# Divergence Theorem Verification: Surface Integral

• Master1022
In summary, the conversation discusses a question about the validity of a step in the method of calculating a surface integral. The method involves calculating the normal surface vectors and mapping them onto integration variables. The person asking the question is unsure about the use of the equation ## x^2 + y^2 = 3 ## and how the mapping is done to the xy plane. The experts clarify that the equation can only be used on the circle bounding the disk and that the mapping is coincidental. They also provide a general method for calculating the vector differential surface element and advise against normalizing the normal surface vectors.
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.

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.)

Master1022
Master1022 said:
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.

jambaugh said:
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.

jambaugh said:
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.

jambaugh said:
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.

jambaugh said:
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} ##

jambaugh said:
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

## 1. What is the Divergence Theorem?

The Divergence Theorem, also known as Gauss's Theorem, is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the volume integral of the divergence of the same vector field over the region enclosed by the surface. In simpler terms, it states that the net flow of a vector field through a closed surface is equal to the amount of sources or sinks inside the surface.

## 2. How is the Divergence Theorem used in surface integrals?

The Divergence Theorem is used in surface integrals to simplify the calculation of the surface integral by converting it into a volume integral. This is done by using the divergence operator to relate the surface integral to the volume integral, making it easier to evaluate the integral using the properties of the divergence.

## 3. What is the purpose of verifying the Divergence Theorem in surface integrals?

Verifying the Divergence Theorem in surface integrals is important to ensure the accuracy of calculations and to check for any errors. It also helps to understand the underlying principles of the theorem and its application in different scenarios.

## 4. What are the steps involved in verifying the Divergence Theorem in surface integrals?

The steps involved in verifying the Divergence Theorem in surface integrals include:

1. Identifying the vector field and the closed surface in question.
2. Calculating the surface integral of the vector field over the closed surface.
3. Calculating the volume integral of the divergence of the same vector field over the region enclosed by the surface.
4. Comparing the results of the two calculations to check if they are equal, thus verifying the Divergence Theorem.

## 5. What are some common applications of the Divergence Theorem in real-world scenarios?

The Divergence Theorem has many practical applications, some of which include:

• Calculating the flow of fluid through a pipe or a container.
• Calculating the electric flux through a closed surface due to a point charge.
• Studying the behavior of electromagnetic fields in different media.
• Analyzing the flow of heat in a solid object.

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