Exemplos do teorema da divergência em 3D

Conhecimentos prévios

• Teorema da divergência 3D
  • Fluxo em três dimensões
    • Divergência
    • Integrais triplas

O teorema da divergência (rápida recapitulação)

• $\blueE{\textbf{F}}(x, y, z)$start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, comma, z, right parenthesis é um campo vetorial tridimensional.
• $\redE{V}$start color #bc2612, V, end color #bc2612 é um volume tridimensional (pense em uma bolha no espaço).
• $\redE{S}$start color #bc2612, S, end color #bc2612 é a superfície de $\redE{V}$start color #bc2612, V, end color #bc2612.
• $\greenE{\hat{\textbf{n}}}$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f é uma função que dá vetores unitários normais sobre a superfície de $\redE{S}$start color #bc2612, S, end color #bc2612.

Bolando uma Estratégia

Exemplo 1: integral de superfície através de um cubo.

Escolha 1 resposta:

Escolha 1 resposta:

• (Escolha A)  

  $\displaystyle \int_0^1 \int_0^1 \int_0^1 \nabla \times (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy\,dz$integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, times, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y, d, z

  A

  $\displaystyle \int_0^1 \int_0^1 \int_0^1 \nabla \times (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy\,dz$integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, times, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y, d, z
• (Escolha B)  

  $\displaystyle \int_0^1 \int_0^1 \int_0^1 \nabla \cdot (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy\,dz$integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y, d, z

  B

  $\displaystyle \int_0^1 \int_0^1 \int_0^1 \nabla \cdot (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy\,dz$integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y, d, z
• (Escolha C)  

  $\displaystyle 6 \int_0^1 \int_0^1 \nabla \times (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy$6, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, times, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y

  C

  $\displaystyle 6 \int_0^1 \int_0^1 \nabla \times (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy$6, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, times, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y
• (Escolha D)  

  $\displaystyle 6 \int_0^1 \int_0^1 \nabla \cdot (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy$6, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y

  D

  $\displaystyle 6 \int_0^1 \int_0^1 \nabla \cdot (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) \;dx\,dy$6, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, d, x, d, y

Verificar

[Resposta]

$\nabla \cdot (2y\hat{\textbf{i}} + 3y^2 \hat{\textbf{j}} + 4z\hat{\textbf{k}}) =$del, dot, left parenthesis, 2, y, start bold text, i, end bold text, with, hat, on top, plus, 3, y, squared, start bold text, j, end bold text, with, hat, on top, plus, 4, z, start bold text, k, end bold text, with, hat, on top, right parenthesis, equals

Verificar

[Resposta]

Exemplo 2: integral de superfície através de um cilindro

$\displaystyle \nabla \cdot \left[ \begin{array}{c} x^3 \\ y^3 \\ x^3 + y^3 \end{array} \right] =$

Verificar

[Resposta]

$\displaystyle \iiint_{\redE{C}} \left( \nabla \cdot \left[ \begin{array}{c} x^3 \\ y^3 \\ x^3 + y^3 \end{array} \right] \right) \; \redE{dV} =$

Verificar

[Resposta]

Exemplo 3: área de superfície de uma integral de volume

Escolha 1 resposta:

Escolha 1 resposta:

• (Escolha A)  

  $\blueE{\textbf{F}} = \greenE{\hat{\textbf{n}}}$start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, equals, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f

  A

  $\blueE{\textbf{F}} = \greenE{\hat{\textbf{n}}}$start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, equals, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f
• (Escolha B)  

  $\blueE{\textbf{F}} = \greenE{\hat{\textbf{n}}} \times ( x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}})$start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, equals, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, times, left parenthesis, x, start bold text, i, end bold text, with, hat, on top, plus, y, start bold text, j, end bold text, with, hat, on top, plus, z, start bold text, k, end bold text, with, hat, on top, right parenthesis

  B

  $\blueE{\textbf{F}} = \greenE{\hat{\textbf{n}}} \times ( x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}})$start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, equals, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, times, left parenthesis, x, start bold text, i, end bold text, with, hat, on top, plus, y, start bold text, j, end bold text, with, hat, on top, plus, z, start bold text, k, end bold text, with, hat, on top, right parenthesis

Verificar

[Resposta]

Escolha 1 resposta:

Escolha 1 resposta:

• (Escolha A)  

  $\displaystyle \iiint_\redE{B} \nabla \cdot \greenE{\hat{\textbf{n}}} \,\redE{dV}$\iiint, start subscript, start color #bc2612, B, end color #bc2612, end subscript, del, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, V, end color #bc2612

