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Thread: [SOLVED] Surface

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Surface

    Which surface is created by the intersection of the cylinder $\displaystyle x^2+y^2=1$ and the plane $\displaystyle z=y+3$

    That is, the cylinder cuts out a surface from the plane.
    Last edited by Spec; Sep 18th 2008 at 12:13 PM.
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  2. #2
    MHF Contributor

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    Drawing a sketch is very helpful (in this case, it is not very difficult to do).

    First we choose a coordinate system in the plane $\displaystyle z=y+3$, i.e. an origin $\displaystyle O'$ and an orthonormal basis $\displaystyle (\vec{u},\vec{v})$, so we can write the equation of the intersection using it. For instance (as shows the sketch, it is not only a simple but also an appropriate choice), $\displaystyle O'=(0,0,3)$, $\displaystyle \vec{u}=(1,0,0)$ and $\displaystyle \vec{v}=\frac{1}{\sqrt{2}}(0,1,1)$.

    Then, for any $\displaystyle M=(x,y,z)$ in the intersection, $\displaystyle z=y+3$, hence $\displaystyle M=O'+x\vec{u}+\sqrt{2}y\vec{v}$, so the coordinates of $\displaystyle M$ in the new system are $\displaystyle X=x$ and $\displaystyle Y=\sqrt{2}y$. The equation $\displaystyle x^2+y^2=1$ becomes $\displaystyle X^2+\frac{Y^2}{2}=1$. In other words, the intersection is an ellipse with minor and major axes respectively equal to $\displaystyle 1$ and $\displaystyle \sqrt{2}$, centered at $\displaystyle O'$. The major axis is in direction $\displaystyle \vec{v}$ and the minor one in direction $\displaystyle \vec{u}$.
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  3. #3
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    That's very nice Laurent. I used your equation in the form of:

    $\displaystyle f[\text{t$\_$}]\text{:=}\{0,0,3\}+\text{Cos}[t] \{1,0,0\}+\sqrt{2} \text{Sin}[t] \frac{1}{\sqrt{2}}\{0,1,1\}$;

    Then did a parametric plot of $\displaystyle f(t)$ (in red) over the range $\displaystyle [0,2\pi]$

    and then plotted it with the cylinder and plane so defined:
    Attached Thumbnails Attached Thumbnails [SOLVED] Surface-cylinder-plane.jpg  
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