Question

Dealing with a problem involving Stokes Theorem with a vector field in R3 and a line integral in R2; a little confused about something that pops up because of it.

Answer 1

Posted below in a moment, just one second.

http://i.imgur.com/HMz7sqa.png

Answer 2

What should I do about that z in the integrand/is that okay?

Answer 3

z = 0 if you're living in xy plane

Answer 4

So does the integrand just become -1?

e.g. in any situation where I took the Curl of a vector field and got some result where there was a nonzero component parallel to my normal, and \[\triangledown \times F \cdot n \neq 0 \](Lol redundant)-

Answer 5

I can just eliminate it or plug in the value that it constantly is?

Answer 6

Yes which ever way u prefer

Answer 7

Alright, so if I had a situation where my normal vector was, say, \[j\] and my Curl vector was \[x i + 12y \ j - 3k,\]The dot product and thus integrand would be 12y, and I could *not* progress or go further in the integral without simplifying, or I could just factor it out and treat it like a constant? (Now continuing with this problem)

Answer 8

\[\int\limits_{-3}^{3}\int\limits_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}(-1) \ dx \ dy \]Is this general setup correct for cartesian?

Answer 9

`12y` is not a constant so you need to setup bounds and evaluate the integral the usual way

Answer 10

yes but "-1" is constant here, simply multiply the area of circle by the curl

Answer 11

Well I mean, if your line integral was lying in the xz plane and thus the normal unit vector was j hat, could you treat y like a constant if you got y in your grad cross F dot n, or even factor it out like a constant?

Answer 12

y is 0 if your path is in xz plane

Answer 13

(Will do, just talking about that last part first)-Oh, or parallel to the xz plane, just any plane parallel or including the xz plane, could you treat it as a constant.

I know I can do this, heh, but I'm really unfamiliar with it here; so when I have this area element dA, how do I systematically do the thing you're talking about, multiplying by the area of the object?

Answer 14

\[ \large \begin{align} \int\limits_{-3}^{3}\int\limits_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}(-1) \ dx \ dy &= -\int\limits_{-3}^{3}\int\limits_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} 1 \ dx \ dy \\~\\
&= - (\text{Area of circle of radius 3})\\~\\
&= -(\pi\cdot 3^2)
\end{align} \]

Answer 15

Is that what you talking about ?

Answer 16

Yeah, no, that makes sense, I'm just being dumb and not thinking in terms of what integration actually is. Lemme check to see if I got the right answer.

Answer 17

My book says it's supposed to be \[-\frac{5}{6}\]

Answer 18

Going to look for a mistake, but I don't yet see where I made one; no transformation this time.

Answer 19

GUHHHH >:C I MISREAD THE PROBLEM, the last term in the vector field is +x^2, not -z^2

Answer 20

it wont change the answer because we don't care about z component

Answer 21

It will change the Curl vector, though, won't it? It will make some things not disappear and some other things appear.

Answer 22

We now have a nonzero j component in the Curl vector, if I think right

Answer 23

try it, you will get the same value for curl

Answer 24

j component or k component ?

Answer 25

you're changing k component from -z^2 to x^2 right ?

Answer 26

J component, since it's a determinant, it changes a good bit :E

Yeah

Answer 27

Lemme write out and show my setup comparing the two.

Answer 28

how does changing z component of force affect the things in xy plane ?

Answer 29

you will get the same "-1" for integrand, work it and see

Answer 30

I see what you mean, but yeah, it does change the curl vector, which is what I was saying, heh. http://i.imgur.com/ZrNImYc.png

Now I just need to find my mistakes otherwise.

Answer 31

yeah i see :O

Answer 32

Going to triple check that I read both the problem and the answer key correctly.

Answer 33

>:CCCCCCCCCCCC

Answer 34

lol

Answer 35

I switched an entire problem, I did number two, which there is no answer for in the answer key! I'll close this one and do it entirely separately, MAN

Answer 36

:/

Answer 37

Lul, my second to last comment made it light-hearted enough for me. Opening up the new question in a minute. It might be really similar IIRC, so I may not even open the new one, will see momentarily.

Answer 38

ok