MHB Combining Stoke's Theorem and the Divergence Theorem?

[Vishak95]

Hi,

Could someone please guide me with this question? I'm unsure as to what the curl has to do with finding flux..

PS. This isn't actually assessed work, it's from a past question paper that I am using to revise.

View attachment 2637

Thanks heaps!

 

[Ackbach]

I can tell you what the curl has to do with things. You're told that $\mathbf{F}=x^2 \hat{\mathbf{i}}+(y-2xy)\hat{\mathbf{j}}-z\hat{\mathbf{k}},$ and that $\mathbf{F}=\nabla\times\left(yz\hat{\mathbf{i}}+x^2 y\hat{\mathbf{k}}\right)$. Let $\mathbf{G}=yz\hat{\mathbf{i}}+x^2 y\hat{\mathbf{k}}$, so that $\mathbf{F}=\nabla\times\mathbf{G}$. You're asked to compute
$$\iint_{S} \mathbf{F}\cdot d\mathbf{A}=\iint_{S}(\nabla\times\mathbf{G})\cdot d\mathbf{A},$$
which is the flux of $\mathbf{F}$. By Stoke's Theorem, you have that
$$\iint_{S}(\nabla\times\mathbf{G})\cdot d\mathbf{A}=\oint_{C}\mathbf{G}\cdot d\mathbf{r}.$$
Here $C$ is the contour that forms the "boundary" of $S$. Stoke's Theorem says that you can convert the surface integral of a curl into the closed line integral of the surface's boundary. To get a handle on $C$, set $z=0=2x-x^2-y^2$, and see what that curve looks like in the $x-y$ plane. The business of saying "oriented away from the $x-y$ plane" tells you in which direction to perform the line integral.

 

[Vishak95]

I can tell you what the curl has to do with things. You're told that $\mathbf{F}=x^2 \hat{\mathbf{i}}+(y-2xy)\hat{\mathbf{j}}-z\hat{\mathbf{k}},$ and that $\mathbf{F}=\nabla\times\left(yz\hat{\mathbf{i}}+x^2 y\hat{\mathbf{k}}\right)$. Let $\mathbf{G}=yz\hat{\mathbf{i}}+x^2 y\hat{\mathbf{k}}$, so that $\mathbf{F}=\nabla\times\mathbf{G}$. You're asked to compute
$$\iint_{S} \mathbf{F}\cdot d\mathbf{A}=\iint_{S}(\nabla\times\mathbf{G})\cdot d\mathbf{A},$$
which is the flux of $\mathbf{F}$. By Stoke's Theorem, you have that
$$\iint_{S}(\nabla\times\mathbf{G})\cdot d\mathbf{A}=\oint_{C}\mathbf{G}\cdot d\mathbf{r}.$$
Here $C$ is the contour that forms the "boundary" of $S$. Stoke's Theorem says that you can convert the surface integral of a curl into the closed line integral of the surface's boundary. To get a handle on $C$, set $z=0=2x-x^2-y^2$, and see what that curve looks like in the $x-y$ plane. The business of saying "oriented away from the $x-y$ plane" tells you in which direction to perform the line integral.

Thanks for the response. That clears most of my doubts up.

Just a further query though, the form of Stokes' Theorem in my textbook looks like this:

$$\int \int (CurlF)\cdot n dA = \oint F \cdot dr$$

Could you please explain why the n is not needed in dealing with this problem :P

 

[Ackbach]

Thanks for the response. That clears most of my doubts up.

Just a further query though, the form of Stokes' Theorem in my textbook looks like this:

$$\int \int (CurlF)\cdot n dA = \oint F \cdot dr$$

Could you please explain why the n is not needed in dealing with this problem :P

I think if you look, you will see that (note carefully the bold and non-bold letters) what you have for Stoke's Theorem is
$$\iint(\text{curl}\;\mathbf{F})\cdot \mathbf{n} \, dA=\dots,$$
whereas I wrote
$$\iint(\nabla\times\mathbf{G})\cdot d\mathbf{A}=\dots$$
The explanation is that $\mathbf{n}\,dA=d\mathbf{A}$. So the normal vector is absolutely needed here, it's just that some people include it in $d\mathbf{A}$, thinking of that differential area vector as including the normal vector in its definition.

 

[Vishak95]

I think if you look, you will see that (note carefully the bold and non-bold letters) what you have for Stoke's Theorem is
$$\iint(\text{curl}\;\mathbf{F})\cdot \mathbf{n} \, dA=\dots,$$
whereas I wrote
$$\iint(\nabla\times\mathbf{G})\cdot d\mathbf{A}=\dots$$
The explanation is that $\mathbf{n}\,dA=d\mathbf{A}$. So the normal vector is absolutely needed here, it's just that some people include it in $d\mathbf{A}$, thinking of that differential area vector as including the normal vector in its definition.

Ohh okay, thank you.

Just to get my fundamentals correct, would be I be correct in saying this?

$$\int \int \boldsymbol{F}\cdot \boldsymbol{n}dS = \int \int \int div(\boldsymbol{F})dV = \int \int (curl\boldsymbol{F})\cdot \boldsymbol{n }dA = \oint \boldsymbol{F}\cdot dr$$

 

[Ackbach]

Ohh okay, thank you.

Just to get my fundamentals correct, would be I be correct in saying this?

$$\int \int \boldsymbol{F}\cdot \boldsymbol{n}dS = \int \int \int div(\boldsymbol{F})dV = \int \int (curl\boldsymbol{F})\cdot \boldsymbol{n }dA = \oint \boldsymbol{F}\cdot dr$$

Here's what you have - the three basic Fundamental Theorems of Calculus, in increasing numbers of dimensions:
\begin{align*}
\int_a^bf'(x) \, dx&=f(b)-f(a) \quad \text{Fundamental Theorem of Calculus} \\
\iint_S (\nabla\times\mathbf{G}) \cdot d\mathbf{A}&= \oint_C \mathbf{G}\cdot d\mathbf{r} \quad \text{Stoke's Theorem} \\
\iiint_V (\nabla\cdot\mathbf{F}) \, dV&= \iint_S \mathbf{F}\cdot d\mathbf{A} \quad \text{Divergence Theorem},
\end{align*}
where that last surface integral in the divergence theorem is a closed integral. For some reason our $\LaTeX$ rendering does not allow the \oiint symbol.

A couple of comments about these three theorems:

1. Note that all three theorems look like $$\int_{\text{interior}}(\text{derivative of something}) \, dH=\int_{\text{boundary}}(\text{something}) \,dG, $$
   where $dH$ has one more dimension than $dG$.
2. Green's Theorem is a special case of Stoke's Theorem.

I'm afraid you can't combine Stoke's Theorem with the Divergence Theorem. I've thought myself sometimes: couldn't you combine the Divergence Theorem with Stoke's Theorem? Here's the problem: Stoke's Theorem has an open surface with a closed boundary. But the Divergence Theorem has a volume with a closed surface. A closed surface has no contour for a boundary, so that you can't go from Divergence to Stoke's all in one equation. In addition, in order to use both theorems all together (referencing the equations above), you'd have to have $\mathbf{F}=\nabla\times\mathbf{G}$. But $\nabla\cdot(\nabla\times\mathbf{G})=0$, a vector calculus identity. So even if you were able to combine the equations into one like this:
$$\iiint_V (\nabla \cdot(\nabla\times\mathbf{G})) \, dV
=\iint_S (\nabla\times\mathbf{G})\cdot d\mathbf{A}
=\oint_C \mathbf{G}\cdot d\mathbf{r},$$
the initial volume integral is zero. In other words, this equation couldn't tell you anything.