Calculating surface area of intersection between solid cylinder and plane

Question

I wanted to calculate the surface of $$\{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2<1, x+y+z=1\}$$

but to calculate it, I need a parametrization.

My first attempt was to just put: $y=\sqrt{1-x^2}, z = 1 - x - \sqrt{1-x^2}$

so I get the parametrization $$(x,\sqrt{1-x^2}, 1 - x - \sqrt{1-x^2}),$$ but I also have to consider $$(x,-\sqrt{1-x^2}, 1 - x + \sqrt{1-x^2}),$$

is this correct?

Answer 1

For a surface in 3D space, parametrization needs two parameters. (a curve needs one parameters.) You can use (x, y) as two parameters.

The surface area is, $$S=\iint \limits_D \sqrt {f_x^2+f_y^2+1} dxdy$$

Here function $f$ is defined as, $$z=f(x,y)=1-x-y$$ and the domain is the circle $$x^2+y^2<1$$

This gives your the answer, $$\sqrt3 \pi$$

Answer 2

You are intersecting the cylinder $x^2+y^2\leq1$ with the plane $x+y+z=1$. If $\theta$ is the angle between the $z$-axis and the normal of the plane then $$\cos\theta=(0,0,1)\cdot{1\over\sqrt{3}}(1,1,1)={1\over\sqrt{3}}\ .$$ If you project a piece of surface $S$ orthogonally onto the $(x,y)$-plane under such circumstances then the area $\omega(S)$ is multiplied by the factor $\cos\theta$. It follows that the area of the ellipse in question is $\>{\displaystyle{1\over\cos\theta}}=\sqrt{3}\>$ times the area of the unit disk in the $(x,y)$-plane, hence is equal to $\sqrt{3}\pi$.

Answer 3

The given plane of intersection has direction cosines $( \frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})$, surface normal direction along major diagonal of a unit cube.

Ratio of true and projected areas is:

$$ \dfrac{{\dfrac{\sqrt3 \cdot {\sqrt2}^2}{4}}} {{\dfrac{1^2}{2}}} = {\sqrt3}$$

So, the slant area of cut ellipse is $ {\sqrt3} \pi.$