# What are the limits of integration for this surface integral?

#### JD_PM

Problem Statement
Relevant Equations

I want to compute:
 $$\oint_{c} F \cdot dr$$

I have done the following:

 $$\iint_{R} (\nabla \times v) \cdot n \frac{dxdy}{|n \cdot k|} = \iint (9z-1) dxdy$$

I don't know what limits the surface integral will have. Actually, I am not sure what's the surface.

May you shed some light?

Thanks

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#### andrewkirk

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It is a line integral, not a surface integral, so it must be a single, not a double integral.
The curve is the intersection of two surfaces. The surface given by the first equation is a vertical, infinite prism with elliptic cross section, the centre of the ellipse being (x=2, y=1). The surface given by the second equation is a plane. The intersection of the two will be a simple, closed curve, roughly elliptic in shape. We need to integrate around that curve.

There are a variety of ways to do the integration. One would be to use the highest and lowest points of the curve (z coordinate) to split the curve into two parts, then integrate over each part using z as integration variable, and add the results.

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#### Orodruin

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It is a line integral, not a surface integral
Stokes' theorem.

#### Orodruin

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I don't know what limits the surface integral will have. Actually, I am not sure what's the surface.

May you shed some light?
Did you try to compute $\nabla \times \vec F$? What did you get? It does not matter what the surface is as long as it has $C$ as its boundary (this follows from the divergence theorem and $\nabla\cdot(\nabla\times \vec v) = 0$). In other words, it is up to you to choose the exact surface, but you should choose it to make life easy for yourself.

#### LCKurtz

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View attachment 241115

I want to compute:
 $$\oint_{c} F \cdot dr$$

I have done the following:

 $$\iint (9z-1) dxdy$$

I don't know what limits the surface integral will have. Actually, I am not sure what's the surface.
Actually, I get an integrand of $9z-3$ so one of us has an arithmetic mistake. Hard to say who because you didn't show your work. But more to the point, did you finish the problem or have you abandoned this thread?

#### JD_PM

did you finish the problem or have you abandoned this thread?
I did not abandon it. I will post more details today.

#### JD_PM

Actually, I get an integrand of $9z-3$ so one of us has an arithmetic mistake.
After some thought I came to a different result of $\int (\nabla \times \vec F) \cdot da$:

$$\nabla \times \vec F = (2z - 1) \hat{i} + 3 \hat{j}$$

$$\vec n = 3 \hat{i} + \hat{j} + \hat{k}$$

$$\hat{n} = \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}$$

As Orodruin pointed out, $\int (\nabla \times \vec F) \cdot da$ depends only on the boundary line, so we can pick out whatever surface we want (in our favour of course). Let's pick the circle as the surface:

Using polar coordinates we get:

$\int (\nabla \times \vec F) \cdot da = \int_{s} 6z ds = \int_{0}^{2\pi} \int_{0}^{3} 6r(1-3rcos\theta - rsin\theta) drd\theta = 54\pi$

How do you see it?

I've been trying to check it evaluating $\oint F \cdot dl$ but that integral gets messy...

EDIT

I've been thinking that my mistake here may be that I did not multiply by the unit normal vector.

$$\iint (\nabla \times \vec F) \cdot da = \iint (\nabla \times v) \cdot \hat{n} \frac{dxdy}{|\hat{n} \cdot k|} = \int_{s} 6z ds = \int_{0}^{2\pi} \int_{0}^{3} 6r(1-3rcos\theta - rsin\theta) drd\theta = 54\pi$$

But I get the same result using the formula involving the unit normal vector.

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#### LCKurtz

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After some thought I came to a different result of $\int (\nabla \times \vec F) \cdot da$:

$$\nabla \times \vec F = (2z - 1) \hat{i} + 3 \hat{j}$$
Show your work. I think the $3$ should be $3z$.

$$\vec n = 3 \hat{i} + \hat{j} + \hat{k}$$

$$\hat{n} = \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}$$

As Orodruin pointed out, $\int (\nabla \times \vec F) \cdot da$ depends only on the boundary line, so we can pick out whatever surface we want (in our favour of course). Let's pick the circle as the surface:
Your curve is not a circle. It is the intersection of an elliptical cylinder with the slanted plane. The projection of the enclosed surface on the xy plane is the given ellipse, which you might note, is not centered at the origin.

Using polar coordinates we get:

$\int (\nabla \times \vec F) \cdot da = \int_{s} 6z ds = \int_{0}^{2\pi} \int_{0}^{3} 6r(1-3rcos\theta - rsin\theta) drd\theta = 54\pi$

How do you see it?

I've been trying to check it evaluating $\oint F \cdot dl$ but that integral gets messy...

