Begin2.DVI
Chapter8 Vector Calculus II In this chapter we examine in more detail the operations of gradient, divergence and curl as well as ...
Another way to visualize a vector field is to create a bundle of curves in space, where each curve in the bundle has the propert ...
and then integrate both sides to obtain ∫ 6 −x 2 3 −x dx= ∫ 4 −y 4 +y^2 dy (8.3) Use a table of integrals and evaluate the integ ...
in a vector field and try to determine how the vector field punctures this surface. Let the surface be divided into nsmall areas ...
ratio of VolumeFlux turns out to measure a point characteristic of the vector field called the divergence. Symbolically, the div ...
Treating the vector function F as a function of , one can expand F in a Taylor’s series about = 0,to obtain F=F(x 0 , y 0 ...
where the derivatives are to be evaluated at = 0 .The volume of the sphere of radius centered at the point (x 0 , y 0 , z 0 )i ...
The proof of the Gauss divergence theorem begins by first verifying the integrals ∫∫∫ V ∂F 1 ∂x dV = ∫∫ S F 1 ˆe 1 ·eˆndS ∫∫∫ V ...
An integration in the z-direction produces ∫∫∫ V ∂F 3 ∂z dzdx dy = ∫∫ R F 3 (x, y, z ) z 2 (x,y) z 1 (x,y) dx dy = ∫∫ R F 3 (x, ...
in the y−direction from y= 0 to the circle y= √ 4 −x^2 to form a slab. The slab is then summed in the x−direction from x= 0 to x ...
On S 1 , z = 4, eˆn=ˆe 3 , dS =dx dy and ∫∫ S 1 F·dS= ∫ 2 0 ∫√ 4 −x 2 0 16 dy dx = 16 ∫ 2 0 √ 4 −x^2 dx = 16 π. On S 2 , y = 0 ...
evaluated. The geometry suggests a change to cylindrical coordinates. In cylindrical coordinates the following relations are sat ...
Figure 8-5. Tube of field lines. The sides of the tube are composed of field lines, and at any point on a field line the directi ...
Figure 8-6. Vector field V =√xkx (^2) +y 2 ˆe 1 +√xky (^2) +y 2 ˆe 2 ,k > 0 constant. Observe that the magnitude of the vect ...
where the line integral is taken in a counterclockwise direction around the simple closed curve Cwhich encloses the region R. To ...
The remaining part of the right-hand side can then be expressed ∫∫ R ∂N ∂x dx dy = ∫y 2 y 1 ∫x 2 (y) x 1 (y) ∂N ∂x dx dy = ∫y 2 ...
On the curve C 1 ,where y= 0, dy = 0,the first integral reduces to ∫ C 1 Mdx +N dy = ∫a 0 x^2 dx = a^3 3 On the curve C 2 ,where ...
Solution of Differential Equations by Line Integrals The total differential of a function φ=φ(x, y )is dφ = ∂φ ∂x dx + ∂φ ∂y dy. ...
Figure 8-9. Arbitrary paths connecting points (x 0 , y 0 )and (x, y ). Consequently, one can write ∫x x 0 M(x, y 1 (x)) dx +N(x, ...
Since the line integral is independent of the path of integration, it is possible to select any convenient path of integration f ...
«
5
6
7
8
9
10
11
12
13
14
»
Free download pdf