. SupposeCis orientedcounterclockwise when viewed from above. Let~F(x,y,z) = x3+ 2y,siny+z,x+ sinz2 . Evaluate theline integral C~F d~ ll use Stokes Theorem . To do this, we need to think of an oriented surfaceSwhose(oriented) boundary isC(that is, we need to think of a surfaceSand orient it so that the givenorientation ofCmatches). Then, Stokes Theorem says that C~F d~r= Scurl~F d~S. Let s computecurl~Ffirst. (It s worthwhile to do this first because, if we find out it s~0, then we know the integralwill be 0 without any more work.)

5 In this case, curl~F= 1, 1, 2 .Now, let s think of a surface whose boundary is the given curveC. We are told thatCis the intersectionof a plane and a cylinder (left picture), so one surface we could use is the part of the plane inside thecylinder (right picture):xyzxyzLet s call thisSand figure out how it should be oriented. We want to orientSso that, if a penguinwalks along the given curveC(going counterclockwise when viewed from above) on the positive sideofS, he keeps the surface on his left. This means that we want the top side ofSto be the positive side, so we should orientSwith normals pointing , using Stokes Theorem , we have changed the original problem into a new one:Evaluate the flux integral S~G d~S, whereSis the part of the planez=x+ 4 inside thecylinderx2+y2= 4, oriented with normals pointing upward, and~Gis the vector field~G(x,y,z) = 1, 1, 2 .To do this new problem, let s follow the same three steps we used in #4(a) on the worksheet FluxIntegrals.

6 First, we parameterizeS. Since the plane has equationz=x+4, we can usexandyas our we letx=uandy=v, thenz=u+ 4. This gives the parameterization~r(u,v) = u,v,u+ 4 . Sincewe are only interested in the part of the plane inside the cylinderx2+y2= 4, we wantx2+y2< terms ofuandv, this saysu2+v2<4, so the regionRin theuv-plane describing the possibleparameter values is a disk:3R-22u-22vNext, we need to see what orientation this parameterization describes. To do this, we compute~ru ~rv:~ru= 1,0,1 ~rv= 0,1,0 ~ru ~rv= 1,0,1 This always points upward, which matches the orientation we want. So, the flux integral is S~G d~S= R~G(~r(u,v)) (~ru ~rv)dA= R 1, 1, 2 1,0,1 dA= R 1dAAlthough we could evaluate this double integral by converting it to an iterated integral, there is aneasier way remember that R1dAgives the area ofR(see #2(a) on the worksheet DoubleIntegrals ). Therefore, S~G d~S= R1dA= (area ofR)= 4 the oriented curve parameterized by~r(t) = cost,sint,8 cos2t sint ,0 t <2 , and let~Fbe the vector field~F(x,y,z) = z2 y2, 2xy2,e zcosz.

7 Evaluate C~F d~ line integral is very difficult to compute directly, so we ll use Stokes Theorem . Thecurl of the given vector field~Fis curl~F= 0,2z,2y 2y2 .To use Stokes Theorem , we need to think of a surface whose boundary is the given curveC. First, let stry to understandCa little better. We are given a parameterization~r(t) ofC. In this parameterization,x= cost,y= sint, andz= 8 cos2t sint. So, we can see thatx2+y2= 1 andz= 8 x2 other words,Cmust be the intersection of the surfacex2+y2= 1 (which is a cylinder) and thesurfacez= 8 x2 y(which we don t need to visualize particularly well, beyond noticing that it sthe graph of a functionf(x,y) = 8 x2 y). So, one surface we could use is the part of the surfacez= 8 x2 yinside the cylinderx2+y2= 1 (right picture).4xyzxyzLet s call this surfaceSand figure out how it should be oriented. The original curve was parameterizedusingx= cost,y= sint, so when viewed from above, it was oriented counterclockwise.

8 Therefore, wewant to orientSso that is top is the positive side (a penguin walking on the top along the boundary,going counterclockwise when viewed from above, keeps the surface on his left). So, we should orientSwith normals pointing , using Stokes Theorem , we have changed the original problem into a new one:Evaluate the flux integral S~G d~S, whereSis the part of the surfacez= 8 x2 yinside thecylinderx2+y2= 1, oriented with normals pointing upward, and~Gis the vector field~G(x,y,z) = 0,2z,2y 2y2 .To do this new problem, let s follow the same three steps we used in #4(a) on the worksheet FluxIntegrals .First, we parameterizeS. Since the surface has equationz= 8 x2 y, we can parameterize it as~r(u,v) = u,v,8 u2 v . Since we are only interested in the part of the surface inside the cylinderx2+y2= 1, we wantx2+y2<1; in terms ofuandv, this saysu2+v2<1, so the regionRin theuv-plane describing the possible parameter values is a disk:R-11u-11vNext, we need to see what orientation this parameterization describes.

9 To do this, we compute~ru ~rv:~ru= 1,0, 2u ~rv= 0,1, 1 ~ru ~rv= 2u,1,1 This always points upward, which matches the orientation we want. So, the flux integral is S~G d~S= R~G(~r(u,v)) (~ru ~rv)dA= R 0,2(8 u2 v),2v 2v2 2u,1,1 dA= R(16 2u2 2v2)dA5 Since our regionRis a disk, let s do this integral in polar coordinates. The diskRcan be describedas 0 r <1, 0 <2 , so S~G d~S= 2 0 10(16 2r2) r dr d = 2 0 10(16r 2r3)dr d = 2 0(8r2 12r4 r=1r=0)d = 2 0152d =15 the curve of intersection of2x2+ 2y2+z2= 9withz=12 x2+y2, oriented counterclockwisewhen viewed from above, and let~F(x,y,z) = 3y,2yz,xz3+ sinz2 . Evaluate C~F d~ , we ll use Stokes Theorem . The curl of the given vector field is curl~F= 2y, z3, 3 .We ll start by thinking of an oriented surfaceSwhose (oriented) boundary is the given curveC. Thecurve is the intersection of an ellipsoid with a cone:xyzIt appears from the picture that the curve lies in a plane parallel to thexy-plane.

10 To verify this, let slook at the equations 2x2+ 2y2+z2= 9 andz=12 x2+y2defining the curve. The second equationcan be rewritten asx2+y2= 4z2. Plugging this into the first equation, 9z2= 9, soz= 1. Since thecone is only defined forz 0, we know the intersection is wherez= 1, in which casex2+y2= are many surfaces whose boundary is the given curveC. From what we know so far, there areseveral surfaces that might come to mind:6xyzxyzxyzxyzThe part of the coneinside the part of the el-lipsoid lying abovethe part of the el-lipsoid lying belowthe part of theplanez= 1 lying in-sidex2+y2= of these should be oriented so that the yellow side is the positive side. The last is probablythe simplest, so let s use that. To have the yellow side be the positive side, we want the normals topoint upward. Thus, we have rewritten the original problem as:Evaluate the flux integral S~G d~S, whereSis the part of the planez= 1 lying insidex2+y2= 4,oriented with normals pointing upward, and~Gis the vector field~G(x,y,z) = 2y, z3, 3.