# Surface Integral (and Greens theorem) confusion

In summary, the conversation discusses the evaluation of a surface integral over a cylinder bound by certain conditions. There is confusion about which method to use and how to parametrize the surface. The conversation also mentions the divergence theorem and the use of cylindrical coordinates. There is further discussion about the dot product and the position vector.
Hello all,

Evaluate

$$\int\int r. da$$

over the whole surface of the cylinder bound by

$$x^{2} + y^{2} = 1, z=0$$ and $$z=3.$$

$$\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$$

Sorry for the awkward formatting, this site is giving me trouble.
Anyways,
it seems to me that since I have 3 dimensions I'm asked to use I must resort to Green's theorem,
and I will eventually have a triple integral with x=0 to 1, y=0 to 1 and z=0 to 3.

What's my tau (volume unit) though? I have the area condition which only includes x and y, and the z condition on its own...

help!

Last edited:
Or maybe I need to parametrize the equation... that would work, but I still don't know how r relates to that...

Hello all,

Evaluate

$$\int\int r. da$$

over the whole surface of the cylinder bound by

$$x^{2} + y^{2} = 1, z=0$$ and $$z=3.$$

$$\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$$

Sorry for the awkward formatting, this site is giving me trouble.
Anyways,
it seems to me that since I have 3 dimensions I'm asked to use I must resort to Green's theorem,
and I will eventually have a triple integral with x=0 to 1, y=0 to 1 and z=0 to 3.

What's my tau (volume unit) though? I have the area condition which only includes x and y, and the z condition on its own...

help!

Don't you need to rewrite
$$\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$$
to

$$\vec{r} = x \cdot \frac{x}{\|x\|} + y \frac{y}{\|y\|} + z \cdot \frac{z}{\|z\|} = x \cdot \frac{x}{\sqrt{x \cdot x}} + y \frac{y}{\sqrt{y\cdot y}} + z \cdot \frac{z}{\sqrt{z \cdot z}} = |x|+|y| + |z|$$

Last edited:
Hello all,

Evaluate

$$\int\int r. da$$

over the whole surface of the cylinder bound by

$$x^{2} + y^{2} = 1, z=0$$ and $$z=3.$$

$$\vec{r} = x \hat{x} + y \hat{y} + z \hat{z}$$

Sorry for the awkward formatting, this site is giving me trouble.
Anyways,
it seems to me that since I have 3 dimensions I'm asked to use I must resort to Green's theorem,

Green's theorem?? Is this a line integral around a closed curve in the plane?

and I will eventually have a triple integral with x=0 to 1, y=0 to 1 and z=0 to 3.

Those limits would describe a rectangular block, which a cylinder isn't.

What's my tau (volume unit) though? I have the area condition which only includes x and y, and the z condition on its own...

help!

Why do you want the unit of volume? You are given a surface integral, not a volume integral.

You need to parameterize the top, bottom, and lateral surfaces in terms of two parameters and work each piece separately. Since you are using an area vector, presumably you have an orientation given for the surface. I'll give you a hint to get you started. For the lateral surface, try parameterizing it in terms of z and θ.

Sorry I meant to say the divergence theorem, which is why I would need the unit of volume.

My problem is, I don't see, regardless of what I do, how I'm going to do the dot product of something with 2 dimensions (da) with something with 3 dimensions (r)...

If I parametrize to cos(theta) and sin(theta) (radii are 1), I don't see how I've solved my problem..

thanks for the help by the way!

Sorry I meant to say the divergence theorem, which is why I would need the unit of volume.

My problem is, I don't see, regardless of what I do, how I'm going to do the dot product of something with 2 dimensions (da) with something with 3 dimensions (r)...

If I parametrize to cos(theta) and sin(theta) (radii are 1), I don't see how I've solved my problem..

thanks for the help by the way!

The da in your problem is the area vector. It is the unit normal multiplied by the scalar area element. How to approach this problem depends on the intent of your textbook author for that section. You can work the problem directly by parameterizing the surfaces and using the fact that if the surface is parameterized by R(u,v), then

$$d\vec a = \pm \vec R_u \times \vec R_v$$

where the sign is chosen to agree with the orientation. Or you can apply the divergence theorem. Some texts ask you to work both sides of the divergence theorem as an exercise for practice. Which leads to the obvious question: Have you written out what the divergence theorem says for this particular problem? You might be surprised...

I found a good example in Paul's online notes (example 3), but, could someone explain the change that occurs between him declaring r as the parametrization of the cylinder, and then r-theta a few steps later.
The sin is now on the i'th component... and negative?

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx

I don't have time to work through that example right now, but I would point out to you that your question involves a flux integral and the vector da is not the same as the scalar da.

I've run into another problem. When parametrizing everything, I obviously need my surface to be in cylindrical coordinates, but I can't very well parametrize r = x+y+z in cylindrical because there is no axis...

