Question

I need Amistre or someone. Vector Calc question.

Answer 1

Ask away...?

Answer 2

IKnow how to evaluate the actual integral (the flux/surface integral) I just need helping discerning whether it is orientation preserving or reversing (one is the negative of the other) This is the problem:
Let S be a part of the cone \[z=\sqrt{x^2+y^2}\]
such that its projection onto the x-y plane is the square \[0 \le x \le 1; 0 \le y \le 1\]
The normal vector n is chosen so that\[ \vec{n} \cdot \vec{k} < 0\]
Find the flux of the vector field \[\vec{F}(x,y,z)=z \vec{i} +z \vec{j} +z^2 \vec{k}\] across S.

Answer 3

MATLAB is your friend :-D

Answer 4

*brain explodes* I hope amistre helps you.

Answer 5

I used the parameterization of:
\[\Phi = x \vec{i} + y\vec{j}+\sqrt{x^2+y^2} \vec{k}\]
Took derivatives with respect to x and y and formed the cross product. I find that to be:
\[\frac{-x}{\sqrt{x^2+y^2}} \vec{i} -\frac{y}{\sqrt{x^2+y^2}}\vec{j}+\vec{k}\]

Answer 6

Now my question: Since the k is positive and it tells me in the problem that n dot k is negative, is it orientation reversing or preserving? It only alters my answer by a negative sign but my teacher is making this question be about finding the orientation, the computation is too easy...

Answer 7

I think it is preserving btw.

Answer 8

its indeed over my head at the moment, but this might be useful:

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx

Answer 9

I tried that one but couldn't find exactly what I needed to answer my question :/

Answer 10

hmmm, what is the n vector to begin with? or is my thinking a bit off

Answer 11

its the normal vector to the vector field at all points

Answer 12

http://bcs.whfreeman.com/marsdenvc5e/pages/bcs-main.asp?s=&n=00010&i=01010.01&v=&o=&ns=0&t=&uid=0&rau=0
These power points are free. This is the text book I'm using. Its section 7.6 or 7.7 I think the latter however.

Answer 13

ill check the pps, but im curious; is this the derivatives? or is the vector field more like derivatives of a surface?

\[\vec{F}(x,y,z)=z \vec{i} +z \vec{j} +z^2 \vec{k}\]
\[\vec{t}=\vec{F}(x,y,z)'=1 \vec{i} +1 \vec{j} +2z \vec{k}\]
\[\vec{n}=\vec{F}(x,y,z)''=0 \vec{i} +0 \vec{j} +2 \vec{k}\]

Answer 14

Those derivatives are fine. I'm just not sure if you can find unit normal vectors that way is all.

Answer 15

hmm, unit would have to divide the n by 2 to get <0,0,1>

but isnt that just a unit of k if we did?

Answer 16

Exactly, but I am wary because it says:
\[\hat{n} \cdot \hat{k} <0\]
But if
\[\hat{n}=<0,0,1>;\hat{k}=<0,0,1> \rightarrow \hat{n} \cdot \hat{k}=1 \]
And 1 is not strictly less than 0.

Answer 17

yeah, which makes me ponder if we have to "reverse" it .... but im reading up on it :)

Answer 18

I might think you're right. Because if k points up (in the cone) and n points negative (so the dot product is negative) then it would point out. This means orientation reversing.

Answer 19

thats my intuition, so i got like a 13.45% chance of being right ;)

Answer 20

Haha, he gave us another one where n dot j was psoitive and they both point IN the cone (its laid over a different axis). So I think since the dot product is switched that could work.

Answer 21

Thanks for the mental support though xDD :)

Answer 22

if you can work out what Tx and Ty are on the surface; then the normal of course is Tx x Ty

and the unit is well, you know ....

this is what i find in the power points, not that i can decipher it that well

Answer 23

and this defines the Tu x Tv .... parts