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MIT 18 02 - Topic 45: More surface integrals, divergence theorem

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18.02A Topic 45: More surface integrals, divergence theorem.Author: Jeremy OrloffRead: SN: V10Closed Surfaces (with outward normals)Divergence: F = hM, N, P i ⇒ divF = Mx+ Ny+ Pz.Divergence Theorem (Also called Gauss’ Theorem.)This is like the flux form of Green’s Theorem.Take S a closed surface with interior D and F a vector field continuously differentiable onall of D. ThenZZSF · n dS =ZZZDdivF dVwhere n is the outward normal.Notation: Sometimes we write dS for n dS.Example: Verify the theorem for F = xi + yj + zk andS = sphere of radius a.F · n = a ⇒RRF · n dS = flux = a· area of S = a 4πa2= 4πa3.divF = 3 ⇒RRRDdivF dV = 3 × vol. = 4πa3. QEDExample: Let S1= part of the paraboloid z = 1 − x2− y2above thexy-plane and S2= the unit disk in the xy-plane.Take F = hyz, xz, xyi and find the flux of F upward through S1using the divergence theorem.Write F = Mi + Nj + P k, where M = yz, N = xz, P = xy ⇒ divF = Mx+ Ny+ Pz= 0.The divergence theorem says: flux =ZZS1+S2F · n dS =ZZZDdivF dV =ZZZD0 dV = 0⇒ZZS1F · n dS +ZZS2F · n dS = 0 ⇒ZZS1F · n dS = −ZZS2F · n dS.Therefore to find what we want we only need to compute the flux through S2.But S2is in the xy-plane, so dS = dx dy, n = −k ⇒ F · n dS = −xy dx dy on S2.Since S2is the unit disk, symmetry givesZZS2−xy dx dy = 0 ⇒ZZS1F · n dS = −ZZS2F · n dS = 0.118.02A topic 45 2Proof of the divergence theoremFirst we look at a small box. Then just like for Green’s Theorem we handle the generalvolume by adding together a lot of boxes.Let F = hM, N, P i, P0= (x0, y0, z0), S the box as shown.The box has limits: x: x0to x0+ ∆x, y: y0to y0+ ∆y, z: z0to z0+ ∆z.Look at the flux through the top and bottom.Top: n = k ⇒ F·n dS = P dS = P dx dy ⇒ flux =Zx0+∆xx0Zy0+∆yy0P (x, y, z0+ ∆z) dy dx.Bottom: n = −k ⇒ F·n dS = −P dS = −P dx dy ⇒ flux =Zx0+∆xx0Zy0+∆yy0−P (x, y, z0) dy dx.⇒ net flux through top and bottom =Zx0+∆xx0Zy0+∆yy0P (x, y, z0+ ∆z) − P (x, y, z0) dy dx=Zx0+∆xx0Zy0+∆yy0Zz0+∆zz0∂P∂z(x, y, z) dz dy dx.In brief, net flux through top and bottom =RRRDPZdV .Likewise, net flux through front and back =RRRDMxdV .Likewise, net flux through left and right =RRRDNydV .⇒ net flux out of the box =ZZZDMx+ Ny+ PzdV . QEDExample: Let S be the part of the plane shown. Compute the upward flux of F = y j twoways. (i) directly; (ii) using the divergence theorem.(i) The plane has equation x + y/2 + z/3 = 1⇒ N = h1, 1/2, 1/3i. ⇒ n dS = h3, 3/2, 1i dx dy.(We know this because we know the k component in the vector must be 1.)Thus F · n dS =32y dx dy⇒ flux =ZZR32y dx dy =Z10Z2−2x032y dy dx = . . . = 1.(ii) To use the divergence theorem we need a closed surface.We use the tetrahedron with faces S, S1, S2, S3. Computing flux:S1: n = −j ⇒ F · n = −y = 0 on S1⇒ flux = 0.S2: n = −i ⇒ F · n = 0 on S2⇒ flux = 0.S3: n = −k ⇒ F · n = 0 on S3⇒ flux = 0.Now using all these zeros and the divergence theorem we haveZZSF · n dS =ZZS+S1+S2+S3F · n dS =ZZZDdivF dV=ZZZD1 dV = volume =13base × height = 1.1x2y3zSS2S1S3xy12REnd of topic 45


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MIT 18 02 - Topic 45: More surface integrals, divergence theorem

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