I'm helping a calculus student. It's been awhile since I've done some vector calculus. I know Stokes' Theorem in terms of differential forms and manifolds with boundary, but I've forgotten much of Stokes' Theorem in the undergraduate setting.
The problem is (paraphrased)
Consider the vector field $F(x,y,z) = [2z-y, x+z, 3x-2y]^T$. Let $\Sigma$ be the surface of the paraboloid defined by $z \ge 0$ and $z = 9-x^2-y^2$. Let $\Sigma$'s unique outward-pointing normal vector be $n$. Verify Stokes' Theorem $$\int_C F^T dr = \int \int_{\Sigma} (\nabla \times F)^Tn dS \tag{A}$$
Some notes:
I think we have $F: \mathbb R^3 \to \mathbb R^3$ (or $F: \mathbb R^3 \to \mathbb R^6$ if you think of tangent bundle), but no range is given. The domain isn't given either. It could be $\mathbb R^3 \setminus 0$ for all we know, but I think, in undergraduate, a map with 3 variables usually has domain $\mathbb R^3$ unless the problem says otherwise.
I think paraboloids are surfaces, so I guess 'the surface of the paraboloid' is in a similar meaning as 'game of tennis' rather than 'colour of the phone'. Or 'the surface of the paraboloid defined ... $-y^2$' means 'the section of the surface, which is the paraboloid $z=9-x^2-y^2$, defined by intersecting the paraboloid with $z \ge 0$'
No $C$ is mentioned. I guess this is up to the students to determine.
Questions
Question : Is this a correct computation for the line integral?
For C: The boundary of $\Sigma$ is $B = \{x^2+y^2=9\} \times \{0\}$. Therefore, we choose $C$ to be the projection of $B$ onto $z=0$, which is the circle $C = \{x^2+y^2=9\}$. Equivalently, I think, we choose $C$ to be the boundary of the projection of $S$ onto $z=0$, where the projection is, I think, the disc $D = \{x^2+y^2 \le 9 \}$.
- I guess $B$ is the manifold boundary of $\Sigma$. I mostly just pretend that I remember what is meant by 'boundary' in undergraduate.
For F: For $z=0$, $F=F(x,y,z)$ becomes the projection of $F=F(x,y,0) = [-y,x,3x-2y]^T$ onto $z=0$ which is $F=[-y,x]^T$ (I think this new $F$ is the pullback $\iota^{*}F$ for $\iota: C \to S$)
For the bounds: I think $0$ to $2 \pi$, where we parametrise $C$ as the image of the curve $r: [0, 2 \pi] \to \mathbb R^2$, $r(t)=[x(t),y(t)]^T=[3\cos(t),3\sin(t)]^T$
Hence, the line integral is $$\int_C F^T dr = \int_0^{2 \pi} [-y(t),x(t)] [dx(t), dy(t)]^T$$ with $x(t) = 3 \cos(t)$, $y(t)= 3 \sin(t)$, $dx(t) = -3 \sin(t) dt$, $dy(t) = 3 \cos(t)$, $dr(t) = [dx(t),dy(t)]^T$.
Therefore, the line integral is $18 \pi$, as computed here.
Question : Is this another correct computation for the line integral? I use Green's theorem $$\int_C F^T dr = \int \int_{R} (\nabla \times F)^T[0,0,1]^T dA$$
The $F$ in the line integral is $F=[-y,x]^T$. The $F$ on the double integral, I think, can be either $[-y,x,0]^T$ or $[-y,x,3x-2y]^T$ because, regardless, $(\nabla \times F)^T[0,0,1]^T = 2$.
The $C$ is the still the circle. I choose $R=D$ from Question (1).
Hence, the line integral is equal to the double integral $$\int_C F^T dr = \int \int_{D} 2 dx dy = 2 \int \int_{D} dx dy = 2 \ \text{Area}(D)$$
Therefore, the line integral is again $18 \pi$.
Question : Which of the two computations for the surface integral are correct?
For both computations:
For curl: $curl F = \nabla \times F = [-3,-1,2]^T$, as computed here.
For n (First method): Form the explicit surface $\{z=f\}$, where $f$ is $f: \mathbb R^2 \to \mathbb R$, $f(x,y) = x^2+y^2-9$. Let $\nabla f = [f_x, f_y]^T$. Then $n$ is normalised version of either $[-f_x, -f_y, 1]^T = [-2x,-2y,1]^T$ or $[f_x, f_y, -1]^T = [2x,2y,-1]^T$. I guess $[-f_x, -f_y, 1]^T$ is outward-pointing.
