Calculating surface area of intersection between solid cylinder and plane 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? 
 A: 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 $\sqrt{3}\pi$.
A: 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, $$\sqrt3 \pi$$
A: 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.$
A: You're given a cylinder, so why not use cylindrical coordinates?
Parameterize the surface (call it $S$) by
$$\begin{align}\vec r(u,v)&=(x(u,v),y(u,v),1-x(u,v)-y(u,v))\\&=(u\cos v,u\sin v,1-u\cos v-u\sin v)\end{align}$$
with $0\le u\le1$ and $0\le v\le2\pi$.
Take the normal vector to $S$ to be
$$\vec n=\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}=(u,u,u)$$
which has norm $\|\vec n\|=\sqrt{u^2+u^2+u^2}=\sqrt3\,u$.
The area of $S$ is given by the surface integral,
$$\iint_S\mathrm dx\,\mathrm dy=\sqrt3\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\frac{\sqrt3}2\int_0^{2\pi}\mathrm dv=\boxed{\sqrt3\,\pi}$$