EDIT

I've been thinking that my mistake here may be that I did not multiply by the unit normal vector.

$$\iint (\nabla \times \vec F) \cdot da = \iint (\nabla \times v) \cdot \hat{n} \frac{dxdy}{|\hat{n} \cdot k|} = \int_{s} 6z ds = \int_{0}^{2\pi} \int_{0}^{3} 6r(1-3rcos\theta - rsin\theta) drd\theta = 54\pi$$

But I get the same result using the formula involving the unit normal vector.
Your polar coordinates are not going over the given ellipse. You are working the integral $\iint_S 6z~dS$ and I am not agreeing with that integrand yet. No point in evaluating it until we agree on the integrand.

#### JD_PM

Show your work. I think the $3$ should be $3z$.

Your curve is not a circle. It is the intersection of an elliptical cylinder with the slanted plane. The projection of the enclosed surface on the xy plane is the given ellipse, which you might note, is not centered at the origin.

Your polar coordinates are not going over the given ellipse. You are working the integral $\iint_S 6z~dS$ and I am not agreeing with that integrand yet. No point in evaluating it until we agree on the integrand.
I will update later today

#### JD_PM

Your curve is not a circle. It is the intersection of an elliptical cylinder with the slanted plane. The projection of the enclosed surface on the xy plane is the given ellipse.
After some thought I saw that the projection of the enclosed surface on the $xy$ plane is the given ellipse. So we indeed need to get a unit normal vector of that ellipse:

$$n = \nabla (\frac{(x-2)^2}{4} + (y-1)^2) = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$

$$\hat{n} = \frac{\frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}$$

I guess there has to be a way of simplifying $\hat{n}$ but don't see it :(

So now we just have to set up the integral:

$$\iint (\nabla \times \vec F) \cdot d \vec a = \iint[(2z - 1)\hat{i} + 3z\hat{j}][\frac{\frac{x-2}{2}\hat{i} + 2(y-1)\hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}]da$$

Now we gotta make a change of variables so as to go from an ellipse to a circle centred at the origin.

Rearranging the curve we get:

$$\frac{(x-2)^2}{36} + \frac{(y-1)^2}{9} = 1$$

Using the following change of variables and using it:

$$x - 2 = 6u$$

$$y - 1 = 3v$$

$$u^2 + v^2 = 1$$

Don't forget the Jacobian! :)

$$\frac{\partial(x, y)}{\partial (u, v)} = 18$$

$$-18\iint[(36u+6v+13)\hat{i} + (36u+9v+12)\hat{j}][\frac{\frac{6u}{2}\hat{i} + 2(3v)\hat{j}}{\sqrt{\frac{(6u)^2}{4}+ 4(3v)^2}}]dudv$$

There's just one step left: apply the last change of variables (polar coordinates):

$$u = rcos\theta$$
$$v = rsin\theta$$

Before doing that please let me know of your thoughts on what I've done at this point. (I am really having a good time with this problem btw xD)

#### Orodruin

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After some thought I saw that the projection of the enclosed surface on the xyxyxy plane is the given ellipse
This is a (really) bad idea. Taking the cylinder surface you will not close the surface and you certainly will not be able to parametrise it using the x and y coordinates. You need to take a surface that has your curve as its boundary, nothing else.

#### JD_PM

This is a (really) bad idea. Taking the cylinder surface you will not close the surface and you certainly will not be able to parametrise it using the x and y coordinates. You need to take a surface that has your curve as its boundary, nothing else.
But isn't the curve the ellipse itself? What I have in mind is an elliptic cylinder being cut by the plane. Then I just took the projection of that cut (an ellipse) on the $xy$ plane.

Why is this wrong?

#### Orodruin

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But isn't the curve the ellipse itself?
No, your curve is not in the xy-plane. Your curve is the intersection of a cylinder with an elliptic cross section and a plane. The mantle area of the cylinder (which is what you are computing the normal for) has nothing to do with the eventual surface you want to use for your surface integral. You are also not after a curve integral, you are writing down a surface integral.

#### JD_PM

Mmm,I think I misunderstood the problem. Actually I thought that we had this situation (example coming from Marsden):

But instead of $C$ being the intersection of the shown cylinder and the shown plane we got $C$ being the intersection of the elliptic cylinder $\frac{(x-2)^2}{36} + \frac{(y-1)^2}{9} = 1$ and the plane $3x + y + z = 1$.

If I am not mistaken, your point is that $C$ is not the intersection curve but the elliptic cylinder $\frac{(x-2)^2}{36} + \frac{(y-1)^2}{9} = 1$ itself, right?

