Question

Calcule a integraL curvinea abaixo, onde C é o quadrado ABCD, tal que A=(1,0,1), B= (1,1,1), C=(0,1,1) e D = (0,0,1)

Answer 1

Resposta:

8

Explicação passo-a-passo:

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Seja f uma função escalar e seja C uma curva suave simples. Suponha que podemos dividir C em caminhos menores, C₁, C₂, C₃ e C₄, conectados entre si, então a integral de linha de f ao longo de C é:

$\boxed{\mathsf{\displaystyle \oint_Cf\cdot ds =\sum_{i=1}^4\int_{C_i}f\cdot ds}}$

• Integral ao longo de C₁ :

1. Parametrização:

$\mathsf{A\rightarrow B \qquad (1,0,1)\rightarrow (1,1,1)}\\\\\mathsf{r(t)=(1-t)\,r_o+t\,r_1\qquad t \in [0,1]}\\\\\mathsf{r(t)=(1-t)\,(1,0,1)+t\,(1,1,1)}\\\\\therefore \mathsf{r(t)=(1,t,1)}$

2. Derivada de r:

$\mathsf{r'(t)=(0,1,0)}$

3. Calcule o módulo:

$\mathsf{|r'(t)|=\sqrt{0^2+1^2+0^2}}\\\\\therefore \mathsf{|r'(t)|=1}$

4. Substitua na integral original:

$\mathsf{\displaystyle \int_{C_1}x+y+z\,ds=\int_0^11+t+1\,dt}$

$=\mathsf{\displaystyle \int_0^1t+2\,dt=\bigg[\dfrac{t^2}{2}+2t\bigg]_0^1}\\\\=\mathsf{\dfrac{1}{2}+2-0}\\\\=\mathsf{\dfrac{5}{2}}$

• Integral ao longo de C₂ :

1. Parametrização:

$\mathsf{B\rightarrow C \qquad (1,1,1)\rightarrow (0,1,1)}\\\\\mathsf{r(t)=(1-t)\,r_o+t\,r_1\qquad t \in [0,1]}\\\\\mathsf{r(t)=(1-t)\,(1,1,1)+t\,(0,1,1)}\\\\\therefore \mathsf{r(t)=(1-t,1,1)}$

2. Derivada de r:

$\mathsf{r'(t)=(-1,0,0)}$

3. Calcule o módulo:

$\mathsf{|r'(t)|=\sqrt{(-1)^2+0^2+0^2}}\\\\\therefore \mathsf{|r'(t)|=1}$

4. Substitua na integral original:

$\mathsf{\displaystyle \int_{C_2}x+y+z\,ds=\int_0^11-t+1+1\,dt}$

$=\mathsf{\displaystyle \int_0^13-t\,dt=\bigg[3t-\dfrac{t^2}{2}\bigg]_0^1}\\\\=\mathsf{3-\dfrac{1}{2}-0}\\\\=\mathsf{\dfrac{5}{2}}$

• Integral ao longo C₃ :

1. Parametrização:

$\mathsf{C\rightarrow D \qquad (0,1,1)\rightarrow (0,0,1)}\\\\\mathsf{r(t)=(1-t)\,r_o+t\,r_1\qquad t \in [0,1]}\\\\\mathsf{r(t)=(1-t)\,(0,1,1)+t\,(0,0,1)}\\\\\therefore \mathsf{r(t)=(0,1-t,1)}$

2. Derivada de r:

$\mathsf{r'(t)=(0,-1,0)}$

3. Calcule o módulo:

$\mathsf{|r'(t)|=\sqrt{0^2+(-1)^2+0^2}}\\\\\therefore \mathsf{|r'(t)|=1}$

4. Substitua na integral original:

$\mathsf{\displaystyle \int_{C_3}x+y+z\,ds=\int_0^10+1-t+1\,dt}$

$=\mathsf{\displaystyle \int_0^12-t\,dt=\bigg[2t-\dfrac{t^2}{2}\bigg]_0^1}\\\\=\mathsf{2-\dfrac{1}{2}-0}\\\\=\mathsf{\dfrac{3}{2}}$

• Integral ao longo de C₄ :

1. Parametrização:

$\mathsf{D\rightarrow A \qquad (0,0,1)\rightarrow (1,0,1)}\\\\\mathsf{r(t)=(1-t)\,r_o+t\,r_1\qquad t \in [0,1]}\\\\\mathsf{r(t)=(1-t)\,(0,0,1)+t\,(1,0,1)}\\\\\therefore \mathsf{r(t)=(t,0,1)}$

2. Derivada de r:

$\mathsf{r'(t)=(1,0,0)}$

3. Calcule o módulo:

$\mathsf{|r'(t)|=\sqrt{1^2+0^2+0^2}}\\\\\therefore \mathsf{|r'(t)|=1}$

4. Substitua na integral original:

$\mathsf{\displaystyle \int_{C_4}x+y+z\,ds=\int_0^1t+0+1\,dt}$

$=\mathsf{\displaystyle \int_0^1t+1\,dt=\bigg[\dfrac{t^2}{2}+t\bigg]_0^1}\\\\=\mathsf{\dfrac{1}{2}+1-0}\\\\=\mathsf{\dfrac{3}{2}}$

Portanto a integral de linha ao longo do caminho C é:

$\boxed{\mathsf{\displaystyle \oint_{C}x+y+z\,ds=\dfrac{5}{2}+\dfrac{5}{2}+\dfrac{3}{2}+\dfrac{3}{2}}=\mathsf{\dfrac{16}{2}}=\mathsf{8}}$

Bons estudos!

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