How to parametrize the following surface in $\mathbb{R}^3$: the intersection of $S=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2\leqslant 1\}$ and $D=\{(x,y,z)\in\mathbb{R}^3:x+y=1\}$.
Any hints are welcome.Thanks!
How to parametrize the following surface in $\mathbb{R}^3$: the intersection of $S=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2\leqslant 1\}$ and $D=\{(x,y,z)\in\mathbb{R}^3:x+y=1\}$.
Any hints are welcome.Thanks!
Our intersection is the set of $(x,1-x,z)$ such that $$\begin{align}1 &\ge x^2+(1-x)^2+z^2\\ &= 2x^2-2x+1+z^2\\ &= 2(x^2-x)+1+z^2\\ &= 2\left(x^2-x+\frac14\right)-\frac12+1+z^2\\ &= 2\left(x-\frac12\right)^2+\frac12+z^2,\end{align}$$ or equivalently, the set of $(x,1-x,z)$ such that $$\cfrac{\left(x-\frac12\right)^2}{\left(\frac12\right)^2}+\cfrac{z^2}{\left(\frac1{\sqrt{2}}\right)^2}\le 1.\tag{$\clubsuit$}$$
Do you know how to parameterize an elliptical region such as the one given by $(\clubsuit)$?
Let $x=t$, $z=s$ and $y=1-t$. Then you just have to put the appropriate bounds on $s$ and $t$ such that $t^2+(1-t)^2+s^2\leq 1$ (for example I think $t$ should be in $[0,1]$).