Cálculo 3... 30 pontos!
O cálculo da integral de linha consiste em uma das importantes ferramentas fornecidas pelo cálculo...
Continuação em anexo.
Anexos:
![](https://pt-static.z-dn.net/files/d38/2d1306de72768865226a78e7e5db3624.png)
Lukyo:
Acho que tem um erro no enunciado. Se a curva é uma hélice, z(t) não pode ser constante...
Soluções para a tarefa
Respondido por
3
Calcular a integral de linha
![I=\displaystyle\int_{C}{(x^{2}+y^{2}-z)}\,d\mathbf{s} I=\displaystyle\int_{C}{(x^{2}+y^{2}-z)}\,d\mathbf{s}](https://tex.z-dn.net/?f=I%3D%5Cdisplaystyle%5Cint_%7BC%7D%7B%28x%5E%7B2%7D%2By%5E%7B2%7D-z%29%7D%5C%2Cd%5Cmathbf%7Bs%7D)
sendo
a curva parametrizada
![C:~\begin{array}{cc} \left\{ \begin{array}{l} x=\cos t\\ y=\mathrm{sen\,}t\\ z=t \end{array} \right.&~~~~\text{com }t\in \mathbb{R}. \end{array} C:~\begin{array}{cc} \left\{ \begin{array}{l} x=\cos t\\ y=\mathrm{sen\,}t\\ z=t \end{array} \right.&~~~~\text{com }t\in \mathbb{R}. \end{array}](https://tex.z-dn.net/?f=C%3A%7E%5Cbegin%7Barray%7D%7Bcc%7D+%5Cleft%5C%7B+%5Cbegin%7Barray%7D%7Bl%7D+x%3D%5Ccos+t%5C%5C+y%3D%5Cmathrm%7Bsen%5C%2C%7Dt%5C%5C+z%3Dt+%5Cend%7Barray%7D+%5Cright.%26amp%3B%7E%7E%7E%7E%5Ctext%7Bcom+%7Dt%5Cin+%5Cmathbb%7BR%7D.+%5Cend%7Barray%7D)
Calculando o vetor tangente ![C'(t): C'(t):](https://tex.z-dn.net/?f=C%27%28t%29%3A)
![C(t)=(\cos t,\;\mathrm{sen\,}t,\;t)\\ \\ \\ C'(t)=(-\mathrm{sen\,}t,\;\cos t,\;1) C(t)=(\cos t,\;\mathrm{sen\,}t,\;t)\\ \\ \\ C'(t)=(-\mathrm{sen\,}t,\;\cos t,\;1)](https://tex.z-dn.net/?f=C%28t%29%3D%28%5Ccos+t%2C%5C%3B%5Cmathrm%7Bsen%5C%2C%7Dt%2C%5C%3Bt%29%5C%5C+%5C%5C+%5C%5C+C%27%28t%29%3D%28-%5Cmathrm%7Bsen%5C%2C%7Dt%2C%5C%3B%5Ccos+t%2C%5C%3B1%29)
Para o cálculo da integral de linha, vamos precisar apenas do módulo do vetor tangente:
![\|C'(t)\|=\|(-\mathrm{sen\,}t,\;\cos t,\;1)\|\\ \\ \|C'(t)\|=\sqrt{(-\mathrm{sen\,}t)^{2}+(\cos t)^{2}+1^{2}}\\ \\ \|C'(t)\|=\sqrt{\mathrm{sen^{2}\,}t+\cos^{2}\,t+1}\\ \\ \|C'(t)\|=\sqrt{1+1}=\sqrt{2}\,,~~~~\forall~t\in\mathbb{R}. \|C'(t)\|=\|(-\mathrm{sen\,}t,\;\cos t,\;1)\|\\ \\ \|C'(t)\|=\sqrt{(-\mathrm{sen\,}t)^{2}+(\cos t)^{2}+1^{2}}\\ \\ \|C'(t)\|=\sqrt{\mathrm{sen^{2}\,}t+\cos^{2}\,t+1}\\ \\ \|C'(t)\|=\sqrt{1+1}=\sqrt{2}\,,~~~~\forall~t\in\mathbb{R}.](https://tex.z-dn.net/?f=%5C%7CC%27%28t%29%5C%7C%3D%5C%7C%28-%5Cmathrm%7Bsen%5C%2C%7Dt%2C%5C%3B%5Ccos+t%2C%5C%3B1%29%5C%7C%5C%5C+%5C%5C+%5C%7CC%27%28t%29%5C%7C%3D%5Csqrt%7B%28-%5Cmathrm%7Bsen%5C%2C%7Dt%29%5E%7B2%7D%2B%28%5Ccos+t%29%5E%7B2%7D%2B1%5E%7B2%7D%7D%5C%5C+%5C%5C+%5C%7CC%27%28t%29%5C%7C%3D%5Csqrt%7B%5Cmathrm%7Bsen%5E%7B2%7D%5C%2C%7Dt%2B%5Ccos%5E%7B2%7D%5C%2Ct%2B1%7D%5C%5C+%5C%5C+%5C%7CC%27%28t%29%5C%7C%3D%5Csqrt%7B1%2B1%7D%3D%5Csqrt%7B2%7D%5C%2C%2C%7E%7E%7E%7E%5Cforall%7Et%5Cin%5Cmathbb%7BR%7D.)
