1/12/2002, INTEGRAL THEOREMS REVIEW Math 21a, O. KnillSUMMARY. We review the fundamental theorem of line integrals, Green’s theorem, Stokes theorem andGauss theorem. These four results generalize the fundamental theorem of calculus and are all of the formRdRF =RRdF, where dR is the boundary of R and dF is a derivative of F .INTEGRATION.Line integral:RγF · ds =RbaF (r(t)) · r0(t) dtSurface integralR RSfdS =RbaRdcf(X(u, v))|Xu(u, v) × Xv(u, v)| dudvFlux integral:R RSF · dS =RbaRdcF (X(u, v)) · Xu(u, v) × Xv(u, v) dudvDouble integral:R RRf dA =RbaRdcf(x, y) dxdy.Triple integral:R R RRf dV =RbaRdcRpof(x, y, z) dxdydz.AreaR RR1dA =R RR1 dxdySurface areaR R1dS =R R|Xu× Xv|dudvVolumeR R RB1 dV =R R RB1 dxdydzDIFFERENTIATION.Derivative: f0(t) =˙f(t) = d/dtf(t).Partial derivative: Ux(x, y, z) =∂U∂x(x, y, z).Gradient: grad(U) = (Ux, Uy, Uz)Curl in 2D: curl(F ) = curl((P, Q)) = Qx− PyCurl in 3D: curl(F ) = curl(P, Q, R) = (Ry− Qz, Pz− Rx, Qx− Py)Div: div(F ) = div(P, Q, R) = Px+ Qy+ Rz.IDENTITIES.div(curl(F )) = 0curl(grad(U)) =(0, 0, 0)LINE INTEGRAL THEOREM. If γ : t 7→ r(t), t ∈ [a, b] is a curve and U is a function either in 3D or the plane.ThenZγ∇U · ds = r(b) − r(a)where ∇U = grad(U) is the gradient of U .CONSEQUENCES.1) If the curve is closed, then the line integralRγ∇U · ds is zero.2) If F = ∇U, the line integral between two points P and Q does not depend on the chosen path.REMARKS.1) The theorem holds in any dimension. In one dimension, itreduces to the fundamental theorem of calculusRbaU0(x) dx = U(b) − U(a)2) The theorem justifies the name conservative for gradient vector fields.3) In physics, U is the potential energy and ∇U a force. The theorem says that for such forces we haveenergy conservation.PROBLEM. Let U(x, y, z) = x2+ y4+ z. Find the line integral of the vector field F (x, y, z) = ∇U (x, y, z)along the path r(t) = (cos(5t), sin(2t), t2) from t = 0 to t = 2π.SOLUTION. r(0) = (1, 0, 0) and r(2π) = (1, 0, 4π2) and U (r(0)) = 1 and U(r(2π)) = 1 + 4π2. Applying thefundamental theorem givesRγ∇U ds = U(r(2π)) − U(r(0)) = 4π2.GREEN’S THEOREM. If R is a region with boundary γ and F = (P, Q) is a vector field, thenZ ZRcurl(F ) dA =ZγF · dswhere curl(F )(x, y) = Qx− Py.REMARKS.1) The theorem can be used to calculate two dimensional integrals by a one dimensional integral along a curve.Sometimes, however, it is used to evaluate line integrals by evaluating 2D integrals.2) The curve is oriented in such a way that the region is to your left.3) The region has to has piecewise smooth boundaries (i.e. it should not look like the Mandelbrot set).4) If γ : t 7→ r(t) = (x(t), y(t)), the line integral isRba(P (x(t), y(t)), Q(x(t), y(t)) · (x0(t), y0(t)) dt.5) Green’s theorem was found by George Green (1793-1841) in 1827 and by Michel Ostogradski (1801-1861).CONSEQUENCES.1) If curl(F ) = 0 everywhere in space or the plane, then the line integral along a closed curve is zero. If twocurves connect two points then the line integral along those curves agrees.2) Taking F (x, y) = (−y, 0) gives an area formula Area(R) =R−y dx. Similarly Area(A) =Rxdy.PROBLEM. Find the line integral of the vector field F (x, y) = (x4+ sin(x) + y, x + y3) along the pathr(t) = (cos(t), 