INTEGRAL THEOREM PROBLEMS Math21a, O. KnillSUMMARY. This is a collection of problems on line integrals, Green’s theorem, Stokes theorem and the diver-gence theorem. Some of them are more challenging.LINE INTEGRALS GREEN THEOREM.The curve ~r(t) = hcos3(t), sin3(t)i is called a hypocycloid. It bounds aregion R in the plane.a) Calculate the line integral of the vector field F (x, y) = (x, y) alongthe curve.b) Find the area of the hypocycloid.a) Because curl(~F ) = 0 the result is zero by Green’s theorem.b) Use the vector field~F (x, y) = h0, xi which has curl(~F ) = 1. Theline integral isR2π0~F (~r(t)) · ~r0(t) dt =R2π0cos3(t)3 sin2(t) cos(t) dt=R2π03 cos4(t) sin2(t) dt = 3π/8. (To compute the integral, use that8 cos4(t) sin2(t) = cos(2t) sin2(2t) + sin2(2t)).LENGTH OF CURVE AND LINE INTEGRALS.Assume C : t 7→ ~r(t) is a closed path in space and~F (~r(t)) is the unittangent vector to the curve (that is a vector parallel to the velocityvector which has length 1).a) What isRC~F dr?b) Can~F be a gradient field?Answer:a)~F (~r(t)) = ~r0(t)/|~r0(t)|. By definition of the line integral,ZC~F (~r(t)) · ~r0(t) dt =Zba~r0(t)|~r0(t)|· ~r0(t) dt =Zba|~r0(t)| dt ,which is the length of the curve.b) No: If~F were a gradient field, then by the fundamental theorem ofline integrals, we would have that the line integral along a closed curveis zero. But because this is the length of the curve, this is not possible.SURFACE AREA AND FLUX.Assume S : (u, v) 7→ ~r(u, v) is a closed surface in space and~F (~r(u, v)) isthe unit normal vector on S (which points in the direction of ru× rv).a) What isR RS~F ·~dS?b) Is it possible that~F is the curl of an other vector field~G?c) Is it possible that div(~F ) = h0, 0, 0i everywhere inside the surface.Answer:a)~F (~r(u, v)) = (~ru× ~rv)/|~ru× ~rv|. By definition of the flux integral,R RS~F ·~dS =R RR~F (~r(u, v)) ·~ru×~rv=R RR(~ru×~rv/|~ru×~rv|)·~ru×~rv=R RR|~ru× ~rv| dudv which is the area of the surface.b) No, if~F were the curl of an other field G, then the flux of~F throughthe closed surface would be zero. But since it is the area, this is notpossible.c) From the divergence theorem follows that div(~F ) is nonzero some-where inside the surface.GREENS THEOREM AND LAPLACIAN.Assume R is a region in the plane and let ~n denote the unit normalvector to the boundary C of R. For any function u(x, y), we use thenotation ∂f/∂u = grad(u)·~n which is the directional derivative of u intothe direction ~n normal to C. We also use the notation ∆u = uxx+ uyy.Show thatZC∂u/∂n · dr =Z ZR∆u dAAnswer: Define~F (x, y) = h−B, Ai if ∂u/∂n = hA, Bi. The left integralis the line integral of~F along C. The right integral is the double integralover ∆u = curl(~F ).STOKES THEOREM OR DIVERGENCE THEOREMFindR RScurl(~F ) ·~dS, where S is the ellipsoid x2+ y2+ 2z2= 10 and~F (x, y, z) = hsin(xy), ex, −yzi.Answer. The integral is zero because the boundary of S is empty. Thisfact can be seen using Stokes theorem. It can also been seen by diver-gence theoremZ ZScurl(~F ) ·~dS =Z Z Zdivcurl(~F ) dV .using div(curl)(~F ) = 0.AREA OF POLYGONS.If Pi= (xi, yi), i = 1, . . . , n are the edges of a polygon in the plane, thenits area is A =Pi(xi− xi+1)(yi+1+ yi)/2.The proof is an application of Green’s theorem. The line integralof the vector field F (x, y) = (−y, 0) through the side Pi, Pi+1is(xi− xi+1)(yi+1+ yi)/2, because (xi+1− xi) is the projected area ontothe x-axis and (yi+1+ yi)/2 is the average value of the vector field onthat side. Because curl(F )(x, y) = 1 for all (x, y), the result followsfrom Greens theorem.The result can also be seen geometrically: (xi− xi+1)(yi+1+ yi)/2 is thesigned area of the trapezoid (xi, 0), (xi+1, 0), (xi+1, yi+1). In the picture,we see two of them. The second one is taken negatively.VOLUME OF POLYHEDRA.Verify with the divergence theorem: If Pi= (xi, yi, zi) are the edgesof a polyhedron in space and Fj= {Pi1, ...Pikj} are the faces, thenV =PjAjzjwhere Ajis the area of the xy-projection (*) of thepolygon Fjand zj= (zi1+ ... + zkj)/kjis the average z value of theface Fj.Solution. The vector field~F (x, y, z) = z has divergence 1. The fluxthrough a face F is |Fj|(zi1+ ... + zkj)/kj. Gauss theorem assures thatthe volume is the sum of the fluxes Ajzjthrough the faces.(*) The projection of a polygon is the ”shadow” when projecting from spacealong the z-axis onto the xy-plane. A triangle (1, 0, 1), (1, 1, 0), (0, 1, 2) forexample would be projected to the triangle (1, 0), (1, 1), (0, 1).STOKES AND GAUSS TOGETHER.Can you derive div(curl(~F )) = 0 using Gauss and Stokes theorem?Consider a sphere S of radius r around a point (x, y, z). It bounds aball G. Consider a vector field~F . The flux of curl(F ) through S iszero because of Stokes theorem. Gauss theorem tells that the integral off = div(curl(~F )) over G is zero. Because S was arbitrary, f must vanisheverywhere.FUNDAMENTAL THEOREM AND STOKES.Can you derive the identity curl(grad(~F )) = 0 from integral theorems?To see that the vector field~G = curl(grad(~F )) = 0 is identically zero, itis enough to show that the flux of~G trough any surface S is zero. ByStokes theorem, the flux trough S isRCgrad(~F )·dr. By the fundamentaltheorem of line integrals, this is zero.VOLUME COMPUTATION WITH GAUSS.Calculate the volume of the torus T (a, b) enclosed by the surface~r(u, v) = h(a + b cos(v)) cos(u), (a + b cos(v)) sin(u), b sin(v)i using Gausstheorem and the vector field~F (x, y, z) = hx, y, 0i/2.The vector field~F has divergence 1. The parameterization of the torusgives~ru× ~rv= b(a + b cos(v))hcos(u) cos(v), cos(v) sin(u), sin(v)i .The flux of this vectorfield through the boundary of the torus isR2π0R2π0b(a + b cos(v))2cos(v) dudv = 2π2ab2.GAUSS OR STOKES?You know that the flux of the vector field~G = curl(~F )(x, y, z) through5 faces of a cube D is equal to 1 each. What is the flux of the samevector field~G through the 6’th face?Solution: the problem is best solved with the divergence theorem: be-cause the flux of~G through the entire surface is zero, the flux throughthe 6’th face must cancel the sum of the fluxes 5 through the other 5surfaces. The result is −5.WORK COMPUTATION USING STOKES.Calculate the work of the vector field~F (x, y, z) = hx − y + z, y −z + x, z − x + yi, along the path C which connects the points(1, 0, 0) → (0, 1, 0) → (0, 0, 1) → (1, 0, 0) in that order.Answer. The line integral over each part is each 1. The total
View Full Document
Unlocking...