  A

  $\displaystyle \iiint_\redE{B} \nabla \cdot \greenE{\hat{\textbf{n}}} \,\redE{dV}$\iiint, start subscript, start color #bc2612, B, end color #bc2612, end subscript, del, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, V, end color #bc2612
• (Escolha B)  

  $\displaystyle \iiint_\redE{B} \nabla \cdot (\greenE{\hat{\textbf{n}}} \cdot \greenE{\hat{\textbf{n}}}) \,\redE{dV}$\iiint, start subscript, start color #bc2612, B, end color #bc2612, end subscript, del, dot, left parenthesis, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, right parenthesis, start color #bc2612, d, V, end color #bc2612

  B

  $\displaystyle \iiint_\redE{B} \nabla \cdot (\greenE{\hat{\textbf{n}}} \cdot \greenE{\hat{\textbf{n}}}) \,\redE{dV}$\iiint, start subscript, start color #bc2612, B, end color #bc2612, end subscript, del, dot, left parenthesis, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, right parenthesis, start color #bc2612, d, V, end color #bc2612

Verificar

[Resposta]

Escolha 1 resposta:

Escolha 1 resposta:

• (Escolha A)  

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = (x + y + z) (\hat{\textbf{i}} + \hat{\textbf{j}} + \hat{\textbf{k}})$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, left parenthesis, x, plus, y, plus, z, right parenthesis, left parenthesis, start bold text, i, end bold text, with, hat, on top, plus, start bold text, j, end bold text, with, hat, on top, plus, start bold text, k, end bold text, with, hat, on top, right parenthesis

  A

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = (x + y + z) (\hat{\textbf{i}} + \hat{\textbf{j}} + \hat{\textbf{k}})$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, left parenthesis, x, plus, y, plus, z, right parenthesis, left parenthesis, start bold text, i, end bold text, with, hat, on top, plus, start bold text, j, end bold text, with, hat, on top, plus, start bold text, k, end bold text, with, hat, on top, right parenthesis
• (Escolha B)  

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = \dfrac{1}{\sqrt{3}}x \hat{\textbf{i}} + \dfrac{1}{\sqrt{3}}y \hat{\textbf{j}} + \dfrac{1}{\sqrt{3}}z \hat{\textbf{k}}$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, start fraction, 1, divided by, square root of, 3, end square root, end fraction, x, start bold text, i, end bold text, with, hat, on top, plus, start fraction, 1, divided by, square root of, 3, end square root, end fraction, y, start bold text, j, end bold text, with, hat, on top, plus, start fraction, 1, divided by, square root of, 3, end square root, end fraction, z, start bold text, k, end bold text, with, hat, on top

  B

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = \dfrac{1}{\sqrt{3}}x \hat{\textbf{i}} + \dfrac{1}{\sqrt{3}}y \hat{\textbf{j}} + \dfrac{1}{\sqrt{3}}z \hat{\textbf{k}}$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, start fraction, 1, divided by, square root of, 3, end square root, end fraction, x, start bold text, i, end bold text, with, hat, on top, plus, start fraction, 1, divided by, square root of, 3, end square root, end fraction, y, start bold text, j, end bold text, with, hat, on top, plus, start fraction, 1, divided by, square root of, 3, end square root, end fraction, z, start bold text, k, end bold text, with, hat, on top
• (Escolha C)  

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, x, start bold text, i, end bold text, with, hat, on top, plus, y, start bold text, j, end bold text, with, hat, on top, plus, z, start bold text, k, end bold text, with, hat, on top

  C

  $\displaystyle \greenE{\hat{\textbf{n}}}(x, y, z) = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, x, start bold text, i, end bold text, with, hat, on top, plus, y, start bold text, j, end bold text, with, hat, on top, plus, z, start bold text, k, end bold text, with, hat, on top

Verificar

[Resposta]

$\nabla \cdot \greenE{\hat{\textbf{n}}} =$del, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, equals

Verificar

[Resposta]

$\displaystyle \iiint_\redE{B} \nabla \cdot \greenE{\hat{\textbf{n}}} \,\redE{dV} =$\iiint, start subscript, start color #bc2612, B, end color #bc2612, end subscript, del, dot, start color #0d923f, start bold text, n, end bold text, with, hat, on top, end color #0d923f, start color #bc2612, d, V, end color #bc2612, equals

Verificar

[Resposta]

Resumo

• O teorema da divergência é útil sempre que o volume interior de uma região for mais fácil de descrever do que sua superfície.
• Também ajuda se a divergência do campo vetorial relevante o transformar em uma função mais simples.