Surface integrals of vector fields, normal - does scaling matter?

[laser1]

Homework Statement

desc

Relevant Equations

N.A.

Source: https://tutorial.math.lamar.edu/Problems/CalcIII/SurfIntVectorField.aspx

Alright so my confusion lies in the following step: Consider the side x = 0. Okay, from the formula (I am just going to insert an image here)

Okay, so gradient of f would just be <1, 0, 0>, which is simply i, which agrees with the solution. However, what if I said that the plane x = 0 is equivalent to the plane 2x = 0? Then the gradient would be <2, 0, 0>, which changes the problem! There must be a flaw in my logic here, but I can't figure it out. Thanks!

 

[pasmith]

The plane $kx = 0$ has surface element $dy\,dz$. Its unit normal is $\pm \mathbf{e}_x$, so $d\mathbf{S} = \pm\mathbf{e}_x\,dy\,dz$.

What appears in your image implies $dS = \|\nabla f\|\,dA$, which is a change of variable from $(y,z)$ to something different. Working out the detail there should cancel the 2 you introduced into $\nabla f = 2\mathbf{e}_x$.

 

[laser1]

What appears in your image implies $dS = \|\nabla f\|\,dA$

Oh, is this not correct then? Edit: I am not fully sure what you are saying

 

[pasmith]

The normal of a level surface of $f$ is parallel to $\nabla f$, ie. there is some non-zero scalar field $k$ such that $$\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = k(u,v)\nabla f$$ so that $$d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv = k \nabla f \,du\,dv$$ and $$dS = |k|\|\nabla f\|\,du\,dv.$$ The actual source of your error is to assume $k \equiv 1$.

If you parametrize $f(x,y,z) = Cx = 0$, $C \neq 0$, as $\mathbf{r}(u,v) = u\mathbf{e}_y + v\mathbf{e}_z$ then you find $$d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv = \mathbf{e}_y \times \mathbf{e}_z\,du\,dv = \mathbf{e}_x\,du\,dv = \frac1C \nabla f\,du\,dv$$ so that $k \equiv \frac1C$.

 

[laser1]

The normal of a level surface of $f$ is parallel to $\nabla f$, ie. there is some non-zero scalar field $k$ such that $$\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = k(u,v)\nabla f$$ so that $$d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv = k \nabla f \,du\,dv$$ and $$dS = |k|\|\nabla f\|\,du\,dv.$$ The actual source of your error is to assume $k \equiv 1$.

If you parametrize $f(x,y,z) = Cx = 0$, $C \neq 0$, as $\mathbf{r}(u,v) = u\mathbf{e}_y + v\mathbf{e}_z$ then you find $$d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv = \mathbf{e}_y \times \mathbf{e}_z\,du\,dv = \mathbf{e}_x\,du\,dv = \frac1C \nabla f\,du\,dv$$ so that $k \equiv \frac1C$.

Thanks, that makes sense! What about this other example?

It seems equally as valid to parameterise a surface as <0, 2y, 2z>, where y and z can be any number, to represent x = 0. Yet the results suggest otherwise!

 

[pasmith]

The general case for $f(x,y,z) = x = 0$ is $\mathbf{r} = y(u,v)\mathbf{e}_y + z(y,v)\mathbf{e}_z$, with $$d\mathbf{S} = \mathbf{e}_x \left( \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v}\right)\,du\,dv$$ and $$dS = \left| \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v} \right|\,du\,dv$$ which you should recognise as the change of variable formula for a double integral, $$dy\,dz = \left| \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v} \right|\,du\,dv.$$

 

[laser1]

The general case for $f(x,y,z) = x = 0$ is $\mathbf{r} = y(u,v)\mathbf{e}_y + z(y,v)\mathbf{e}_z$, with $$d\mathbf{S} = \mathbf{e}_x \left( \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v}\right)\,du\,dv$$ and $$dS = \left| \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v} \right|\,du\,dv$$ which you should recognise as the change of variable formula for a double integral, $$dy\,dz = \left| \frac{\partial y}{\partial u}\frac{\partial z}{\partial v} - \frac{\partial z}{\partial u}\frac{\partial y}{\partial v} \right|\,du\,dv.$$

I don't follow, sorry. In one example I get $\textbf{i}$ and in the other example I get $\textbf{4i}$ for the same equation.

 

[laser1]

I don't follow, sorry. In one example I get $\textbf{i}$ and in the other example I get $\textbf{4i}$ for the same equation.

resolved: the parameterisation is different, so the bounds will change accordingly. All is fine! :)