For n (Second method): Alternatively, form the implicit surface $\{h=0\}$, where $h$ is $h: \mathbb R^3 \to \mathbb R$, $h(x,y,z) = x^2+y^2-9-z$. Then $n$ is normalised version of either $\nabla h = [h_x, h_y, h_z]^T = [2x,2y,-1]^T$ or $-\nabla h$. I guess $-\nabla h$ is outward-pointing.
The first computation:
For the bounds and for $dS$: I'm actually not really sure of this. Off the top of my head, I think I kind of just project $\Sigma$ onto $z=0$ to get $D$ from Question (1). (See Question (4).) Therefore, I think $dS = |\nabla h| dA = |\nabla h| dx dy$, where we use the notation '$dS$' (of course in differential geometry, $dS$ and $dA$ are no longer just notation) for $\Sigma$ and '$dA$' for $D$, though I think we should have instead 'dA = du dv' instead of 'dA = dx dy' (see the 'However, upon...' later a few bullet points down).
Hence, the surface integral does not need any of the $\nabla h$, $\nabla f$ above and just uses $n = [0,0,1]^T$: $$\int \int_{\Sigma} (\nabla \times F)^Tn dS = \int \int_{D} (\nabla \times F)^T[0,0,1]^T dA = \int \int_{D} (\nabla \times F)^T [0,0,1]^T dx dy$$ $$= \int \int_{D} [-3,-1,2]^T [0,0,1]^T dx dy = \int \int_{D} 2 dx dy = \int_{0}^{2 \pi} \int_{0}^{3} 2 dx dy$$
Therefore, the surface integral is $18 \pi$, with precisely the same computation as in Question (2).
The second computation:
However, upon closer inspection of my undergraduate calculus textbook (no longer off the top of my head), what is supposed to be done, I think, is something like: $$\int \int_{\Sigma} (\nabla \times F)^Tn dS = \int \int_{P} (\nabla \times F)^Tn |\nabla h| dA$$ where $P$ is our 'parameter domain', so our $dA$ becomes not $dx dy$ but actually $du dv$ (which makes more sense actually). Our surface is given explicitly as $\Sigma = \{z=f(x,y)=9-x^2-y^2 | z \ge 0\}$, but we don't really have an explicit parameter domain $P=\{(u,v)\}$ here to have $\Sigma$ as the image of some $r(u,v)$, $r: P \to \mathbb R^3$. The book says we can $x,y$ as our parameters $u,v$, when our surface is given explicitly, so, for some parameter domain $P$, we have $\Sigma = \{z=f(x,y)=9-x^2-y^2 | (x,y) \in P\}$. Converting '$z \ge 0$' into '$(x,y) \in P$' seems to be done in taking $P=D$, i.e. we take our parameter domain $P$ to be the projection $D$ of the surface $\Sigma$, i.e. the surface $\Sigma$ is parametrised by its projection $D$.
- Edited to add: Oh, I guess this is the 3D version of the same idea for curves: This part of a parabola $L = \{y=x^2 | y \in [0,1] \}$ is explicit and so although one might parametrise $L$ as $L = \{(x,y)=(t,t^2) | t \in [-1,1]\}$ with a new variable $t$, one can also do $L = \{(x,y)=(x,x^2) | x \in [-1,1] =: P_L\}$, and indeed $P_L := [-1,1]$ is both the parameter domain of the map $r: [-1,1] \to \mathbb R^2$, $r(t) = (t,t^2)$ with image $L$ and the projection of $L$ into a dimension one lower than the dimension of $L$.
Therefore, it appears that the intuition of the idea of flattening, a.k.a. projecting, $\Sigma$ onto $z=0$ to get $D$ is justified by the fact that the flattening, a.k.a. projection, $D$ can be used to parametrise $\Sigma$.
Anyway, the surface integral should instead be computed as follows $$\int \int_{\Sigma} (\nabla \times F)^Tn dS = \int \int_{D} (\nabla \times F)^Tn |\nabla h| dA = \int \int_{D} (\nabla \times F)^T \frac{-\nabla h}{|\nabla h|} |\nabla h| dx dy$$ $$= \int \int_{D} [-3,-1,2]^T [-2x,-2y,1]^T dx dy = \int \int_{D} 6x +2y + 2 dx dy = \int_{0}^{2 \pi} \int_{0}^{3} [6x(r,\theta) +2y(r,\theta) + 2] (r dr d\theta)$$
Therefore, the surface integral is $18 \pi$, as computed here.
Question : It appears in this case we can convert a surface integral into a double integral $$\int \int_{D} (\nabla \times F)^T[0,0,1]^T dA = \int \int_{\Sigma} [-3,-1,2]^T dS \tag{B}$$
where $D$ is the projection of $\Sigma$ onto $z=0$ (or $D$ is the region in $z=0$ that is diffeomorphic to the boundary of $\Sigma$ or something).
