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The Divergence Theorem. (Sect. 16.8) I The divergence of a vector field in space. I The Divergence Theorem in space. I The meaning of Curls and Divergences. Applications in electromagnetism: I I I Gauss’ law. (Divergence Theorem.) Faraday’s law. (Stokes Theorem.) The divergence of a vector field in space Definition The divergence of a vector field F = hFx , Fy , Fz i is the scalar field div F = ∂x Fx + ∂y Fy + ∂z Fz . Remarks: I It is also used the notation div F = ∇ · F. I The divergence of a vector field measures the expansion (positive divergence) or contraction (negative divergence) of the vector field. I A heated gas expands, so the divergence of its velocity field is positive. I A cooled gas contracts, so the divergence of its velocity field is negative. The divergence of a vector field in space Example Find the divergence and the curl of F = h2xyz, −xy , −z 2 i. Solution: Recall: div F = ∂x Fx + ∂y Fy + ∂z Fz . ∂x Fx = 2yz, ∂y Fy = −x, ∂z Fz = −2z. Therefore ∇ · F = 2yz − x − 2z, that is ∇ · F = 2z(y − 1) − x. Recall: curl F = ∇ × F. i j k ∂y ∂z = h(0 − 0), −(0 − 2xy ), (−y − 2xz)i ∇ × F = ∂x 2xyz −xy −z 2 We conclude: ∇ × F = h0, 2xy , −(2xz + y )i. C The divergence of a vector field in space Example r Find the divergence of F = 3 , where r = hx, y , zi, and ρ p ρ = |r| = x 2 + y 2 + z 2 . (Notice: |F| = 1/ρ2 .) y z x F = , F = . , y z ρ3 ρ3 ρ3 −3/2 ∂x Fx = ∂x x x 2 + y 2 + z 2 Solution: The field components are Fx = ∂x Fx = x 2 + y 2 + z 2 −3/2 −5/2 3 − x x2 + y2 + z2 (2x) 2 1 y2 1 x2 ∂x Fx = 3 − 3 5 ⇒ ∂y Fy = 3 − 3 5 , ρ ρ ρ ρ 1 z2 ∂z Fz = 3 − 3 5 . ρ ρ 3 (x 2 + y 2 + z 2 ) 3 ρ2 3 3 ∇·F= 3 −3 = − 3 = − . ρ ρ5 ρ3 ρ5 ρ 3 ρ3 We conclude: ∇ · F = 0. C The Divergence Theorem. (Sect. 16.8) I The divergence of a vector field in space. I The Divergence Theorem in space. I The meaning of Curls and Divergences. Applications in electromagnetism: I I I Gauss’ law. (Divergence Theorem.) Faraday’s law. (Stokes Theorem.) The Divergence Theorem in space Theorem The flux of a differentiable vector field F : R3 → R3 across a closed oriented surface S ⊂ R3 in the direction of the surface outward unit normal vector n satisfies the equation ZZ ZZZ F · n dσ = (∇ · F) dV , S V where V ⊂ R3 is the region enclosed by the surface S. Remarks: I The volume integral of the divergence of a field F in a volume V in space equals the outward flux (normal flow) of F across the boundary S of V . I The expansion part of the field F in V minus the contraction part of the field F in V equals the net normal flow of F across S out of the region V . The Divergence Theorem in space Example Verify the Divergence Theorem for the field F = hx, y , zi over the sphere x 2 + y 2 + z 2 = R 2 . ZZ ZZZ F · n dσ = (∇ · F) dV . Solution: Recall: S V We start with the flux integral across S. The surface S is the level surface f = 0 of the function f (x, y , z) = x 2 + y 2 + z 2 − R 2 . Its outward unit normal vector n is n= ∇f , |∇f | ∇f = h2x, 2y , 2zi, p |∇f | = 2 x 2 + y 2 + z 2 = 2R, 1 hx, y , zi, where z = z(x, y ). R |∇f | R Since dσ = dx dy , then dσ = dx dy , with z = z(x, y ). |∇f · k| z We conclude that n = The Divergence Theorem in space Example Verify the Divergence Theorem for the field F = hx, y , zi over the sphere x 2 + y 2 + z 2 = R 2 . 