Gauss's Theorem and Stoke's Theorem and Green's Theorem (PDF)

Gauss's theorem and Stoke’s theorem and Green's theorem Electromagnetics for medical physics ‫الكهرومغناطيسية للفيزياء الطبية‬ Second year ‫المستوى الثاني‬ The second lecture ‫المحاضرة الثانية‬ Dr. Faten Monjed Hussein Alfeel...

Gauss's theorem and Stoke’s theorem and Green's theorem Electromagnetics for medical physics ‫الكهرومغناطيسية للفيزياء الطبية‬ Second year ‫المستوى الثاني‬ The second lecture ‫المحاضرة الثانية‬ Dr. Faten Monjed Hussein Alfeel 1  GAUSS DIVERGENCE THEOREM, STOKES’ THEOREM, and GREEN’S THEOREM 1 THE DIVERGENCE THEOREM OF GAUSS The divergence theorem of Gauss states that if V is the volume bounded by a closed surface S and A is a vector function of position with continuous derivatives, then where nˆ is the positive (outward drawn) normal to S. Gauss divergence theorem can also be stated as following: The surface integral of the normal component of a vector A taken over a closed surface is equal to the integral of the divergence of A taken over the volume enclosed by the surface. Proof: Let S be a closed surface which is such that any line parallel to Figure 1: the coordinate axes cuts S in at most two points. Assume the equations of the lower and upper portions, S1 and S2, to be z = f1(x,y) and z = f2(x,y) respectively. Denote the projection of the surface on the xy plane by R. Consider (2) 2 For the upper portion S2, dydx = cosγ2dS2 = k.n2dS2 since the normal n2 to S2 makes an acute angle γ2 with k. For the lower portion S1, dydx = −cosγ1dS2 = k.n1dS1 since the normal n1 to S1 makes an obtuse angle γ1 with k. Then so from equations (2) and (3): (4) Similarly, by projecting S on the other coordinate planes, (5) (6) Adding equations (4), (5) and (6), Or Proved. Example Q: A vector field 𝐹⃗ = 𝑠𝑖𝑛𝑦𝑖 +𝑥(1+𝑐𝑜𝑠𝑦)𝑗̂ 3 Evaluate the integral ∫𝐶 𝐹⃗. 𝑑𝑟⃗ where C is the circular path given by x²+y²=a². Area of Plane Region by Green’s Theorem 1.1 Physical demonstration of the divergence theorem Let A = velocity v at any point of a moving fluid. From Figure (2 a) below: Volume of fluid crossing dS in ∆t seconds = volume contained in cylinder of base dS and slant height v∆t = (v∆t).ndS = v.ndS∆t Then, volume per second of fluid crossing dS is = v.dS Prom Figure (2 b) 4  Figure 2 above: Total volume per second of fluid emerging from closed surface S And we know that ∇.vdV is the volume per second of fluid emerging from a volume element dV. Then total volume per second of fluid emerging from all volume elements in S: Thus (9) 5 6 2 STOKES’ THEOREM Stokes’ theorem states that if S is an open, two-sided surface bounded by a closed, nonintersecting curve C (simple closed curve) then if A has continuous derivatives where C is traversed in the positive direction. The direction of C is called positive if an observer, walking on the boundary of S in this direction, with his head pointing in the direction of the positive normal to S, has the surface on his left. In other words, Stokes’ theorem may be stated as following: The line integral of the tangential component of a vector A taken around a simple closed curve C is equal to the surface integral of the normal component of the curl of A taken over any surface S having C as its boundary. Surface integral of the component of curl along the normal to the surface S, taken over the surface S and bounded by curve C is equal to : ∮ 𝐹⃗. 𝑑𝑟⃗ = ∬𝑠 𝑐𝑢𝑟⃗𝑙 𝐹⃗. 