Divergence Theorem Verification: Surface Integral

[Master1022]

Homework Statement

Demonstrate the validity of the divergence theorem by calculating the surface integral $\iint_S \vec F \cdot \hat n dS$ over the closed surface defined by $z = 3 - x^2 - y^2$ and the plane $z = 0$ where $\vec F = x \hat i + y \hat j$

Relevant Equations

Gauss' Theorem

Hi,

I just had a quick question about a step in the method of calculating the surface integral and why it is valid. I have already done the divergence step and it yields the correct result.

Method:
Let us calculate the normal: $\nabla (z + x^2 + y^2 - 3) = (2x, 2y, 1)$. Just to double check, am I correct in thinking that I don't need to normalise the normal surface vectors as the magnitude cancel out after being mapped onto the integration variables?

For the z = 0 plane portion, the surface vector will be
$$
\begin{pmatrix}
0 \\
0 \\
-1
\end{pmatrix} $$ and thus the contribution to the surface integral will be 0

Now for the 'downward-facing parabola'
$$ \vec F \cdot \hat n dS = \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2x \\ 2y \\ 1 \end{pmatrix} = 2(x^2 + y^2) $$

The solution suggests that we can use the fact that $x^2 + y^2 = 3$ at $z = 0$ such that
$$ \iint_S \vec F \cdot \hat n dS = \iint_S 2(3) dS = 6 \iint_S dS = 18\pi $$

This is where I have the questions:
1) Why are we allowed to just use $x^2 + y^2 = 3$ when x and y actually vary from 0 to the radius (i.e. the surface exists for $x^2 + y^2 \leq 3$)? I remember making a similar mistake in the past where I just used the equation when the surface had values for lower x and y.

2) How do I know that this has been mapped from the '3D' surface to the $xy$ plane? When I defined $dS$ it was on the 3D surface but now we are working in the plane... I know the purpose of these is to map to the xy plane where the geometry is much simpler (and sometimes an easy shape). I suppose this is the downside of not normlising - it hides these steps?

Any clarification is greatly appreciated.

 

[jambaugh]

To your zeroth question, "it depends". The full method you should learn is to calculate the vector differential surface element $\vec{dS}$. That will depend on your parameterization. For general parameterization of your surface $\vec{r}=\vec{r}(u,v)$ you would then define the differential surface normal as:
$$\vec{dS} =\pm \frac{\partial \vec{r}(u,v)}{\partial u}\times \frac{\partial \vec{r}(u,v)}{\partial v} dudv$$ with the sign chosen so the normal is in the correct direction. You do not normalize as the magnitude is a component of the differential area.

Now in the special case where you are using $(u,v)=(x,y)$ for a surface $z=f(x,y)$ as your parameterization, $\vec{r}=\langle x,y,f(x,y)\rangle$ will yield:
$$\vec{dS}=\pm \langle 1,0,f_x\rangle \times \langle 0,1,f_y\rangle dxdy = \pm\langle -f_x, -f_y ,1\rangle dxdy$$
You took the gradient of the surface defined by $g(x,y,z)=0$ which in this case takes the form:
$g(x,y,z)=z-f(x,y)$ Thus, with this special case the two agree. But note, you could have defined the surface by $\tilde{g}(x,y,z) = 42\cdot g(x,y,z)$. Same surface but different gradient (by a factor of 42) and it will give you the wrong differential normal surface vector. You should learn to start with the general method or always be sure you work with a function in the form of $g$ above (or it's negative). If you are working with a generic surface $\tilde{g}(x,y,z)=const$ then you should normalize and then you must work out the surface area differential separately if it's simple. The general method of doing this is essentially doing the full blown method I showed initially.

For your second question, the fact that you are mapping to the x-y plane is coincidental. You are ultimately mapping to a new parameter space. $(u,v)$ but where the mapping happens to parameterize x as u, y as v (and z as some function of them).

Your concern in question 1 is valid. The constraint $x^2 +y^2 =3$ comes from the intersection of the paraboloid with the x-y plane. It is only true on the circle bounding the disk over which you are integrating. It may coincidentally give the same value for the particular choice of $f$ but you should work it out correctly and verify it equals the other side of Gauss' theorem (I'd switch to polar coordinates myself.)

 

[vela]

Why are we allowed to just use $x^2 + y^2 = 3$ when x and y actually vary from 0 to the radius (i.e. the surface exists for $x^2 + y^2 \leq 3$)?

You're not. The solution is wrong.

