Line Integral in Cylindrical Coordinates

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SUMMARY

The discussion centers on evaluating the surface integral \(\int \text{curl} \, \textbf{A} \cdot \textbf{a}\) for the vector field \(\textbf{A} = y \textbf{i} + z \textbf{j} + x \textbf{k}\) over the surface defined by the paraboloid \(z = 1 - x^2 - y^2\). Participants utilized Stokes' theorem to convert the surface integral into a line integral around the boundary of the paraboloid, specifically a circle with radius \(1 - z\). The challenge lies in parameterizing the curve correctly to facilitate integration, with discussions on cylindrical coordinates and the implications of setting \(s = 1 - z\).

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  • Understanding of Stokes' theorem
  • Familiarity with cylindrical coordinates
  • Knowledge of vector calculus and surface integrals
  • Ability to parameterize curves in three-dimensional space
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Homework Statement



Find the value of the (surface) integral \int curl \textbf{A} \bullet \textbf{a}

if the vector \textbf{A}=y \textbf{i}+z \textbf{j}+x \textbf{k}

and S is the surface defined by the paraboloid z=1-x^2-y^2


Homework Equations



x=s\cos\phi

y=s\sin\phi

d\textbf{l}=ds\mathbf{\hat{s}}+s d\phi\mathbf{\hat{\phi}}+dz\mathbf{\hat{z}}

\mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}

\mathbf{\hat{y}}=\sin\phi\mathbf{\hat{s}}+\cos\phi\mathbf{\hat{\phi}}

The Attempt at a Solution



First I used Stokes' theorem in order to turn the integral into a line integral, the integral of the dot product of \textbf{A} and d\textbf{l}

Then I turned \textbf{A} into cylindrical coordinates using the above x hat and y hat equations. When I took that dot product of \textbf{A} and d\textbf{l}, I came up with an answer of:

(s\cos\phi\sin\phi+z\sin\phi)ds\\\ +\\\ (-s^2\sin^2\phi+z\s\cos\phi)d\phi\\\ +\\\ s\cos\phi dz

I'm trying to reduce it down to one parameter in order to do the integration.

So, s=1-z. Would that imply ds=-dz? Is dz=1?
 
Last edited:
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msd213 said:

Homework Statement



Find the value of the (surface) integral \int curl \textbf{A} \bullet \textbf{a}

if the vector \textbf{A}=y \textbf{i}+z \textbf{j}+x \textbf{k}

and S is the surface defined by the paraboloid z=1-x^2-y^2

You mean \int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a} right?

And is \mathcal{S} really that entire paraboloid or just some part of it?

First I used Stokes' theorem in order to turn the integral into a line integral, the integral of the dot product of \textbf{A} and d\textbf{l}

I'm trying to reduce it down to one parameter in order to do the integration.

So, s=1-z. Would that imply ds=-dz? Is dz=1?

Well, the first step is to figure out what curve you are going to integrate over...whaat does Stokes theorem say about that?

Once you know what curve you are integrating over, you can parameterize the curve and easily determine the limits of integration.
 
gabbagabbahey said:
You mean \int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a} right?

And is \mathcal{S} really that entire paraboloid or just some part of it?



Well, the first step is to figure out what curve you are going to integrate over...whaat does Stokes theorem say about that?

Once you know what curve you are integrating over, you can parameterize the curve and easily determine the limits of integration.

Yes, S is the surface the paraboloid and yes it is \int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a}.

Stokes theorem says that you integrate over the boundary curve, so it would be the circle at the "end" of the paraboloid, with radius 1-z.

I'm still not sure how to parametrize it. If I used, for A:

y=sin(t)

x=cos(t)

z=z

With the boundary's parametrization given as:

y=(1-z)sin(t)

x=(1-z)cos(t)

z=z

I get an answer of zero for the integral. This parametrization is clearly incorrect but I feel as though I was getting somewhere with the cylindrical coordinates. If you think of the paraboloid in terms of a contour plot, at each z height, s is constant (and obviously z is too). So you could have dz = 0 and ds = 0 so d\phi would be your only parameter to integrate over.
 
Last edited:
msd213 said:
Yes, S is the surface the paraboloid and yes it is \int_{\mathcal{S}} (\mathbf{\nabla}\times\textbf{A})\cdot d\textbf{a}.

Stokes theorem says that you integrate over the boundary curve, so it would be the circle at the "end" of the paraboloid, with radius 1-z.

Surely \mathcal{S} isn't the entire parabaloid z=1-x^2-y^2...if it were, x and y would both range from -\infty to \infty and so the radius at the boundary of the surface would be infinite...what is the exact wording of the original question?
 

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