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 <br /> \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = k(u,v)\nabla f so that <br /> d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv =<br /> 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} = <br /> \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 <br /> \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = k(u,v)\nabla f so that <br /> d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv =<br /> 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} =<br /> \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 <br /> 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 <br /> 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, <br /> 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 <br /> 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 <br /> 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, <br /> 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! :)