There are some problems that bothering me, "Euler Charatcteristic, $\chi(M)(= v-e+f)$"
Here are some Questions that I'm stuck.
Find the $\chi(M)$
$Q1)$ $M$ is a rotation of $y=cosx $($-\pi \leq x \leq\pi$) whose axis is $Y-axis$
$Q2$ $M$ is a rotation of $y=2+sinx $($0 \leq x \leq 2\pi$) whose axis is $X-axis$
I'm trying to find $\chi(M) = v+e-f$ by dividing surfaces by traingles or rectangulars(polygonalization) but failed. How could I found that?
Here is my trial of the Q1
I don't know what I was wrong
The answer of $Q1$ and $Q2$ in my textbook are $\chi (M) =1$ and $\chi(M) =0$ respectively.
I can't understand Why the answers are $\chi (M) =1$ and $\chi(M) =0$. 🙁
P.S) Please give me some method or examples for finding the $\chi (M)$ by divide as the triangle or rectangular for surfaces in $R^3$
Any help would be appreciate!
Best Answer
In Q1, it looks like you counted $f$ wrong, it should be $f=2$ and therefore $v-e+f=1$.
Perhaps you mistakenly counted the 2-dimensional round disc at the bottom of the figure as one of the faces? If so, $M$ is a surface of revolution of the graph $y = \cos(x)$ ($-\pi < x < \pi$), and that 2-dimensional disc is not part of the surface.
Added to address the question in the edit: The graph that is being rotated is the graph of the function $y = \cos(x)$ ($-\pi \le x \le \pi$), more explicitly the set $$\{(x,y) \mid y = \cos(x), -\pi \le x \le \pi\} $$ That's the set which you correctly drew in your diagram.
So, for example, when $x=0$ you get $y=1$ and the point $(0,1)$ is part of the graph that is being rotated; however the point $(0,-1)$ is NOT part of the graph that is being rotated.
For another example, when $x=\pi/2$ you get $y=0$ and the point $(\pi/2,0)$ is part of the graph that is being rotated. However, the point $(\pi/2,-1)$ is NOT part of the graph that is being rotated.
In fact, no point with $y=-1$ is on the graph that is being rotated, except for the points $(\pi,-1)$ and $(-\pi,-1)$. Those two points, when rotated, give you a circle in the plane $y=-1$ which is part of the rotated graph. However, the points on the plane $y=-1$ which are on the inside of that circle are NOT part of the rotated graph.
If you had connected the two points $(\pi,-1)$ and $(-\pi,-1)$ by a straight line segment in $y=-1$, then that entire line segment would have been rotated and your rotated graph would have contained the disc in question. But, you did NOT connect those two points by a straight line segment.