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Encontrando os limites de integração em![t: t:](https://tex.z-dn.net/?f=t%3A)
![\bullet\;\;C(0)=(\cos 0,\;\mathrm{sen\,}0,\;0)\\ \\ C(0)=(1,\;0,\;0)=A\\ \\ \\ \bullet\;\;C(2\pi)=(\cos 2\pi,\;\mathrm{sen\,}2\pi,\;2\pi)\\ \\ C(2\pi)=(1,\;0,\;2\pi)=B \bullet\;\;C(0)=(\cos 0,\;\mathrm{sen\,}0,\;0)\\ \\ C(0)=(1,\;0,\;0)=A\\ \\ \\ \bullet\;\;C(2\pi)=(\cos 2\pi,\;\mathrm{sen\,}2\pi,\;2\pi)\\ \\ C(2\pi)=(1,\;0,\;2\pi)=B](https://tex.z-dn.net/?f=%5Cbullet%5C%3B%5C%3BC%280%29%3D%28%5Ccos+0%2C%5C%3B%5Cmathrm%7Bsen%5C%2C%7D0%2C%5C%3B0%29%5C%5C+%5C%5C+C%280%29%3D%281%2C%5C%3B0%2C%5C%3B0%29%3DA%5C%5C+%5C%5C+%5C%5C+%5Cbullet%5C%3B%5C%3BC%282%5Cpi%29%3D%28%5Ccos+2%5Cpi%2C%5C%3B%5Cmathrm%7Bsen%5C%2C%7D2%5Cpi%2C%5C%3B2%5Cpi%29%5C%5C+%5C%5C+C%282%5Cpi%29%3D%281%2C%5C%3B0%2C%5C%3B2%5Cpi%29%3DB)
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Portanto,
![I=\displaystyle\int_{C}{(x^{2}+y^{2}-z)}\,d\mathbf{s}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left[(\cos t)^{2}+(\mathrm{sen\,}t)^{2}-t \right ]\cdot \|C'(t)\|\,dt}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left[\cos^{2}t+\mathrm{sen^{2}\,}t-t \right ]\cdot \sqrt{2}\,dt}\\ \\ \\ =\sqrt{2}\int\limits_{0}^{2\pi}{(1-t)\,dt}\\ \\ \\ =\sqrt{2}\cdot \left.\left(t-\dfrac{t^{2}}{2} \right )\right|_{0}^{2\pi} I=\displaystyle\int_{C}{(x^{2}+y^{2}-z)}\,d\mathbf{s}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left[(\cos t)^{2}+(\mathrm{sen\,}t)^{2}-t \right ]\cdot \|C'(t)\|\,dt}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left[\cos^{2}t+\mathrm{sen^{2}\,}t-t \right ]\cdot \sqrt{2}\,dt}\\ \\ \\ =\sqrt{2}\int\limits_{0}^{2\pi}{(1-t)\,dt}\\ \\ \\ =\sqrt{2}\cdot \left.