Flux of a vector field through a surface

[DottZakapa]

Homework Statement

Consider the vector field F (x, y, z) = (0, z, y) and the surface Σ= (x,y,z)∈R^3 : x=2y^2z^2, 0≤y≤2, 0≤z≤1
oriented so that its normal vector forms an acute angle with the fundamental versor of the x–axis. Compute the flux of F through Σ.

Relevant Equations

flux of F through sigma

Given
$F (x, y, z) = (0, z, y)$ and the surface $\Sigma = (x,y,z)∈R^3 : x=2 y^2 z^2, 0≤y≤2, 0≤z≤1$

i have parametrised as follows

$\begin{cases} x=2u^2v^2\\ y=u\\ z=v\\ \end{cases}$

now I find the normal vector in the following way

$\begin{vmatrix} i & j & k \\ \frac {\partial x} {\partial u} & \frac {\partial y} {\partial u}& \frac {\partial z} {\partial u} \\ \frac {\partial x} {\partial v} & \frac {\partial y} {\partial v}& \frac {\partial z} {\partial v} \\ \end{vmatrix} = \begin{vmatrix} i & j & k \\ 4uv^2 & 1 & 0 \\ 4u^2v & 0 & 1\\ \end{vmatrix} = \vec i(1)-\vec j(4uv^2)+\vec k(- 4u^2v)$

$\Rightarrow N(u,v) = (1,-4uv^2,- 4u^2v)$

Is there anything wrong on the normal vectors signs? what having an acute angle with x translates in?
I don't understand why in the solution the second and third components have negative sign.

 

[Orodruin]

You should double check your determinant computation.

what having an acute angle with x translates in?

What is the cosine of an acute angle? How do you find the cosine between the x-direction and an arbitrary vector?

 

[DottZakapa]

You should double check your determinant computation.
minus in front of $j$

 

[DottZakapa]

What is the cosine of an acute angle? How do you find the cosine between the x-direction and an arbitrary vector?

the cos has to be positive, ok so the $\vec i$ component must be positive, hence the normal vector satisfy the requirement.
correct?

 

[vanhees71]

I don't see an error. I'm not sure, what @Orodruin is after...

 

[Orodruin]

I don't see an error. I'm not sure, what @Orodruin is after...

OP has edited out the error (last edit at 12.01) after my post (made at 11.49). In the original post, the j component had the opposite sign, as indicated in #3 (where OP wrote the response inside the quotes of my post).

 

[Orodruin]

the cos has to be positive, ok so the $\vec i$ component must be positive, hence the normal vector satisfy the requirement.
correct?

Correct, you have also edited out the sign error from your original post, which should also mean that your question regarding the signs is resolved.

Please note that correcting errors by editing the original post can be very confusing to others reading the thread (as evidenced by @vanhees71 in post #5) and therefore generally should not be done. It is better to make a new post in the thread with the correction and also mentioning if your questions have been fully answered or if you still have doubts.

 

[DottZakapa]

Correct, you have also edited out the sign error from your original post, which should also mean that your question regarding the signs is resolved.

Please note that correcting errors by editing the original post can be very confusing to others reading the thread (as evidenced by @vanhees71 in post #5) and therefore generally should not be done. It is better to make a new post in the thread with the correction and also mentioning if your questions have been fully answered or if you still have doubts.

yes, sorry about it, you are absolutely right.
Next time i'll do as you have correctly said.
Thank you, should i edit again the original post, restoring the wrong sign?

 

[vanhees71]

I usually correct typos and errors within the post but making an editorial note. I find this better than to make another posting, which may get not recognized by the readers of the thread and then leading to confusion.