 

[Master1022]

Thank you very much @jambaugh and @vela for responding.

To your zeroth question, "it depends". The full method you should learn is to calculate the vector differential surface element $\vec{dS}$. That will depend on your parameterization. For general parameterization of your surface $\vec{r}=\vec{r}(u,v)$ you would then define the differential surface normal as:
$$\vec{dS} =\pm \frac{\partial \vec{r}(u,v)}{\partial u}\times \frac{\partial \vec{r}(u,v)}{\partial v} dudv$$ with the sign chosen so the normal is in the correct direction. You do not normalize as the magnitude is a component of the differential area.

Now in the special case where you are using $(u,v)=(x,y)$ for a surface $z=f(x,y)$ as your parameterization, $\vec{r}=\langle x,y,f(x,y)\rangle$ will yield:
$$\vec{dS}=\pm \langle 1,0,f_x\rangle \times \langle 0,1,f_y\rangle dxdy = \pm\langle -f_x, -f_y ,1\rangle dxdy$$

Does this look like the right way of attempting the problem:
Let $\vec r = x \hat i + y \hat j + z \hat k = x \hat i + y \hat j + (3 - x^2 - y^2) \hat k$.
Therefore,
$$ \frac{\partial \vec r}{\partial x} =
\begin{pmatrix} 1 \\ 0 \\ -2x \end{pmatrix} $$ and $$
\frac{\partial \vec r}{\partial y} =
\begin{pmatrix} 0 \\ 1 \\ -2y \end{pmatrix} $$

Therefore,

$$\vec{dS} =\pm \frac{\partial \vec{r}}{\partial x}\times \frac{\partial \vec{r}}{\partial y} dxdy = \begin{pmatrix} 2x \\ 2y \\ 1 \end{pmatrix}$$

which agrees with that I had in the other method.

You took the gradient of the surface defined by $g(x,y,z)=0$ which in this case takes the form:
$g(x,y,z)=z-f(x,y)$ Thus, with this special case the two agree.

Okay, yes I understand this.

But note, you could have defined the surface by $\tilde{g}(x,y,z) = 42\cdot g(x,y,z)$. Same surface but different gradient (by a factor of 42) and it will give you the wrong differential normal surface vector. You should learn to start with the general method or always be sure you work with a function in the form of $g$ above (or it's negative). If you are working with a generic surface $\tilde{g}(x,y,z)=const$ then you should normalize and then you must work out the surface area differential separately if it's simple. The general method of doing this is essentially doing the full blown method I showed initially.

Apologies, but does this mean:
- If we are dealing with $\tilde{g}(x,y,z) = k \cdot g(x,y,z)$ for some constant $k$ that I should first normalise (i.e. divide by a constant such that the coefficient of $z$ is 1) and then revert to the above INSTEAD of going straight to the above approach? (Note: I have used $z$ here as that is the nature of most of the problems I face)

From an initial glance, it looks like the final surface vector would have an extra factor of $k^2$ if we don't normalise which is what you alluded to.

For your second question, the fact that you are mapping to the x-y plane is coincidental. You are ultimately mapping to a new parameter space. $(u,v)$ but where the mapping happens to parameterize x as u, y as v (and z as some function of them).

Thank you. One question I have is where the metric coefficients $h_u$ and $h_v$ come into effect? Are those are hidden within the partial derivatives $\frac{\partial \vec{r}(u,v)}{\partial u}$ and $\frac{\partial \vec{r}(u,v)}{\partial v}$

Your concern in question 1 is valid. The constraint $x^2 +y^2 =3$ comes from the intersection of the paraboloid with the x-y plane. It is only true on the circle bounding the disk over which you are integrating. It may coincidentally give the same value for the particular choice of $f$ but you should work it out correctly and verify it equals the other side of Gauss' theorem (I'd switch to polar coordinates myself.)

Agreed.

When re-evaluating the surface integral which was now mapped to be in the $xy$ plane within the circle with radius $\sqrt 3$:
$$ \iint_S \vec F \cdot \hat n dS = 2 \iint_S (x^2 + y^2) dx dy = 2 \int_{\phi = 0}^{2\pi} \int_{r=0}^{\sqrt 3} r^2 \cdot r dr d\phi $$
$$ = 4\pi \frac{(\sqrt 3)^4}{4} = 9\pi $$

I did manage to re-do the divergence integral in two different ways (one obtaining $9 \pi$ and the other obtaining $18 \pi$). I am just trying to see whether I can reconcile the differences before posting again