Begin2.DVI
Example 8-6. Find the solution of the exact differential equation (2 xy −y^2 )dx + (x^2 − 2 xy )dy = 0 Solution After verifying ...
In Green’s theorem the functions M and N are arbitrary. Therefore, in the special case M=−yand N= 0 one obtains ∫ C ©− y dx = ∫∫ ...
Bto Cto A. On the straight-line C 1 where y= 0 and dy = 0 the first line integral has the value of zero. On the cycloid one find ...
Let M(u, v ) = 1 2 ( x ∂y ∂u −y ∂x ∂u ) and N(u, v ) = 1 2 ( x ∂y ∂v −y ∂x ∂v ) and apply Green’s theorem to the integral (8.30) ...
The Curl of a Vector Field Let F =F(x, y, z) = F 1 (x, y, z)ˆe 1 +F 2 (x, y, x )ˆe 2 +F 3 (x, y, z)ˆe 3 denote a continuous ve ...
where μ 0 = ∫ 2 π 0 F 0 ·dξ, F 0 =F(x 0 , y 0 , z 0 ) μ 1 = ∫ 2 π 0 dF d ·d ξ μ 2 = ∫ 2 π 0 1 2! d^2 F d^2 ·dξ ··· where ...
The curl F can be expressed by using the operator ∇in the determinant form curl F =∇× F = ∣∣ ∣∣ ∣∣ ˆe 1 ˆe 2 ˆe 3 ∂ ∂x ∂ ∂y ∂ ...
Figure 8-12. Comparison of two dimensional velocity fields. In comparison, consider the two-dimensional velocity field V =yˆe 1 ...
The curl of this velocity field is curl V =∇× V = ∣∣ ∣∣ ∣∣ ˆe 1 ˆe 2 ˆe 3 ∂ ∂x ∂ ∂y ∂ ∂z −ω 0 y −ω 0 x 0 ∣∣ ∣∣ ∣∣= 2ω^0 ˆe^3. ...
Example 8-10. Illustrate Stokes theorem using the vector field F=yz ˆe 1 +xz^2 ˆe 2 +xy ˆe 3 , where the surface Sis a portion ...
The left-hand side of Stokes theorem can be expressed ∫∫ S curlF·dS= ∫∫ S curlF·ˆendS = ∫∫ S [ x(1 − 2 z) sin θcos φ+ (z^2 −z ...
Figure 8-13. Surface Sbounded by a simple closed curve Con the surface. Let z=z(x, y )define the surface S and consider the proj ...
Now on the surface Sdefined by z=z(x, y )one finds F 1 =F 1 (x, y, z ) = F 1 (x, y, z (x, y )) is a function of x and y so that ...
The surface integral portion of Stokes theorem can therefore by expressed as ∫∫ S curl F d S= ∫∫ S (2 xy − 3 x^2 )ˆe 3 ·ˆendS ...
The integral relation (8.51) follows from the divergence theorem. In the diver- gence theorem, substitute F =H ×C , where C ...
For arbitrary C , this integral implies the relation (8.53). That is, one can factor out the constant vector C as long as this ...
Figure 8-14. Multiply-connected region. Such a region can be converted to a simply connected region by introducing cuts Γi, i = ...
one can express equation (8.60) in the form ∫∫ V ( ψ∇^2 φ+∇ψ·∇ φ ) dV = ∫∫ S ψ∇φ·dS = ∫∫ S ψ∂φ ∂n dS (8 .61) This result is kno ...
3. The Laplacian operator ∇^2 =∇·∇ in rectangular Cartesian coordinates is given by ∇^2 = ( ∂ ∂x ˆe 1 + ∂ ∂y ˆe 2 + ∂ ∂z ˆe 3 ) ...
(c) ∇^2 A= ∂^2 ∂x^2 (x (^2) ˆe 1 +xy ˆe 2 +y (^2) ˆe 3 ) + ∂^2 ∂y^2 (x (^2) eˆ 1 +xy ˆe 2 +y (^2) ˆe 3 ) + ∂^2 ∂z^2 (x (^2) ˆe ...
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