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Thread: Calculus 3

  1. #1
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    Exclamation Calculus 3

    Curve $\displaystyle \mathcal{K}$ is defined as an intersection of surfaces $\displaystyle z^2 = 2x^2 + y^2$ and $\displaystyle z = x + 1$. The curve's projection on plane $\displaystyle z = 0$ is positive oriented.


    • Parametrise the curve $\displaystyle \mathcal{K}$.
    • Find the Frenet-Serret basis of the curve $\displaystyle \mathcal{K}$ at point $\displaystyle A(2, 1, 3)$.
    • Find the $\displaystyle \int_{\mathcal{K}}\vec{F}\mathrm{d}\vec{r}$ of the vector field $\displaystyle F(x, y, z) = (z, x - 1, -2y)$ along the curve $\displaystyle \mathcal{K}$.

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  2. #2
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    Re: Calculus 3

    Since z= x+ 1, $\displaystyle z^2= (x+ 1)^2= x^2+ 2x+ 1= 2x^2+ y^2$ so $\displaystyle x^2- 2x- 1+ y^2= 0$. From that $\displaystyle x^2- 2x+ 1+ y^2= 2$ and then $\displaystyle (x- 1)^2+ y^2= 2$.

    Now an obvious parameterization is $\displaystyle x- 1= \sqrt{2}cos(\theta)$, so $\displaystyle x= \sqrt{2}cos(\theta)+ 1$, $\displaystyle y= \sqrt{2}sin(\theta)$, and $\displaystyle z= x+ 1= \sqrt{2}cos(\theta)+ 2$.

    At least try the rest of the problem!
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  3. #3
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    Re: Calculus 3

    Ok, I know the second part. But now, how to setup the integral and find it? @HallsofIvy
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