What's going on? It appears we have the fact that for an explicit surface $\Sigma$, we can take $P=D$ to parametrise $\Sigma$. This fact suggests, if not outright implies, that we can compute a surface integral over $\Sigma$ as the double integral over $\Sigma$'s projection $(P=)D$.
I know Stokes' theorem is a generalisation of not only the fundamental theorem of calculus and the fundamental theorem of line integrals but also Green's theorem and that the theme of all of these is that, whenever applicable, the integral of something depends only on the values of the 'antidervative' ($f(x)$ for $f'(x)$, $f(x)$ for $\nabla f(x)$, $F(x)$ for $curl F(x)$, etc) values at the boundary points. However, it seems that whenever Stokes' theorem is applicable for a surface integral over $\Sigma$, we can convert the integral into a double integral over $D$, a flattening of the $\Sigma$, bypassing any need to obtain the 'boundary' $B$ of $\Sigma$ or 'boundary $C$ of $\Sigma$'s projection $D$. It actually seems more practical to compute the double integral over $D$ than the line integral over $B$ or $C$. (See Question (5).)
I think Stokes' theorem is actually more of something with an equation looks like $(B)$ and then $(A)$ is a corollary of $(B)$. (You might end up proving $(A)$ and $(B)$ are equivalent forms of Stokes' theorem.) Something that would have an equation like $(B)$ might be: Under the same assumptions of Stokes' theorem in $(A)$, you can evaluate the surface integral over $\Sigma$ into the double integral over the projection $D$ of $\Sigma$. The fact that Green's Theorem is used in proving the Stokes' Theorem in (A) suggests there's indeed a form of Stokes' Theorem like in $(B)$ that is simply converting a surface integral into a double integral. Of course, this form $(B)$ would not be in line with the theme mentioned in the preceding paragraph.
I think $$\int_C F^T dr = \int_B F^T dr = \int \int_D (curl F)^T [0,0,1]^T dA = \int \int_{\Sigma} (curl F)^T n dS$$ or something. Here, $\Sigma$ and $D$ have respective boundaries $B$ and $C$. The projections of $\Sigma$ and $B$ are respectively $D$ and $C$.
Question: (A follow-up to Question (4)) If I'm right about the existence of some kind of form $(B)$ for Stokes' theorem or right about something in my bullet points in Question (4), then while I see the theoretical point of Stokes' Theorem in form $(A)$, I don't see the practical point. What is the practical point?
I don't think form $(A)$ is practical for computing line integrals over a boundary curve $B$ of a surface $\Sigma$ because we could instead compute line integral over projection $C$ of the boundary curve $B$ of the surface $\Sigma$, a.k.a. the boundary curve $C$ of the projection $D$ of the surface $\Sigma$, which we would do using Green's theorem.
I don't think form $(A)$ is practical for computing surface integrals over a surface $\Sigma$ because if I convert the surface integral, under form $(A)$ into a to line integral over the boundary curve $B$, why wouldn't I then just use Green's Theorem to further convert the line integral over $B$ (into a line integral over $C$) into a double integral over $D$?
- Edited to add: EuYu might not have explicitly said this in the comments, but maybe my mistake here is assuming is assuming Green's Theorem is applicable. In general, $B$ and $C$ are not curves in $\mathbb R^2$ or even curves that lie on a single plane. Also, in general, $B \ne C$, which was kinda unlike the case above, where $B = \{x^2+y^2 = 9\} \times \{0\} = \{x^2+y^2=9\} = C$ and furthermore even if $B \ne C$ but $C$ is the projection of $B$, I don't believe the line integral is the same. A simple example (for non-closed curves and 1-dimension lower): $C=[0,1]$, the closed (in the sense of containing $\{0,1\}$ rather than in the sense of having the same start and end points) unit interval in $\mathbb R^1$ and $B = ([0,\frac12] \times \{0\}) \cup ([\frac12,1] \times \{1\})$ as a union of intervals from different $\mathbb R^1$'s (Actually $B$ isn't smooth or even continuous, but $B$ is at least piecewise continuous). We have $C$ as the projection of $B$ onto $\mathbb R^1$ (assuming we treat $[0,1]$ as identical to $[0,1] \times \{0\}$), but the line integral of the scalar constant function $f:D \to \mathbb R$, $f(D)=\{2\}$, where $D$ is either $B$ or $C$ is surely different over $B$ (I think it's 1.5) from the one over $C$ (I think it's 2).