𝑛𝑑𝑠 𝐹⃗ line integral of the vector point function taken along closed curve C. where n is unit external normal vector to surface dS. 7 8 9 Proof: Let S be a surface which is such that its projections on the xy, Figure 3: yz and xz planes are regions bounded by simple closed curves, as indicated in the adjoining figure. Assume S to have representation z = f(x,y) or x = g(y,z) or y = h(x,z), where f,g,h are single-valued, continuous and differentiable 10 where Cis the boundary of S. Since , (12) If z = f(x,y) is taken as the equation of S, then the position vector to any point of S is But is a vector tangent to S and thus perpendicular to n, so that Substituting equation (14) into equation (12), we obtain (15) Or (16) Now on S,1(x,y,z) = A1(x,y,f(x,y)) = F(x,y); hence and equation (16) becomes (17) Then (18) where R is the projection of S on the xy plane. By Green’s theorem for the plane the last integral equals ¸Γ Fdx where Γ is the boundary of R. Since at each point (x,y) of Γ the value of F is the same as the value of A1 at each point (x,y,z) of C, and since dx is the same for both curves, we must have (19) 11  The theorem is also valid for surfaces S which may not satisfy the restrictions imposed above. For assume that S can be subdivided into surfaces S1,S2,.....Sk with boundaries C1,C2,.....Ck which do satisfy the restrictions. Then Stokes’ theorem holds for each such surface. Adding these surface integrals, the total surface integral over S is obtained. Adding the corresponding line integrals over C1,C2,.....Ck, the line integral over is obtained. 3 GREEN’S THEOREM IN THE PLANE If R is a closed region of the xy plane bounded by a simple closed curve C and if M and N are continuous functions of x and y having continuous derivatives in R, then (25) where C is traversed in the positive (counterclockwise) direction. Unless otherwise stated we shall always assume ¸ to mean that the integral is described in the positive sense. Green’s theorem in the plane is a special case of Stokes’ theorem. Also, it is of interest to notice that Gauss’ divergence theorem is a generalization of Green’s theorem in the plane where the (plane) region R and its closed boundary (curve) 12 C are replaced by a (space) region V and its closed boundary (surface) S. For this reason the divergence theorem is often called Green’s theorem in space. Green’s theorem in the plane also holds for regions bounded by a finite number of simple closed curves which do not intersect. Proof: Let the equations of the curves AEB and AFB (see Fig. 4) be Figure 4: y = Y1(x) and y = Y2(x) respectively. If R is the region bounded by C, we have Then (27) Similarly let the equations of curves EAF and EBF be x = X1(y) and x = X2(y) respectively. Then 13 Then (29) Adding equations (27) and (29) EX Use Green’s Theorem to evaluate∫𝐶 𝑦𝑥 2 𝑑𝑥 − 𝑥 2 𝑑𝑦 where C is shown. first let’s notice that if we walk along the path in the direction indicated then our left hand will be over the enclosed area and so this path does have the positive orientation and we can use Green’s Theorem to evaluate the integral. From the integral we have, 14 Remember that P is multiplied by x and Q is multiplied by y and don’t forget to pay attention to signs. It is easy to get in a hurry and miss a sign in front of one of the terms. Using Green’s Theorem the line integral becomes, D is the region enclosed by the curve. Since D is just a half circle it makes sense to use polar coordinates for this problem. The limits for D in polar coordinates are, after first converting to polar coordinates of course. 15 Use Green’s Theorem to evaluate where C is shown below. first let’s notice that if we walk along the path in the direction indicated then our left hand will be over the enclosed area and so this path does have the positive orientation and we can use Green’s Theorem to evaluate the integral. From the integral we have, Remember that P is multiplied by x and Q is multiplied by y and don’t forget to pay attention to signs. It is easy to get in a hurry and miss a sign in front of one of the terms. It is also easy to get in a hurry and just assume that PP is the first term in the integral and Q is the second. That is clearly not the case here so be careful! Using Green’s Theorem the line integral becomes, D is the region enclosed by the curve. We’ll leave it to you to verify that the equation of the line along the top of the region is given by y=3−xy=3−x. Once we have this equation the region is then very easy to get limits for. They are, 16 17