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

[JD_PM]

Homework Statement

Please see image

Relevant Equations

Please see image

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:

Then we'd get:

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

 

[Orodruin]

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.

 

[JD_PM]

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]

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 $y$

 

[JD_PM]

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]

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.