Normal vector in surface integral of vector field

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Homework Help Overview

The discussion revolves around the orientation of the normal vector in the context of surface integrals of vector fields. Participants are exploring the relationship between the orientation of the normal vector and the signs of the partial derivatives dz/dx and dz/dy.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand why the partial derivatives are negative when the normal vector is oriented upward. There are requests for examples and further explanations regarding the implications of the normal vector's orientation.

Discussion Status

Some participants have suggested trying examples to clarify the concepts. There is an ongoing exploration of the definitions and calculations related to the normal vector and its gradient, with multiple interpretations being examined.

Contextual Notes

There are indications that participants may be working under specific constraints related to their homework, and some are questioning the assumptions made in the problem setup.

foo9008
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Homework Statement


when the normal vector n is oriented upward , why the dz/dx and dz/dy is negative ? shouldn't the k = positive , while the dz/dx and dz/dy is also positive?

Homework Equations

The Attempt at a Solution


is the author wrong ? [/B]
 

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Try a few examples and you'll see.
Remember when vectors are perpendicular
and when lines are perpendicular
 
BvU said:
Try a few examples and you'll see.
Remember when vectors are perpendicular
and when lines are perpendicular
Can you give some example?
 
z = x
 
BvU said:
z = x
i still don't understand what do you mean, can you explain further?
 
The surface ##\sigma## is defined by ##\phi(x,y,z)=0## where ##\phi(x,y,z)=z-f(x,y)##, and the normal is the gradient of ##\phi(x,y,z)##. What do you get when you calculate that?
 
vela said:
The surface ##\sigma## is defined by ##\phi(x,y,z)=0## where ##\phi(x,y,z)=z-f(x,y)##, and the normal is the gradient of ##\phi(x,y,z)##. What do you get when you calculate that?
why it will become like this ?
 
Why don't you simply fill it in and see ? Asking 'why' forever doesn't make sense. Make a sketch for the simplest case if 4.5.1 is too complicated.
 
foo9008 said:
why it will become like this ?
I'm sure this is covered in your textbook.
 

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