Exercises

Compute the flux through the boundary ¶S of the solid S. Then use the divergence theorem to calculate the flux.

1. F( x,y,z) = á 0,0,1 ñ 2. F( x,y,z) = á 1,0,0 ñ S is [ 0,1] ×[ 0,1] ×[0,1] S is [ 0,1] ×[ 0,1]×[ 0,1] 3. F( x,y,z) = á 0,0,z ñ 4. F( x,y,z) = á y,x,z2 ñ S is the solid between S is the solid between z = 1 - x2 - y2 and z = 0 z = 1 - x2 - y2 and z = 0 5. F( x,y,z) = á x,y,z ñ 6. F( x,y,z) = á y,z,x ñ S is the unit sphere S is the unit sphere

Evaluate using the divergence theorem.

7. ó õ ó õ ¶S  x2dx + ydy + zdz 8. ó õ ó õ ¶S  ( x2-y2) dx + 2xydy S is the solid bounded between S is the solid bounded between z = 2 - x2 - y2 and z = x2 + y2 z = 2 - x2 - y2 and z = x2 + y2 9. ó õ ó õ ¶S  xdx+ydy+zdz 10. ó õ ó õ ¶S  zdx+xdy+ydz S is the tetrahedron with vertices S is the tetrahedron with vertices ( 0,0,0) , ( 1,0,0) , ( 0,1,0) , ( 0,0,1) ( 0,0,0) , ( 1,0,0) ,( 0,1,0) , ( 0,0,1) 11. ó õ ó õ unit  sphere  x3dx+y3dy+z3dz 12. ó õ ó õ unit  sphere  x3dx+y3dy

Verify Gauss' Law for the inverse square field by calculating the flux of F through the boundary of the surface.

F( x,y,z) =    k á x,y,z ñ   ( x2+y2+z2) 3/2

13. S is [ -1,1] ×[ -1,1] ×[-1,1] 14. S is [ -2,1] ×[-1,2] ×[ -1,1] 15. S is the interior of the 16. S is the interior of the cylinder x2+y2 = 1 for z in [ -1,1] cylinder x2+y2 = 1 for z in [ 1,2]

Verify Gauss' Law for the inverse square field by calculating the flux of F through the boundary of the surface.

F( x,y,z) =   k1 á x,y,z-1 ñ   ( x2+y2+( z-1) 2) 3/2   +   k2 á x,y,z+1 ñ   ( x2+y2+( z+1) 2) 3/2

17. S is [ -2,2] ×[ -2,2] ×[0,2] 18. S is [ -2,1] ×[ -1,2]×[ -3,1] 19. S is the interior of the 20. S is the interior of the cylinder x2+y2 = 1 for z in [ -2,2] cylinder x2+y2 = 1 for z in [ 0,2]

21. Explain why if F( x,y,z) and G( x,y,z) are differentiable in each component and if a, b are numbers, then

div( aF+bG) = a  div( F) +b  div( G)

then div( F) = 0 except at the origin, where F is undefined.

22. Suppose that F( x,y,z) is differentiable in each component and that div( F) = 0 everywhere except at the point ( a,b,c) .

23. Show that if F is an inverse square field of the form

F( x,y,z) =    k á x,y,z ñ   ( x2+y2+z2) 3/2

then div( F) = 0 except at the origin, where it is undefined.

24. Use exercises 21 - 23 to explain why the divergence of a sum of inverse square fields is zero except at the points where a field is undefined.

25. Green's First Formula: Use the Divergence theorem to prove that if f( x,y,z) and g( x,y,z) are sufficiently smooth, then

ó õ ó õ ¶S  f  Ñg · dS = ó õ ó õ ó õ S  Ñf · Ñg  dV+ ó õ ó õ ó õ S  f  Dg  dV

where Dg is the Laplacian of g which is defined

Dg =  ¶2g ¶x2  +  ¶2g ¶y2  +  ¶2g  ¶z2

26. Green's Second Formula: Use the result in exercise 25 to show that

ó õ ó õ ¶S  ( f Ñg-g Ñf) · dS = ó õ ó õ ó õ S  ( f  Dg  -g Df) dV

27. Let c = á c₁,c₂,c₃ ñ be a constant vector. By letting F = f c, show that

ó õ ó õ ¶S  f( x,y,z) dS = ó õ ó õ ó õ S  Ñf dV

(Note: the integrals are vector-valued in this result).

28. Let c = á c₁,c₂,c₃ ñ be a constant vector and show that

div( F × c) = curl( F) · c

Then use the divergence theorem to show that

ó õ ó õ ¶S  c · ( F × dS)  = -  ó õ ó õ ó õ S  curl( F ) · c  dV

Explain why the fact that this is true for every c implies the vector-valued result

ó õ ó õ ¶S  F × dS = ó õ ó õ ó õ S  curl( F)   dV

Exercises 29 - 32 use Gauss' theorem and the divergence theorem to explore Maxwell's equations.

(1) div( E) =  1 e0  r(x,y,z,t) (2) div( B) = 0 (3) curl( E) = -   ¶B ¶t (4) curl( B) = m0J + e0m0  ¶E ¶t

29. Use the divergence theorem to calculate the flux of a magnetic field through a closed surface.

30. Light in a vacuum satisfies Maxwell's equations when the source terms-i.e., charge and current densities-vanish. That is, r(x,y,z,t) = 0 and J( x,y,z,t) = 0. Also, e₀m₀ = c^-2, where c is the speed of light in a vacuum. What is the total flux of the magnetic and electric fields through any closed surface through which the light passes?

31. The Poynting vector is given by S = m₀^-1( E×B) . Show that

div( S) = m0-1curl( E) · B  -  m0-1curl( B) · E

Express the total flux of S through a closed surface containing the charges and currents using Maxwell's equations (3) and (4).

32. Consider light in a vacuum-i.e., r = 0 and J = 0 (see exercise 30). Light in a vacuum propogates in the direction of the Poynting vector S = m₀^-1( E×B) . Use the result in exercise 31 with J = 0 and e₀m₀ = c^-2 to show that the Flux of S through any solid W satisfies

Flux  of  S =   ¶( Energy) ¶t

where the energy contained in a solid W through which the light is passing is defined

Energy =  1 2m0 ó õ ó õ ó õ W  ( | B| 2+c-2| E| 2) dV