Page 433 - Advanced engineering mathematics

12.9 Stokes’s Theorem 413

In this,
∂ 2 ∂ 2 ∂ 2
∇· ∇= + + .
∂x 2 ∂y 2 ∂z 2
Since
∂E
∇× H = σE + ,
∂t
we have finally
∂ ∂E
∇(∇· E) − (∇· ∇)E =−μ σE + .
∂t ∂t
In practice, it is often the case that Q = 0. In this event,
Q =∇ · D =∇ · (cE) = ∇· E = 0,

hence,
∇· E = 0.
We can then further conclude that
2
∂E ∂ E
(∇· ∇)E = μσ + μ .
∂t ∂t 2
This is Maxwell’s equation for the electric intensity field. By a similar analysis we obtain
Maxwell’s equation for the magnetic intensity field:
2
∂H ∂ H
(∇· ∇)H = μσ + μ .
∂t ∂t 2
In the case of a perfect dielectric, σ = 0, and Maxwell’s equations become
2
2
∂ (E) ∂ (H)
(∇· ∇)E = μ and (∇· ∇)H = μ .
∂t 2 ∂t 2
If, instead of σ = 0, we have = 0, then we have
∂E ∂H
(∇· ∇)E = μσ and (∇· ∇)H = μσ .
∂t ∂t
These are vector forms of the three-dimensional heat equation.

SECTION 12.9 PROBLEMS

In each of Problems 1 through 5, use Stokes’s theorem 5. F = xyi + yzj + xyk with the part of the plane

to evaluate F · dR or (∇× F) · ndσ, whichever 2x + 4y + z = 8 in the first octant.
C
appears easier. 2
6. Calculate the circulation of F=(x − y)i+ x yj +axzk
2
2
2 2 2 2 2 counterclockwise about the circle x + y = 1. Here
1. F= yx i− xy j+ z k with the hemisphere x + y +
z = 4, z ≥ 0. a is any positive number. Hint: Use Stokes’s theo-
2
rem with any smooth surface having the circle as
2
2. F = xyi + yzj + xzk with the paraboloid z = x + y 2 boundary.
2
2
for x + y ≤ 9.

7. Use Stokes’s theorem to evaluate F · Tds,where C
C
2
2
3. F = zi + xj + yk with the cone z = x + y for
is the boundary of the part of the plane x + 4y + z = 12
0 ≤ z ≤ 4.
lying in the first octant, and
2
2
2
4. F = z i + x j + y k with the part of the paraboloid
2
2
z = 6 − x − y above the x, y - plane. F = (x − z)i + (y − x)j + (z − y)k.
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October 14, 2010 14:53 THM/NEIL Page-413 27410_12_ch12_p367-424