Lecture 20: 7.3 Stokes' Theorem. Let S Be a Surface with Unit Normal N And

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Lecture 20: 7.3 Stokes’ theorem. Let S be a surface with unit normal n and positively oriented boundary C, i.e. if you walk in the direction of the curve on the side of the normal then the surface should be on your left. Stokes’ theorem says Z ZZ F · ds = curl F · n dS C S if F is a smooth vector ﬁeld on S. If S is a domain in the x-y plane then Stoke’s theorem reduces to Green’s theorem. Z Z ∂Q ∂P In fact F·ds = P dx + Qdy, if F= P i+Qj and curl F·n = − , if n = k. C C ∂x ∂y R 3 3 3 Ex. Find the integral C −y dx + x dy − z dz, where C is the intersection of the cylinder x2 +y2 = 1 and the plane x+y+z = 1 and the orientation of C corresponds to a counterclockwise motion in the x-y plane. Sol. Let F = −y3i + x3j − z3k. The integral is by Stokes theorem equal to the surface integral of curl F·n over some surface S with the boundary C and with unit normal positively oriented with respect to the orientation of the boundary. We have curl F = ... = (3x2 + 3y2)k. We take S to be the region in the plane h(x, y, z) = x + y + z =√ 1 with boundary C. A unit normal to S is given√ by n = ∇h/|∇h| = (i+j+k)/ 3 and it has the correct orientation since n·k = 1/ 3 > 0.

We therefore get Z ZZ ZZ √ F · ds = curl F · n dS = 3(x2 + y2)/ 3 dS C S S √ Writing dS = dxdy/|n · k| = 3dxdy we get

ZZ Z 2π Z 1 Z 2π 3 1 3 3π 2 2 2 4 3(x + y ) dxdy = 3r rdr dθ = r dθ = 2π = x2+y2≤1 0 0 0 4 0 4 2

Sol. 2. Directly calculating the line integral. Parameterizing the curve C we can write x = cos t, y = sin t and z = 1 − x − y = 1 − cos t − sin t, 0 ≤ t ≤ 2π and write

Z Z 2π dx dy dz −y3dx + x3dy − z3dz = − y3 + x3 − z3 dt C 0 dt dt dt Z 2π = sin4 t + cos4 t + (1 − cos t − sin t)3(sin t − cos t) dt = ...a lot more work 0

1 2

Interpretation of curl. Furthermore, Stokes Theorem can alternatively be used to deﬁne the curl: The component of curl F in the direction of a unit vector n is deﬁned to be the limit as ε→0 of the line integral of F around a small circle Cε of radius ε perpendicular to n, divided by the area of the disc Sε enclosed by Cε: Z ZZ F · ds = curl F · n dS = curl F · n Area(Sε) Cε Sε where curl F · n is evaluated at some point on Sε. It follows that

R F · ds curl F · n = lim Cε ε→0 Area(Sε)

R z z z Ex. Show that C ye dx + xe dy + xye dz = 0 for a closed curve C. Sol. F = ∇(xyez) so curl F = 0 and by Stokes’s theorem the integral vanishes.

R 2 2 2 2 2 Ex. Find F·ds, where F = (−yi+xj)/(x +y ) and Ca is the circle x +y = a Ca in the x-y plane going counterclockwise. Sol. curl F = ... = 0. Hence one would have thought that by Stokes theorem the line integral would vanish. Wrong! because F is not continuous.

However, if we parameterize x = a cos t and y = a sin t,0 ≤ t < 2π, we get

Z Z 2π −y dx x dy F · ds = 2 2 + 2 2 dt Ca 0 x + y dt x + y dt Z 2π Z 2π −a sin t(−a sin t) a cos t(a cos t) 2 2 = 2 + 2 dt = sin t + cos t dt = 2π 0 a a 0

The reason Stokes’ theorem failed to hold in this case was that the vector ﬁeld F is singular when (x, y) = (0, 0), i.e. along the z-axis. 3

Proof of Stokes’ theorem for a graph. We have seen that Stokes’ theorem for a surface S in the x−y plane reduces to Green’s theorem. We will now show that Stokes’ theorem for a surface S that can be written as a graph z = f(x, y), (x, y) ∈ D, over a region D in the plane, also reduces to Green’s theorem. If T is the tangent vector to the boundary curve C the Stokes’ theorem can be written: Z ZZ F · T ds = curl F · n dS C S where ds is the arc length and dS the surface area element. The surface integral can then be written as an integral over D and the integral over the boundary curve can be written as an integral over the projection of the curve in the x−y plane. Then one can use Green’s theorem in the plane to show that these things are equal.

Since the surface can be written h(x, y, z) = z − f(x, y) a normal is given by N = ∇h = −fxi −fyj + k and the unit normal is given by n = N/|N|. The surface measure is dS = dxdy/|k · n|, where k·n = k·N/|N|= 1/|N|, so dS = |N| dxdy and ZZ ZZ G · n dS = −G1fx − G2fy + G3 dxdy, if G = G1i + G2j + G3k S D If we apply to F this to G = curl F we get ZZ ZZ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 curl F·n dS = − − fx − − fy + − dxdy. S D ∂y ∂z ∂z ∂x ∂x ∂y

If we parameterize the boundary x = x(t), y = y(t) and z = f(x, y) we have dz dx dy = f + f , by the chain rule, and dt x dt y dt Z Z b Z b dx dy dz dx dy F·ds = F1 +F2 +F3 dt = F1 +fxF3 + F2 +fyF3) dt C a dt dt dt a dt dt

This can now be considered as a line integral in the plane: Z Z F · ds = P dx + Q dy, where C ∂D P (x, y) = F1(x, y, f(x, y)) + fx(x, y)F3(x, y, f(x, y)),

Q(x, y) = F2(x, y, f(x, y)) + fy(x, y)F3(x, y, f(x, y)) We can therefore apply Greens formula in the plane. ∂P ∂F ∂F ∂F ∂F = 1 + 1 f + f 3 + f f 3 + f F ∂y ∂y ∂z y x ∂y x y ∂z xy 3 ∂Q ∂F ∂F ∂F ∂F = 2 + 2 f + f 3 + f f 3 + f F ∂x ∂x ∂z x y ∂x x y ∂z xy 3 so by Green’s theorem Z Z ZZ ∂Q ∂P F · ds = P dx + Q dy = − dxdy C ∂D D ∂x ∂y ZZ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 = − − fx − − fy + − dxdy, D ∂y ∂z ∂z ∂x ∂x ∂y 4

Some more calculations of surface integrals. Ex. Let S be the part of the hyperboloid x2 + y2 − z2 = 1 with 0 ≤ z ≤ 1. A parametrization of the surface is given by X(u, v) = (cos u − v sin u)i + (sin u + v cos u)j + vk, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1. a) Find the area element dS expressed in terms of the parametrization du dv. ZZ b) Find the surface integral z dS. S

Sol. a) Xu = (− sin u − v cos u)i + (cos u − v sin u)j and Xv = − sin ui + cos uj + k;  i j k  Xu×Xv =−sin u−v cos u cos u−v sin u 0 =(cos u−v sin u)i+(sin u+v cos u)j−vk − sin u cos u 1 √ 2 Hence dS = |Xu × Xv| dudv = 1 + 2v dudv.

ZZ Z 1Z 2π Z 1 2 3/2 1 3/2 p p (1 + 2v ) 3 − 1 b) z dS = v 1 + 2v2 dudv = 2π v 1 + 2v2 dv = π =π S 0 0 0 3 0 3