5 sin(t) + log(1 + sin(t))), where t runs from t = 0 to t = π.SOLUTION. curl(F ) = 0 imples that the line integral depends only on the end points (0, 1), (0, −1) of thepath. Take the simpler path r(t) = (−t, 0), t = [−1, 1], which has velocity r0(t) = (−1, 0). The line integral isR1−1(t4− sin(t), −t) · (−1, 0) dt = −t5/5|1−1= −2/5.REMARK. We could also find a potential U(x, y) = x5/5 − cos(x) + xy + y5/4. It has the property thatgrad(U) = F . Again, we get U(0, −1) − U (0, 1) = −1/5 − 1/5 = −2/5.STOKES THEOREM. If S is a surface in space with boundary γ and F is a vector field, thenZ ZScurl(F ) · dS =ZγF · dsREMARKS.1) Stokes theorem reduces to Greens theorem if F is z independent and S is contained in the z-plane.2) The orientation of γ is such that if you walk along γ and have your head in the direction, where the normalvector Xu× Xvof S points, then you have the surface to your left.3) Stokes theorem was found by Andr´e Amp`ere (1775-1836) in 1825. It was rediscovered by George Stokes(1819-1903).CONSEQUENCES.1) The flux of the curl of a vector field does not depend on the surface S, only on the boundary of S. This isanalogue to the fact that the line integral of a gradient field only depends on the end points of the curve.2) The flux of the curl through a closed surface like the sphere is zero: the boundary of such a surface is empty.PROBLEM. Compute the line integral of F (x, y, z) = (x3+ xy, y, z) along the polygonal path γ connecting thepoints (0, 0, 0), (2, 0, 0), (2, 1, 0), (0, 1, 0).SOLUTION. The path γ bounds a surface S : X(u, v) = (u, v, 0) parameterized by R = [0, 2] × [0, 1]. By Stokestheorem, the line integral is equal to the flux of curl(F )(x, y, z) = (0, 0, −x) through S. The normal vectorof S is Xu× Xv= (1, 0, 0) × (0, 1, 0) = (0, 0, 1) so thatR RScurl(F ) dS =R20R10(0, 0, −u) · (0, 0, 1) dudv =R20R10−u dudv = −2.GAUSS THEOREM. If S is the boundary of a region B in space with boundary S and F is a vector field, thenZ Z ZBdiv(F ) dV =Z ZSF · dSwhere div(F ) is the divergence of F .REMARKS.1) Gauss theorem is also called divergence theorem.2) Gauss theorem can be helpful to determine the flux of vector fields through surfaces.3) Gauss theorem was discovered in 1764 by Joseph Louis Lagrange (1736-1813), later it was rediscovered byCarl Friedrich Gauss (1777-1855) and by George Green.CONSEQUENCES.1) For divergence free vector fields F , the flux through a closed surface is zero. Such fields F are also calledincompressible or source free.PROBLEM. Compute the flux of the vector field F (x, y, z) = (−x, y, z2) through the boundary S of therectangular box [0, 3] × [−1, 2] × [1, 2].SOLUTION. By Gauss theorem, the flux is equal to the triple integral of div(F ) = 2z over the box:R30R2−1R212z dxdydz = (3 − 0)(2 − (−1))(4 − 1) = 27.1/10/2002, INTEGRAL THEOREMS: PEOPLE & GOSSIP Math 21a, O. KnillSUMMARY. These two pages give some information about people who were involved in the discovery of theintegral theorems.AMP`ERE. Andr´e Marie Amp`ere (1775-1836) made contributions to the theoryof electricity and magnetism. Stokes theorem was found by Amp`ere in 1825.Gossip:• While
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