1 R Solution: Recall: n = hx, y , zi, dσ = dx dy , with z = z(x, y ). R ZZ z ZZ 1 F · n dσ = hx, y , zi · hx, y , zi dσ. R S S ZZ ZZ ZZ 1 2 2 2 F · n dσ = x + y + z dσ = R dσ. R S S S The integral on the sphere S can be written as the sum of the integral on the upper half plus the integral on the lower half, both integrated on the disk R = {x 2 + y 2 6 R 2 , z = 0}, that is, ZZ ZZ R F · n dσ = 2R dx dy . S R z The Divergence Theorem in space Example Verify the Divergence Theorem for the field F = hx, y , zi over the sphere x 2 + y 2 + z 2 = R 2 . ZZ ZZ R F · n dσ = 2R dx dy . Solution: z S R Using polar coordinates on {z = 0}, we get ZZ Z 2π Z R R2 √ r dr dθ. F · n dσ = 2 R2 − r 2 S 0 0 The substitution u = R 2 − r 2 implies du = −2r dr , so, ZZ F · n dσ = 4πR 2 Z u −1/2 R2 S ZZ 0 F · n dσ = 2πR 2 S R 2 1/2 2u 0 (−du) = 2πR 2 2 ZZ ⇒ Z R2 u −1/2 du 0 F · n dσ = 4πR 3 . S The Divergence Theorem in space Example Verify the Divergence Theorem for the field F = hx, y , zi over the sphere x 2 + y 2 + z 2 = R 2 . ZZ F · n dσ = 4πR 3 . Solution: S ZZZ We now compute the volume integral ∇ · F dV . The V divergence of F is ∇ · F = 1 + 1 + 1, that is, ∇ · F = 3. Therefore ZZZ ZZZ 4 3 ∇ · F dV = 3 dV = 3 πR 3 V V ZZZ We obtain ∇ · F dV = 4πR 3 . V We have verified the Divergence Theorem in this case. C The Divergence Theorem in space Example r across the boundary of the region ρ3 between the spheres of radius R1 > R0 > 0, where r = hx, y , zi, p 2 and ρ = |r| = x + y 2 + z 2 . Find the flux of the field F = Solution: We use the Divergence Theorem ZZ ZZZ F · n dσ = (∇ · F) dV . S V ZZZ Since ∇ · F = 0, then (∇ · F) dV = 0. Therefore V ZZ F · n dσ = 0. S The flux along any surface S vanishes as long as 0 is not included in the region surrounded by S. (F is not differentiable at 0.) C The Divergence Theorem. (Sect. 16.8) I The divergence of a vector field in space. I The Divergence Theorem in space. I The meaning of Curls and Divergences. Applications in electromagnetism: I I I Gauss’ law. (Divergence Theorem.) Faraday’s law. (Stokes Theorem.) The meaning of Curls and Divergences Remarks: The meaning of the Curl and the Divergence of a vector field F is best given through the Stokes and Divergence Theorems. I 1 I ∇ × F = lim F · dr, S →{P } A(S) C I where S is a surface containing the point P with boundary given by the loop C and A(S) is the area of that surface. ZZ 1 ∇ · F = lim F · ndσ, R →{P } V (R) S where R is a region in space containing the point P with boundary given by the closed orientable surface S and V (R) is the volume of that region. The Divergence Theorem. (Sect. 16.8) I The divergence of a vector field in space. I The Divergence Theorem in space. I The meaning of Curls and Divergences. Applications in electromagnetism: I I I Gauss’ law. (Divergence Theorem.) Faraday’s law. (Stokes Theorem.) Applications in electromagnetism: Gauss’ Law Gauss’ law: Let q : R3 → R be the charge density in space, and E : R3 → R3 be the electric field generated by that charge. Then ZZZ ZZ q dV = k E · n dσ, R S that is, the total charge in a region R in space with closed orientable surface S is proportional to the integral of the electric field E on this surface S. The Divergence Theorem relates Zsurface ZZ Z Z integrals with volume integrals, that is, E · n dσ = (∇ · E) dV . S R Using the Divergence Theorem we obtain the differential form of Gauss’ law, 1 ∇ · E = q. k Applications in electromagnetism: Faraday’s Law Faraday’s law: Let B : R3 → R3 be the magnetic field across an orientable surface S with boundary given by the loop C , and let E : R3 → R3 measured on that loop. Then ZZ I d B · n dσ = − E · dr, dt S C that is, the time variation of the magnetic flux across S is the negative of the electromotive force on the loop. The Stokes I Theorem Z Z Zrelates line integrals with surface integrals, that is, E·r= (∇ × E) dσ. C S Using the Divergence Theorem we obtain the differential form of Gauss’ law, ∂t B = −∇ × E.