[SOLVED] Surface

**Spec** · September 18th 2008, 11:44 AM

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

That is, the cylinder cuts out a surface from the plane.

**Laurent** · September 18th 2008, 01:54 PM

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 $z=y+3$ , i.e. an origin $O'$ and an orthonormal basis $(\vec{u},\vec{v})$ , 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), $O'=(0,0,3)$ , $\vec{u}=(1,0,0)$ and $\vec{v}=\frac{1}{\sqrt{2}}(0,1,1)$ .

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

**shawsend** · September 18th 2008, 02:42 PM

That's very nice Laurent. I used your equation in the form of:

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

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

and then plotted it with the cylinder and plane so defined:

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