On Divergence Theorem; Line Integral Definition of Curl (RHB 9.8, 9.7)

Lecture 21: More on Divergence Theorem; Line Integral Deﬁnition of Curl (RHB 9.8, 9.7)

21. 1. Worked Examples of use of Divergence Theorem

Let A = (y x)e + x2ze + ze . Calculate the surface integral I = A dS over the open 1 2 3 RS curved surface− of the cylinder x2 + y2 = a2 bounded by z = 0 and z = 1. ·

Writing the closed surface of the cylinder as the open curved surface SC plus the top face ST and bottom face SB, the divergence theorem tells us

Z A dV = Z A dS + Z A dS + Z A dS V ∇ · SC · ST · SB · But for the given A we ﬁnd A = 1 + 0 + 1 = 0 therefore we ﬁnd ∇ · −

Z A dS = Z A dS Z A dS SC · − ST · − SB · At the top (z = 1) of the cylinder dS = e dS and A dS = z dS = dS 3 ·

. 2 .. Z A dS = πa ST · Similarly the integral over the bottom face where z = 0 gives zero. Therefore our result is

2 Z A dS = πa SC · −

Volume of a body Consider the volume of a body: V = Z dV V Recalling that r = 3 we can write ∇ · 1 V = Z r dV 3 V ∇ · which using the divergence theorem becomes 1 V = Z r dS 3 S ·

2 2 2 2 Example Consider the hemisphere x + y + z a centered on e3 with bottom face at z = 0. Recalling that the divergence theorem holds≤ for a closed surface, the above equation for the volume of the hemisphere tells us 2πa3 1 = Z r dS + Z r dS . 3 3 hemisphere · bottom ·

On the bottom face dS = e3 dS therefore r dS = z dS = 0 since z = 0. Hence the only contribution comes from the− (open) surface of· the hemisphere− and we see

3 2πa = Z r dS . hemisphere ·

81 We can verify this directly by using spherical polars to evaluate the surface integral. As was derived in lecture 19, for a hemisphere of radius a

2 dS = a sin θ dθ dφ er .

On the hemisphere r dS = a3 sin θ dθ dφ · π/2 2π . 3 3 .. Z r dS = a Z sin θ dθ Z dφ = 2πa S · 0 0 as predicted by the divergence theorem.

21. 2. Line Integral Deﬁnition of Curl

Let ∆S be a small planar surface containing the point P , bounded by a closed curve C, with unit normal n and (scalar) area ∆S.

Let A be a vector field defined on ∆S, then the component of A parallel to n is defined to be ∇ ×

1 n ( A) = lim I A dr ∆S 0 ∆S · ∇ × → C ·

NB The integral around C is taken in the right-hand sense with respect to the normal n to the surface – as in the ﬁgure above. This deﬁnition of curl is independent of the choice of basis.

21. 3. Cartesian form of curl A

Let P be a point with cartesian coordinates (x0, y0, z0) situated at the centre of a small rectangle ABCD of size δ1 δ2, area ∆S = δ1 δ2, in the (e e ) plane. × 1− 2 The line integral around C is given by the sum of four terms

B C D A I A dr = Z A dr + Z A dr + Z A dr + Z A dr C · A · B · C · D ·

Since r = xe1 + ye2 + ze3, we have dr = e1 dx along DA and CB, and dr = e2 dy along AB and DC. Therefore

B B C A I A dr = Z A2 dy Z A1 dx Z A2 dy + Z A1 dx C · A − C − D D

82 For small δ1 & δ2, we can Taylor expand the integrands, viz

A A Z A1 dx = Z A1(x, y0 δ2/2, z0) dx D D −

x0+δ1/2 δ ∂A (x, y , z ) " 2 1 0 0 2 # = Z A1(x, y0, z0) + O(δ2) dx x0 δ1/2 − 2 ∂y − B B Z A1 dx = Z A1(x, y0 + δ2/2, z0) dx C C

x0+δ1/2 δ ∂A (x, y , z ) " 2 1 0 0 2 # = Z A1(x, y0, z0) + + O(δ2) dx x0 δ1/2 2 ∂y − so 1 A C 1 A B "Z A dr + Z A dr# = "Z A1 dx Z A1 dx# ∆S D · B · δ1 δ2 D − C

1 x0+δ1/2 ∂A (x, y , z ) " 1 0 0 2 # = Z δ2 + O(δ2) dx δ1δ2 x0 δ1/2 − ∂y −

∂A1(x0, y0, z0) as δ1, δ2 0 → − ∂y →

A similar analysis of the line integrals along AB and CD gives

B D 1 ∂A2(x0, y0, z0) "Z A dr + Z A dr# as δ1, δ2 0 ∆S A · C · → ∂x →

Adding the results gives for our line integral deﬁnition of curl yields ∂A ∂A " 2 1 # e3 ( A) = ( A) = · ∇ × ∇ × 3 ∂x − ∂y (x0, y0, z0) in agreement with our original deﬁnition in Cartesian coordinates.

The other components of curl A can be obtained from similar rectangles in the (e2 e3) and (e e ) planes, respectively. − 1− 3 21. 4. Physical Interpretation of Curl (Dawber 6.2)

Let’s illustrate the meaning of curl in physics by means of a simple example (see ﬁgure).

83 ∂F Let F = aye so that 1 = 0; C is a contour in the (e e ) plane, enclosing an area ∆S. 1 ∂y 6 1− 2 The work done on a test particle in moving it around the closed curve C is

I F dr = circulation of F about C C ·

(Remember that when the line integral of F around a closed curve is non-zero the force is non-conservative which implies curl F is non-zero.) The integral is given by the sum of four terms

B C D A I F dr = Z F dr + Z F dr + Z F dr + Z F dr C · A · B · C · D · B C = Z F1 dx Z F1 dx, A − D the 2nd and 4th terms vanish because F dr = 0 along the vertical lines BC and DA. ·

B C ∂F1 In the example (see ﬁgure), F1 = ay = 0, therefore Z F1 dx = Z F1 dx. ∂y 6 A 6 D Hence I F dr = 0. C · 6 But, from the integral deﬁnition of curl, we know that

I F dr (curlF )3 ∆S C · ≈

Therefore: (curl F )3 = 0 (in our example it is less than zero) A non-zero amount of work is done in moving6 the test particle around the small closed⇐⇒ path. Alternatively one can think of the non-zero circulation of F causing a small test particle to rotate about its centre with axis of rotation in the e direction (using r.h. thumb rule). − 3

Thus, in general, n (curl F ) is a measure of the circulation of the vector field F about an infinitesimal area· with normal n. It can be shown that curl F , when defined in this way, is independent of the shape of the infinitesimal area ∆S.