EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM 1. TOKE THEOREM Let be an oriented surface with positively oriented boundary curve, and let F be a 1 vector field defined on. Then (1.1) ( F) d F ds Note: We need to have the correct orientation on the boundary curve. The easiest way to remember this is to use the right-hand rule: With your hand on the boundary curve, point your thumb in the direction of the normal vector to the surface and your palm towards the surface, then your fingers will curl around in the direction of the positive orientation of the boundary curve. Alternately, if you imagine yourself walking around the boundary curve with your head in the direction of the normal vector, then the surface will be on your left. ee the diagram on p533 for more details. Note also that the boundary curve always consists of closed loops, since it is the boundary of the surface. Questions using tokes Theorem usually fall into three categories: (1) Use tokes Theorem to compute F ds. In these examples it will be easier to compute the surface integral of F over some surface with boundary instead. (i.e. we use the left-hand side of tokes Theorem to help us compute the right-hand side). (2) Use tokes Theorem to compute ( F) d. This time it will be easier to compute the line integral F ds, i.e. we want to use the right-hand side of tokes theorem to help us compute the left-hand side. (3) Verify tokes Theorem. In these types of questions you will be given a surface and a vector field F. The question is asking you to compute both sides of tokes theorem and show that they are the same. 2. EXAMPLE Example 1: Let be the boundary of the part of the plane 2x + y + 2z 2 in the first octant, oriented counterclockwise as viewed from above. Let F(x, y, z) (x + y 2, y + z 2, z + x 2 ). Use tokes theorem to compute F ds. In this case, computing the line integral would require a lot of effort (the boundary has three smooth components and hence you would need to compute three separate integrals). Instead we note that the surface enclosed by the boundary curve is just the plane 2x + y + 2z 2, which has the parametrisation X : D R 3 X(s, t) (s, t, 12 ) (2 2s t), Date: May 1, 29. 1
2 EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM where D {(s, t) : t 2 2s, s 1} (note that X is the standard parametrisation of the surface as the graph of a function). A calculation shows that T s (1,, 1) ( T t, 1, 1 ) 2 T s T t (1, 12 ), 1. (check that the normal vector is positively oriented with respect to the boundary curve). We can also compute F ( 2z, 2x, 2y). Therefore the surface integral becomes 1 2 2s ( F) d (2s + t 2, 2s, 2t) (1, 12 ), 1 dt ds 1 2 2s (s t 1) dt ds, which is then easy to evaluate. Example 2: Let be the part of the paraboloid z 9 x 2 y 2 that lies above the plane z 5, oriented with normal vector pointing upwards, and let F(x, y, z) (yz, x 2 z, xy). Use tokes Theorem to evaluate ( F) d. Using the right-hand rule, we orient the boundary curve in the anticlockwise direction as viewed from above. It is the circle of radius 2 which lies on the plane z 5, and is centred at the origin. We can parametrise it by the equation Therefore the work integral becomes F ds F(c(t)) c (t) dt c(t) (x(t), y(t), z(t)) (2 cos t, 2 sin t, 5). ( 1 sin t, 2 cos 2 t, 4 sin t cos t ) ( 2 sin t, 2 cos t, ) dt 2 sin 2 t + 4 cos 3 t dt, which can be easily evaluated (don t forget the double angle formulas!). Example 3: Verify tokes theorem for the case where is the cylinder {(x, y) : x 2 +y 2 4, 3 z 1} with outwards pointing normal vector, and F (yz, z, y). Note that to give the boundary curves the correct orientation, the top curve 2 is oriented clockwise as viewed from above, and the curve 1 is oriented anti-clockwise as viewed from above. First we compute the left-hand side of tokes theorem in equation (1.1) (the surface integral). A calculation shows that F (, y, z), and the standard parametrisation
of the cylinder is EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM 3 X : D R 3 X(s, t) (2 cos s, 2 sin s, t), where D [, 2π] [ 1, 3]. The normal vector is T s ( 2 sin s, 2 cos s, ) T t (,, 1) T s T t (2 cos s, 2 sin s, ). (check that the normal vector points in the same direction as the orientation of the surface given above). In terms of the parametrisation, we have F(c(t)) (, 2 sin s, t), and so the surface integral becomes ( F) d (, 2 sin s, t) (2 cos s, 2 sin s, ) ds dt D 1 3 16π. 4 sin 2 s ds st Now we compute the right-hand side of equation (1.1). We parametrise the curve 1 by c(t) (2 cos t, 2 sin t, 3) for t 2π (notice that this is oriented anti-clockwise as viewed from above). Then we have c (t) ( 2 sin t, 2 cos t, ) and F(c(t)) ( 6 sin t, 3, 2 sin t), and so the line integral around 1 becomes F ds F(c(t)) c (t) dt 1 12π 12 sin 2 t 6 cos t dt To compute 2 F ds we parametrise the curve 2 by c(t) (2 cos t, 2 sin t, 1) for t 2π (notice that this is oriented clockwise as viewed from above). Then we have c (t) ( 2 sin t, 2 cos t, ) and F(c(t)) ( 2 sin t, 1, 2 sin t), therefore the line integral around 2 becomes F ds F(c(t)) c (t) dt 2 4π 4 sin 2 t 2 cos t dt