#### Orodruin

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No. A curve cannot be a surface so this is definitely not my point. There is no projecting the curve on the $xy$ plane, there is parametrising the surface using the $x$ and $y$ coordinates.

#### JD_PM

No. A curve cannot be a surface so this is definitely not my point. There is no projecting the curve on the $xy$ plane, there is parametrising the surface using the $x$ and $y$ coordinates.
I see that we can parametrize the ellipse like:

$$x - 2 = 2cost$$

$$y - 1 = 2sint$$

$$z = 0$$

And then use $\oint_{C} \vec F \cdot d \vec l$.

Now where I am stuck is in understanding $\iint (\nabla \times \vec F) \cdot d \vec a$. As we know, the value of the function at the boundary (i.e. the perimeter of C) is equal to the integral of the curl of F over the surface. The path C encloses an ellipse, that's why I thought that the normal vector had to be:

 $$n = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$

But now I am pretty confused... how can I get $\iint (\nabla \times \vec F) \cdot d \vec a$, then?

#### LCKurtz

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I have been trying to get you to correct your curl. Your work is pretty good down to where you have $\iint_s 6z~ds$ for which I think you mean $\iint_R 6z~dydx$ where $R$ is the elliptical region in the xy plane. I think you should have $9z-3$ in there instead of $6z$. Once you have that expressed in terms of x and y, you will be ready to make the appropriate change of variables to express the region $R$ in polar-like coordinates. You will need two variables $r$ and $\theta$ and the Jacobian.

#### Orodruin

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I see that we can parametrize the ellipse like:

$$x - 2 = 2cost$$

$$y - 1 = 2sint$$

$$z = 0$$

And then use $\oint_{C} \vec F \cdot d \vec l$.
This is not true. This parametrises the projection of $C$ onto the $xy$ plane, not $C$ itself. The value of $z$ along the curve you are trying to integrate along is generally non-zero. You need to separate the curve that you have from its projection in the $xy$ plane. The only reason to write down its projection in the $xy$ plane is to use $x$ and $y$ to parametrise your surface later with those coordinates later and then the $z$ value at the surface will depend (a) on the surface you have chosen - some choices are easier than others - and (b) on the $x$ and $y$ values as your surface will have the $z$ values as a function of the $x$ and $y$ values.

Now where I am stuck is in understanding $\iint (\nabla \times \vec F) \cdot d \vec a$. As we know, the value of the function at the boundary (i.e. the perimeter of C) is equal to the integral of the curl of F over the surface.
I am going to stop you there. It is not the value of the function on the boundary that is equal to the flux of the curl through the surface - it is the circulation integral around the boundary that is, which is not the same thing. This may sound pedantic but it is important to make the distinction and since the issue seems to be with your understanding of surface and line integrals, it is worth pointing it out.

The path C encloses an ellipse, that's why I thought that the normal vector had to be:

 $$n = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$
Curves do not have a single normal so this makes little sense to compute. This is the normal of the ellipsoidal cylinder, not of $C$. For the circulation integral around $C$, you are interested in the curve tangent, not a vector normal to the curve.

#### JD_PM

I have been trying to get you to correct your curl. Your work is pretty good down to where you have $\iint_s 6z~ds$ for which I think you mean $\iint_R 6z~dydx$ where $R$ is the elliptical region in the xy plane. I think you should have $9z-3$ in there instead of $6z$. Once you have that expressed in terms of x and y, you will be ready to make the appropriate change of variables to express the region $R$ in polar-like coordinates. You will need two variables $r$ and $\theta$ and the Jacobian.
My initial idea was projecting the curve $C$ onto the $xy$ plane. Indeed, the curl is $\nabla \times \vec F = (2z - 1) \hat{i} + 3z \hat{j}$ (do you want me to explicitly write down the determinant?).

Then, I took the normal vector to the surface of the ellipse (which arises from the intersection elliptic-cylinder and plane):

$$n = \nabla (\frac{(x-2)^2}{4} + (y-1)^2) = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$

$$\hat{n} = \frac{\frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}$$

Now I would use $\hat{n}$ to calculate $\iint (\nabla \times \vec F) \cdot d \vec a$:

$$\iint_{R} (\nabla \times v) \cdot \hat{n} \frac{dxdy}{| \hat{n} \cdot k|}$$

But you have used the normal to the given plane instead:

$$\vec n = 3 \hat{i} + \hat{j} + \hat{k}$$

$$\hat{n} = \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}$$

$$\iint_{R} (\nabla \times v) \cdot \hat{n} \frac{dxdy}{|\hat{n} \cdot k|} = [(2z - 1) \hat{i} + 3z \hat{j}] \cdot [ \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}][\frac{dxdy}{1/\sqrt{11}}]=$$

$$\iint_{R} (9z-3)dydx = \iint_{R}(-27x - 9y + 6)dydx$$

Before solving the integral I want to discuss why you're using the normal to the plane and not to the ellipse.

