Vector Calculus Theorems Eric Auld March 2, 2016 If we have a two dimensional vector field hF1(x; y); F2(x; y)i, we can calculate the curl just as if it was a three-dimensional vector field–one with zero third component, and which doesn't depend on z anywhere. When we take the curl of such a field, we will always end up with zeros in the first two components. In other words, the curl of a two-dimensional vector field is a scalar function. I Calculate the curl of (3x − y; x + 5y). I Can you find an "anti-curl" for the constant function 1? A: 1 2 (−y; x) As for the normal curl, in three dimensions... I Find the curl of h−y; 2x; x + zi A: (0; −1; 3). I Remind yourself of the parameterization of the sphere of radius R, and the normal. A: G('; θ) = (R cos θ sin '; R sin θ sin '; R cos ') and n('; θ) = R sin '(x; y; z). Today we do a bunch of generalizations of the Fundamental Theorem of Calculus. From the Fundamental Theorem of b 0 Calculus, we know that a f = f (b) − f (a). That allows us to b 0 ´ calculate a f . ´ We're going to use this same process (finding an antiderivative helps us calculate an integral) in our computations today. But strangely, we will also go the other way: we will calculate f (b) − f (a) (or the analog of this) by taking the derivative and an b 0 "anti-boundary" and calculating a f . ´ In each of these cases, we either take the derivative and the "anti-boundary", or else we take the boundary and the anti-derivative. Green's Theorem F~ · d~r = curl(F~) dA: ˛ ¨ @D D We could start with the integral on the right something of the form G~ dA, and find an "anti-curl" of G~, and integrate along the boundary.˜ ~ Or we could start on the left, with something like C F · d~r, and take the curl and find an "anti-boundary". This´ is like doing the Fundamental Theorem backwards. ~ Notice that if we started with something like C F · d~r, we might also be tempted to use the Fundamental Theorem´ of Line Integrals (if we could find an "anti-gradient" for F~! In other words, a potential function.) Side Note F~ · d~r = curl(F~) dA: ˛ ¨ @D D Why is there a circle on the integral on the left? Because this theorem only works if you're doing a line integral on a loop, in other words, if you're integrating on something that is the boundary of a two dimensional thing. (A boundary can't have boundary). That means if we're supposed to take a line integral over something that has different endpoints, we have no chance of employing Green's Theorem! Example Calculate (3x − y) dx + (x + 5y) dy; ˛ unit circle with positive orientation. We know that this is the boundary of the filled in circle, and we know that the curl of this vector field is 2. Therefore the value of this integral is 2 dA = 2π: ¨ unit disk Example Compute the area enclosed by the ellipse parameterized by the curve s(t) = (a cos t; b sin t). We are trying to find 1 dA: ¨ solid ellipse Our theorem tells us we can take a "boundary and and anti-curl". The boundary of the solid ellipse is just the not-filled-in ellipse. 1 And the anti-curl of 1 is 2 (−y; x). So we need to integrate 1 1 2π (−y; x) · d~r: = (−y; x) · (−a sin t; b cos t) dt 2 ˆ 2 ˆ0 ellipse 1 2π = (−b sin t; a cos t) · (−a sin t; b cos t) dt 2 ˆ0 1 2π = ab sin2 t + ab cos2 t dt = πab: 2 ˆ0 Stokes' Theorem We can do a similar thing in three dimensions. Now the two-dimensional blob we're integrating over is maybe a curvy surface in the plane. A vector surface integral can be computed by integrating along the boundary, or vice-versa: curl F~ · n^ dS = F~ · d~r: ¨ ˛ S @S Or, if we take for granted that we are doing a vector surface integral on the left, and a vector line integral on the right, and we abbreviate curl by d, then we can write dF~ = F~: ¨ ˆ S @S Getting the Orientation of a Surface from an Oriented Curve Given an oriented curve that is the boundary of a surface, you should choose the unit normal that if you are a normal vector walking along the oriented curve, the surface is on your left. Given an oriented surface (so you know where the normal vector is), use this same procedure to determine an orientation of the boundary curve. Example What if I had a sphere with outward pointing normal vectors? A: Trick question: the sphere has no boundary. Example Use Stokes' Theorem to calculate the vector line integral of h−y; 2x; x + zi over the unit circle in the xy plane. (Use the upper unit hemisphere as an "anti-boundary".) Stokes' Theorem tells us that we should take the derivative and the "anti-boundary". In this case the derivative is the curl, and we found that the curl of this function is (0; −1; 3). Therefore we want to integrate (0; −1; 3) · n^ dS ¨ upper hemisphere 2π π=2 = (0; −1; 3) · (sin ')(x; y; z) d'dθ ˆ0 ˆ0 Notice just like there are a lot of anti-derivatives, there are a lot of "anti-boundaries", and any of them would have given the same answer! Example Let S be the surface defined by z = x2 + y 2 for z ≤ 4. Use Stokes' Theorem to evaluate h−3xz2; 0; z3i · n^ dS; ¨ S with upward-pointing normal. (Hint: use h0; xz3; 0i as an anti-curl). (Why is the displayed orientation the correct one?) Stokes' Theorem tells us we have to integrate h0; xz3; 0i over the boundary curve. How would we parameterize it? It is a circle of height 4 and radius 2. So let's use c(t) = (2 cos t; 2 sin t; 4). Now we just need to execute that line integral, which you can do. Divergence Theorem In the Divergence Theorem, we move between a vector surface integral, and a triple integral over a blob in three-dimensional space. F~ · n^ dS = div(F~) dV ; ‹ ˚ @R R where we use the outward-pointing normal. Why is there a loop around the double integral on the left? Because if we're going to find an "anti-boundary", we have to be dealing with a "closed-up" surface, like a sphere. Said succintly: a boundary can't have a boundary. Or: @2 = 0. ~ Notice if we were given a surface integral S F · n^ dS, we might also be tempted to take the boundary and˜ the "anti-curl", using Stokes' theorem. So we have two options in this case. Example Integrate F~ · n^ dS, where F~ = h2xy; 3yez ; x sin zi, and S is the outside of˜ the unit cube (that is, with corners at the origin and (1; 1; 1), with outward pointing normal. Instead of finding a normal for each of the six sides, and calculating six integrals, we just need to calculate the triple integral over the inside of the cube of div F~, which you can do..