4 EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM Therefore the integral around the boundary of becomes F ds F ds + F ds 12π + 4π 16π, 1 2 the same result that we obtained earlier for ( F) d. Therefore we have verified tokes theorem in this case. 3. GAU DIVERGENE THEOREM Let be a solid region in R 3 and let be the surface of, oriented with outwards pointing normal vector. Gauss Divergence Theorem states that for a 1 vector field F the following equation holds (3.1) F d ( F) dv. Note that for the theorem to hold, the orientation of the surface must be pointing outwards from the region. Otherwise we will get a minus sign in equation (3.1). Note also that since is the boundary of then it is always a closed surface, i.e. it has no boundary. Questions about Gauss Divergence Theorem usually fall into the same three categories as tokes theorem. (1) ompute F d. In this case it will be easier to integrate F over the region which is bounded by. (2) ompute ( F) dv. In this case it will be easier to compute the surface integral of F over. (3) Verify Gauss Divergence Theorem. In these types of questions you will be given a region and a vector field F. The question is asking you to compute the integrals on both sides of equation (3.1) and show that they are equal. 4. EXAMPLE Example 1: Use the divergence theorem to calculate F d, where is the surface of the box with vertices (±1, ±2, ±3) with outwards pointing normal vector, and F(x, y, z) (x 2 z 3, 2xyz 3, xz 4 ). Note that the surface integral will be difficult to compute, since there are six different components to parametrise (corresponding to the six sides of the box), and so one would have to compute six different integrals. Instead, using Gauss theorem, it is easier to compute the integral of F over. First we compute F 2xz 3 + 2xz 3 + 4xz 3 8xz 3. Now we integrate this function over the region bounded by : F d ( F) dv 3 2 1 3 2 1 8xz 3 dxdydz which is easy to compute. Example 2: Verify the divergence theorem for the case where F(x, y, z) (x, y, z) and is the solid sphere of radius R centred at the origin.
EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM 5 Firstly we compute the left-hand side of (3.1) (the surface integral). To do this we need to parametrise the surface, which in this case is the sphere of radius R. The standard parametrisation using spherical co-ordinates is X(s, t) (R cos t sin s, R sin t sin s, R cos s). The tangent vectors to the co-ordinate curves are and so the normal vector to the surface is T s (R cos t cos s, R sin t cos s, R sin s) T t ( R sin t sin s, R cos t sin s, ), T s T t ( R 2 sin 2 s cos t, R 2 sin 2 s sin t, R 2 sin t cos t ). (check that this points outwards from the solid region ) In terms of the parametrisation, the vector field F is given by F(X(s, t)) (R cos t sin s, R sin t sin s, R cos s), and so the surface integral becomes π F d (R cos t sin s, R sin t sin s, R cos s) (R 2 sin 2 s cos t, R 2 sin 2 s sin t, R 2 sin t cos t ) ds dt R 3 4πR 3 π sin s ds dt Now we compute the right-hand side of equation (3.1) (the volume integral). First we compute F 1 + 1 + 1 3. Then using spherical co-ordinates, the integral becomes ( F) dv 3 dv π R R 3 π 3ρ 2 sin φ dρdθdφ sin φ dθdφ π 2πR 3 sin φ dφ 4πR 3 Therefore we have obtained the same result for the left-hand side and the right-hand side of the divergence theorem, and so we have verified the theorem in this case.
6 EXAMPLE OF TOKE THEOREM AND GAU DIVERGENE THEOREM 5. THE FUNDAMENTAL THEOREM OF ALULU Note: This section will not be tested, it is only here to help your understanding. If f(x) is a continuous function with continuous derivative f (x) then the Fundamental Theorem of alculus (FTO) states that b a f (x) dx f(b) f(a) In other words we can integrate (which involves computing the area under the graph of f by taking the limit of the Riemann sums) by simply anti-differentiating. This is much easier than computing Riemann sums! In the last few weeks we have studied three different higher-dimensional versions of this: The Fundamental Theorem of alculus for Line Integrals (1-dimensional FTO) f ds f(c(b)) f(c(a)) where is a curve parametrised by c(t) for a t b. tokes Theorem (2-dimensional FTO) ( F) d F ds where is an oriented surface in R 3 that has positively oriented boundary curve. Gauss Divergence Theorem (3-dimensional FTO) F dv F d where is a solid region in R 3 with boundary, oriented with outwards pointing normal vector. All of these theorems can be expressed using the same statement: Integrating the derivative of a function over a region is the same as integrating the function over the boundary. For 1-dimensional curves, the derivative of a function f is f, for 2-dimensional surfaces in R 3 the derivative of a vector field F is F, and for 3-dimensional solid regions in R 3 the derivative of a vector field is F.