In my multi variable/vector calculus textbook I encountered a problem which led to more questions than answers when i solved it. Let me start by stateing the problem:
The problem
Let $\gamma$ be a curve on the cylinder surface $x^2+y^2=1$. $\gamma$ is closed but otherwise it does not intersect with it self. Let $\mathbf{F}=(x+y,2yz,y^2)$ be a vector field. Show that $\oint_\gamma \mathbf{F}\bullet d\mathbf{r}$ only can take on 3 different values depending on how $\gamma$ runs along the cylinder surface. Determine these values and describe the different cases.
My attempt and initial thoughts
My inital thought was: How many types of closed curves can there be on a cylindric surface given the conditions. The intersect between the cylinder and a plane could give $\gamma$ either as a circle or a ellipse and still be closed. A plane $\Pi : ax+by+cz=d$ where c=0 cannot as I see it give a closed curve $\gamma$. I also thought about how different geometric shapes could intersect with the cylinder and give a closed curve, spheres e.t.c. But I decided to start by just examine the case of a plane intersecting the cylinder.
Let $\Pi$ be the plane $ax+by+cz=d$ with $a,b,c\in\mathbb{R}$ and where $x^2+y^2\leq 1$. Let $\gamma$ be the intersect curve between the cylinder $x^2+y^2=1$ and $\Pi$. Then the unit normal of $\Pi$ is $\mathbf{N}=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$.
In preperation I calculated $\text{curl}(\mathbf{F})=\nabla \times\mathbf{F}=(0,0,-1)$
By Stoke´s theorem:
$$\oint_\gamma \mathbf{F}\bullet d \mathbf{r}=\iint_\Pi (\nabla \times\mathbf{F})\bullet\mathbf{N} dS=\frac{-c}{\sqrt{a^2+b^2+c^2}}\iint_\Pi dS=\frac{-c}{\sqrt{a^2+b^2+c^2}}\mu (\Pi)$$
Where $\mu (\Pi)$ is the area of $\Pi$ I then proceed to calculate this area, interpreting the plane as a equipotential surface of $f(x,y)=\frac{d-ax-by}{c}$, where $x^2+y^2\leq 1$.
Then:$$\mu (\Pi)=\iint_{x^2+y^2\leq 1} \sqrt{1+(f^´_x)^2+(f^´_y)^2}dxdy=\iint_{x^2+y^2\leq 1} \sqrt{1+(\frac{-a}{c})^2+(c{-b}{c})^2}dxdy=\iint_{x^2+y^2\leq 1} \frac{\sqrt{a^2+b^2+c^2}}{|c|}dxdy=\frac{\sqrt{a^2+b^2+c^2}}{|c|} \int_{0}^{2\pi}\int_{0}^{1} rdrd\theta=\ldots=\frac{\pi\sqrt{a^2+b^2+c^2}}{|c|}$$
Then: $$W=\oint_\gamma\mathbf{F}\bullet d\mathbf{r}=\frac{-c\pi}{|c|}$$
Thus:
- $c>0$ then $\gamma$ has negative orientation around the z-axis and $W=-\pi$
- $c=0$ then $\gamma$ does not go around the z-axis and $W=0$
- $c<0$ then $\gamma$ has positive orientation around the z-axis and $W=\pi$
Textbook answer sheet
- $\gamma$ has negative orientation around the z-axis and $W=-\pi$
- $\gamma$ does not go around the z-axis and $W=0$
- $\gamma$ has positive orientation around the z-axis and $W=\pi$
Questions
- First of the answer in the textbook is clearly not the same as mine but quite simular, why do they dont have c in their answer?
- Then, the first and the third case makes sense to me given the problem statement, but second as I meantioned earlier does not. Am I wrong if I think that if $c=0$ then $\gamma$ would just be two straight lines on the of the cylinder?
- And in the bigger picture, to me it didn´t seem like I covered all possible curves given the conditions. Did I? And if not which most probabilly is the case, how would I do it? For example consider the intersection of the cylinder and the sphere $(x-\frac{1}{2})^2+y^2+z^2=1$.
This is my first post ever and I hope I have been clear and thorough:)
The given orientation of $\gamma$ induces an orientation along the plane $\Pi$: if $\gamma$ is counterclockwise (around $z$) then $\mathbf{N}=\frac{|c|}{\sqrt{a^2+b^2+c^2}}$ is upward, if $\gamma$ is clockwise then $\mathbf{N}=-\frac{|c|}{\sqrt{a^2+b^2+c^2}}$ is downward. These are precisely the orientations needed in order to apply Stoke's theorem. Finally $|c|$ goes away.
I agree, if $c=0$ we have two parallel lines ($\gamma$ is not a closed curve).
No, you did not cover all possible curves. Your approach covers all plane sections of the cylinder, i.e. ellipses. You may have a sinusoidal simple closed curve which wraps around the cylinder. (hint: such curve is homotopic to a plane section of the cylinder). What about a closed curve on the surface of the cylinder which does not go around the $z$-axis? (hint: this curve is null-homotopic on the cylinder).