Question

Sa se calculeze ∫y(x+y-z+2)dl unde y este segmentul AB,A(0,1,1)la B(1,0,1)

Answer 1

$\displaystyle\\\textnormal{Parametrizarea~segmentului~in~}\mathbb{R}^3~:\gamma(t):\begin{cases}x(t)=(1-t)x_A+t\cdot x_B\\ y(t)=(1-t)y_A+t\cdot y_B,~ t\in[0,1]\\ z(t)=(1-t)z_A+t\cdot z_B \end{cases}\\----------------------------------$

$\displaystyle\\\textnormal{Parametrizam,}~\gamma(t):\begin{cases}x(t)=(t-1)\cdot0+t\cdot1 =t\\ y(t)=(1-t)\cdot 1+t\cdot 0=1-t,~~~~~t\in[0,1] \\ z(t)=(1-t)\cdot 1 + t\cdot 1=1 \end{cases}\\\gamma'(t):\begin{cases}x'(t)=1 \\ y'(t)=-1 \\ z'(t)=0 \end{cases} \Longrightarrow ||\gamma'(t)||=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}=\sqrt{2}.$

$\displaystyle\\\textnormal{Asadar~:~} \int_{\gamma}(x+y-z+2)dl=\int_0^1 (t+(1-t)-1+2)\sqrt{2}~\textnormal{d}t=2\sqrt{2}.$