\left(t-\dfrac{t^{2}}{2} \right )\right|_{0}^{2\pi}](https://tex.z-dn.net/?f=I%3D%5Cdisplaystyle%5Cint_%7BC%7D%7B%28x%5E%7B2%7D%2By%5E%7B2%7D-z%29%7D%5C%2Cd%5Cmathbf%7Bs%7D%5C%5C+%5C%5C+%5C%5C+%3D%5Cint%5Climits_%7B0%7D%5E%7B2%5Cpi%7D%7B%5Cleft%5B%28%5Ccos+t%29%5E%7B2%7D%2B%28%5Cmathrm%7Bsen%5C%2C%7Dt%29%5E%7B2%7D-t+%5Cright+%5D%5Ccdot+%5C%7CC%27%28t%29%5C%7C%5C%2Cdt%7D%5C%5C+%5C%5C+%5C%5C+%3D%5Cint%5Climits_%7B0%7D%5E%7B2%5Cpi%7D%7B%5Cleft%5B%5Ccos%5E%7B2%7Dt%2B%5Cmathrm%7Bsen%5E%7B2%7D%5C%2C%7Dt-t+%5Cright+%5D%5Ccdot+%5Csqrt%7B2%7D%5C%2Cdt%7D%5C%5C+%5C%5C+%5C%5C+%3D%5Csqrt%7B2%7D%5Cint%5Climits_%7B0%7D%5E%7B2%5Cpi%7D%7B%281-t%29%5C%2Cdt%7D%5C%5C+%5C%5C+%5C%5C+%3D%5Csqrt%7B2%7D%5Ccdot+%5Cleft.%5Cleft%28t-%5Cdfrac%7Bt%5E%7B2%7D%7D%7B2%7D+%5Cright+%29%5Cright%7C_%7B0%7D%5E%7B2%5Cpi%7D)
![=\sqrt{2}\cdot \left(2\pi-\dfrac{(2\pi)^{2}}{2} \right )\\ \\ \\ =\sqrt{2}\cdot \left(2\pi-\dfrac{4\pi^{2}}{2} \right )\\ \\ \\ =\sqrt{2}\cdot \left(2\pi-2\pi^{2} \right )\\ \\ =2\pi\sqrt{2}\,\left(1-\pi \right ). =\sqrt{2}\cdot \left(2\pi-\dfrac{(2\pi)^{2}}{2} \right )\\ \\ \\ =\sqrt{2}\cdot \left(2\pi-\dfrac{4\pi^{2}}{2} \right )\\ \\ \\ =\sqrt{2}\cdot \left(2\pi-2\pi^{2} \right )\\ \\ =2\pi\sqrt{2}\,\left(1-\pi \right ).](https://tex.z-dn.net/?f=%3D%5Csqrt%7B2%7D%5Ccdot+%5Cleft%282%5Cpi-%5Cdfrac%7B%282%5Cpi%29%5E%7B2%7D%7D%7B2%7D+%5Cright+%29%5C%5C+%5C%5C+%5C%5C+%3D%5Csqrt%7B2%7D%5Ccdot+%5Cleft%282%5Cpi-%5Cdfrac%7B4%5Cpi%5E%7B2%7D%7D%7B2%7D+%5Cright+%29%5C%5C+%5C%5C+%5C%5C+%3D%5Csqrt%7B2%7D%5Ccdot+%5Cleft%282%5Cpi-2%5Cpi%5E%7B2%7D+%5Cright+%29%5C%5C+%5C%5C+%3D2%5Cpi%5Csqrt%7B2%7D%5C%2C%5Cleft%281-%5Cpi+%5Cright+%29.)
Resposta: alternativa![\text{c. }2\pi\sqrt{2}\,\left(1-\pi \right ). \text{c. }2\pi\sqrt{2}\,\left(1-\pi \right ).](https://tex.z-dn.net/?f=%5Ctext%7Bc.+%7D2%5Cpi%5Csqrt%7B2%7D%5C%2C%5Cleft%281-%5Cpi+%5Cright+%29.)
(se o
)
sendo
Para o cálculo da integral de linha, vamos precisar apenas do módulo do vetor tangente:
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Encontrando os limites de integração em
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Portanto,
Resposta: alternativa
(se o
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