What I think is that both options are correct, because as Orodruin pointed out at 4#, it is up to us to choose the exact surface (i.e we can play either with the plane or the ellipse enclosed by the elliptic cylinder) .
Do you agree? (this is just an understanding checking).

#### JD_PM

This is not true. This parametrises the projection of $C$ onto the $xy$ plane, not $C$ itself. The value of $z$ along the curve you are trying to integrate along is generally non-zero. You need to separate the curve that you have from its projection in the $xy$ plane.
I think I see your point; you're seeking for a parametrization of the ellipse, but not its projection. All I've worked with till this point has been related to parametrizations on the $xy$, $yz$ or $xz$ plane but not wit $xyz$ (setting aside household cases like sphere and cylinder)...

Could you give me a hint on how to do so?

#### Orodruin

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All I've worked with till this point has been related to parametrizations on the xyxyxy, yzyzyz or xzxzxz plane but not wit xyzxyzxyz (setting aside household cases like sphere and cylinder)...
You always need a parametrisation that gives all three coordinates. Otherwise it is not a parametrisation.

#### LCKurtz

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My initial idea was projecting the curve $C$ onto the $xy$ plane. Indeed, the curl is $\nabla \times \vec F = (2z - 1) \hat{i} + 3z \hat{j}$ (do you want me to explicitly write down the determinant?).

Then, I took the normal vector to the surface of the ellipse (which arises from the intersection elliptic-cylinder and plane):

$$n = \nabla (\frac{(x-2)^2}{4} + (y-1)^2) = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$

$$\hat{n} = \frac{\frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}$$

Now I would use $\hat{n}$ to calculate $\iint (\nabla \times \vec F) \cdot d \vec a$:

$$\iint_{R} (\nabla \times v) \cdot \hat{n} \frac{dxdy}{| \hat{n} \cdot k|}$$

But you have used the normal to the given plane instead:

$$\vec n = 3 \hat{i} + \hat{j} + \hat{k}$$

$$\hat{n} = \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}$$

$$\iint_{R} (\nabla \times v) \cdot \hat{n} \frac{dxdy}{|\hat{n} \cdot k|} = [(2z - 1) \hat{i} + 3z \hat{j}] \cdot [ \frac{3 \hat{i} + \hat{j} + \hat{k}}{\sqrt{11}}][\frac{dxdy}{1/\sqrt{11}}]=$$

$$\iint_{R} (9z-3)dydx = \iint_{R}(-27x - 9y + 6)dydx$$
Good. I did it a different way, but now we agree on what the xy integral we need to calculate is.

Before solving the integral I want to discuss why you're using the normal to the plane and not to the ellipse.

What I think is that both options are correct, because as Orodruin pointed out at 4#, it is up to us to choose the exact surface (i.e we can play either with the plane or the ellipse enclosed by the elliptic cylinder) .
Do you agree? (this is just an understanding checking).
You have a plane that cuts through a cylinder. The cylinder cuts it in the curve $C$, giving an elliptical shaped portion of the plane. That elliptical shaped portion of the plane is the obvious choice of a surface $S$ bounded by the curve. It is true you could use some other surface with the elliptical curve as its boundary but it would be challenging to find such a surface simpler than a plane. So, to reiterate, the surface is a piece of a plane and that is what you want the normal to.

Now, all you have left is to work out $\iint_R -27x - 9y + 6~dydx$ where $R$ is the interior of your given ellipse which I will write $$\frac{(x-2)^2}{36}+\frac{(y-1)^2}{9}=1$$
Now, previously you were thinking about parameterizing this ellipse like this$$x=2+6\cos \theta,~y=1+3\sin \theta$$The problem is that parameterizes the ellipse curve in the xy plane, but not the inside area of the ellipse. To do that you need two parameters because you need to vary $r$ and $\theta$. Is that enough of a hint for you to get a good parameterization to finish? And don't forget the Jacobian.

#### JD_PM

I'll update later

#### JD_PM

To do that you need two parameters because you need to vary $r$ and $\theta$.
Isn't that what we do if we apply polar coordinates as I suggested at #10?

There's just one step left: apply the last change of variables (polar coordinates):

$$u = rcos\theta$$
$$v = rsin\theta$$

#### LCKurtz

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$u = r\sin\theta,~v=r\cos \theta$ would describe a circle at the origin. You have an ellipse that is not centered at the origin. I gave you a hint in post #22. Look at it again...

"What are the limits of integration for this surface integral?"

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