applying gauss divergence thereom prove that int. r.nds = 3v
Answers
Answer:
Step-by-step explanation:
\blueE{\textbf{F}}(x, y, z)F(x,y,z)start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, comma, z, right parenthesis is some three-dimensional vector field.
\redE{V}Vstart color #bc2612, V, end color #bc2612 is a three-dimensional volume (think of a blob in space).
\redE{S}Sstart color #bc2612, S, end color #bc2612 is the surface of \redE{V}Vstart color #bc2612, V, end color #bc2612.
\greenE{\hat{\textbf{n}}}
n
^
start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f is a function which gives unit normal vectors on the surface of \redE{S}Sstart color #bc2612, S, end color #bc2612.
Here's what the divergence theorem states:
\displaystyle \underbrace{ \iiint_\redE{V} \text{div}\,\blueE{\textbf{F}}\,\redE{dV} }_{\substack{ \text{Add up little bits} \\\\ \text{of outward flow in $\redE{V}$} }} = \underbrace{ \overbrace{ \iint_\redE{S} \blueE{\textbf{F}} \cdot \greenE{\hat{\textbf{n}}} \,\redE{d\Sigma} }^{\text{Flux integral}} }_{\substack{ \text{Measures total outward } \\\\ \text{flow through $\redE{V}$'s boundary} }}
Add up little bits
of outward flow in V
∭
V
divFdV
=
Measures total outward
flow through V’s boundary
∬
S
F⋅
n
^
dΣ
Flux integral
The intuition here is that both integrals measure the rate at which a fluid flowing along the vector field \blueE{\textbf{F}}Fstart color #0c7f99, start bold text, F, end bold text, end color #0c7f99 is exiting the region \redE{V}Vstart color #bc2612, V, end color #bc2612 (or entering \redE{V}Vstart color #bc2612, V, end color #bc2612, if the values of both integrals are negative). Triply integrating divergence does this by counting up all the little bits of outward flow of the fluid inside \redE{V}Vstart color #bc2612, V, end color #bc2612, while taking the flux integral measures this by checking how much is leaving/entering along the boundary of \redE{V}Vstart color #bc2612, V, end color #bc2612.
Strategizing
The divergence theorem lets you translate between surface integrals and triple integrals, but this is only useful if one of them is simpler than the other. In each of the following examples, take note of the fact that the volume of the relevant region is simpler to describe than the surface of that region.
In general, when you are faced with a surface integral over a closed surface, consider if it would be easier to integrate over the volume enclosed by that surface. If it is, it's a strong signal that the divergence theorem will come in handy.
Answer:
Prove that int r.Nds=3V