What do I do? I can't do the dot product between two things parametrized differently =(

I've run into another problem. When parametrizing everything, I obviously need my surface to be in cylindrical coordinates, but I can't very well parametrize r = x+y+z in cylindrical because there is no axis...

What do I do? I can't do the dot product between two things parametrized differently =(

I have several comments for you.

First, r is not the scalar x + y + z. r is the position vector to (x,y,z) which would be written as xi+yj+zk or <x, y, z> depending on your preference. And in a problem such as this which might involve cylindrical coordinates it would be better to use R instead of r to avoid confusion with the r of cylindrical coordinates.

Second, I gave you the hint above to parameterize the lateral surface using z and θ. Have you expressed x, y, and z in terms of those for that surface? Then you can use the other formula I gave you for

$$d\vec a = \pm \vec R_u \times \vec R_v\ dudv$$

Did you understand all that?

Third, I asked you whether you have actually written what the divergence theorem says for this problem. Have you done that? What does it give you?

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Ok,

so, if I make R equal to the parametrization of my surface then,
R= $$cos \theta \hat{i} + sin \theta \hat{j} + z \hat{k}$$ since the axis of the cylinder is along z.

Next, following

$$\int\int \bar{r}.d\bar{a} = \int\int \bar{r} .\bar{n} da$$

Doing the cross product of R w.r.t. theta and z gives:

$$- cos \theta \hat{i}- sin \theta \hat{j}$$

Now is where I hit my problem...

I realize r is a vector and not a scalar, that is exactly what I'm confused about.

I need to do the dot product r.n. Where r=xi + yk+zk, and n is what I already stated in cylindrical coordinates. But, I can't put r into cylindrical coordinates since there is no axis, so I don't know how to proceed with the dot product.

I haven't done anything with the divergence theorem... I don't know how I would come up with a dV since I am only given the shape in 2-D or 1-D pieces.

Ok,

so, if I make R equal to the parametrization of my surface then,
R= $$cos \theta \hat{i} + sin \theta \hat{j} + z \hat{k}$$ since the axis of the cylinder is along z.

Good so far. That correctly parameterizes the lateral surface as a function of θ and z.

Next, following

$$\int\int \bar{r}.d\bar{a} = \int\int \bar{r} .\bar{n} da$$

Doing the cross product of R w.r.t. theta and z gives:

$$- cos \theta \hat{i}- sin \theta \hat{j}$$

Do you mean to say the cross product of rθ and rz? At this point you need to check the orientation of the surface. You haven't actually said it yet but I assume the surface is probably oriented outward. Does the above vector agree with that orientation or do you need its opposite?

Now is where I hit my problem...

I realize r is a vector and not a scalar, that is exactly what I'm confused about.

I need to do the dot product r.n. Where r=xi + yk+zk, and n is what I already stated in cylindrical coordinates. But, I can't put r into cylindrical coordinates since there is no axis, so I don't know how to proceed with the dot product.

I don't know what you mean "there is no axis". You have the substitution x = cos(θ), y = sin(θ), z = z that you used before. Use the same substitution to put r into cylindrical coordinates.

Now you need to look again at the general formula for flux integrals:

$$\int\int_S \vec F \cdot d\vec S =\pm\int\int_{(u,v)} \vec F(u,v)\cdot\vec r_u \times \vec r_v\ dudv$$

Here your u and v are θ and z, your F is r, and the sign is chosen so that the direction of the cross product agrees with you orientation. So what does this give you for your problem?

I haven't done anything with the divergence theorem... I don't know how I would come up with a dV since I am only given the shape in 2-D or 1-D pieces.

That's a shame because the divergence theorem is the obvious easy way to do this problem. Your three surfaces enclose a volume. So I will ask you once more to write down what the divergence theorem says for your particular problem.

## 1. What is a surface integral?

A surface integral is a mathematical tool used to calculate the flux (flow) of a vector field across a surface. It involves breaking down the surface into small, flat pieces and summing up the contributions from each piece.

## 2. How is a surface integral related to Green's theorem?

Green's theorem is a special case of a surface integral, where the surface is a closed curve in the plane. It relates the line integral around the curve to the double integral over the region enclosed by the curve. This allows us to calculate the line integral using simpler double integrals.

## 3. What are some real-world applications of surface integrals?

Surface integrals are used in many fields of science and engineering, such as fluid dynamics, electromagnetism, and heat transfer. They can be used to calculate the flow of a fluid through a surface, the electric field generated by a charged surface, or the heat flux through a solid object.

## 4. How do I set up a surface integral?

To set up a surface integral, you must first define the surface and the vector field you want to integrate. Then, you need to parameterize the surface and express the vector field in terms of the parameters. Finally, you can use a double integral to calculate the flux through the surface.

## 5. What are some common sources of confusion when dealing with surface integrals?

Some common sources of confusion with surface integrals include understanding the direction of the normal vector to the surface, choosing the correct orientation for the surface and the line integral, and correctly setting up the limits of integration for the double integral.

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