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Why the projection of S on the xy plane cannot be used?

  • Thread starter JD_PM
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Problem Statement
Please see image
Relevant Equations
Please see image
Screenshot (260).png

Screenshot (261).png

I don't understand why we could not use the 1/4 of the circle lying on the xy plane as R. In the exercise it is not explained.

The idea would be taking the arc length. I know it is not easier than making your projection on the xz plane, but just wondering if this is possible. I guess it is not, but why...

EDIT

I've just realized there's a big issue if we select our infinitesimal patch of area to lie on the xy plane in this problem. Let's show it:

The unit normal vector is:

Screenshot (263).png

Then we'd get:
Screenshot (262).png

Note that nk yields zero, so this method is discarded.


Thanks.
 
Last edited:

Orodruin

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The point is that giving ##x## and ##y## does not uniquely identify a point on the surface as they are necessarily related through ##R^2 = x^2 + y^2##. You need to use a parametrisation that is a 1-to-1 map to the surface you are integrating over.
 
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The point is that giving ##x## and ##y## does not uniquely identify a point on the surface as they are necessarily related through ##R^2 = x^2 + y^2##. You need to use a parametrisation that is a 1-to-1 map to the surface you are integrating over.
So do you mean that the reason is the following?

As the provided surface has no explicit value for z (##R^2 = x^2 + y^2##) then we cannot integrate through a region where z varies.

Indeed, in the provided solution, the region has a constant value for ##z##.
 

BvU

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I don't understand why we could not use the 1/4 of the circle lying on the xy plane as R
Isn't it much simpler:
your ##\ A\ ## depends on ##z## so you somehow need to keep ##z## as integration variable
your ##\ A\cdot n\ ## does not depend on ##y## so you can afford to lose ##y##
 
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your ##\ A\ ## depends on ##z## so you somehow need to keep ##z## as integration variable
your ##\ A\cdot n\ ## does not depend on ##y## so you can afford to lose ##y##
I get that but actually I am still thinking why the projection of S on the xy plane cannot be used?...
 

LCKurtz

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So do you mean that the reason is the following?

As the provided surface has no explicit value for z (##R^2 = x^2 + y^2##) then we cannot integrate through a region where z varies.
No, that isn't what @Orodruin means.

Indeed, in the provided solution, the region has a constant value for ##z##.
No, it doesn't. ##z## varies for ##0## to ##5##. The fact that your normal vector is independent of ##z## doesn't mean that ##z## doesn't vary. You need to understand Orodruin's post #2. A parameterization of a surface requires two independent variables. Maybe it would help you to think of it this way. Given the equation of the surface, what additional information do you need to locate a point on the surface? I am thinking of a point on the surface and I tell you the ##x## and ##z## coordinates of the point. Can you figure it out? The answer is yes because you can get ##y## from the equation of the surface by knowing ##x##. But suppose I give you just the ##x## and ##y## coordinates. Then you can't tell me the coordinates of the point because you have no way of knowing ##z##. That is one reason you can't use just ##x## and ##y## for the parameterization of the surface.
 

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