Question

Posting a Divergence Theorem problem shortly.

Answer 1

http://i.imgur.com/1wE8H6D.png @Kainui

Answer 2

Is n just the surface normal of the given solid, or what exactly is n?

Answer 3

Okay, and dS is just the surface differential, right?

Answer 4

Okay, so this is just a section of a cone.

Answer 5

How do I find the normal vector in this situation?

Answer 6

Oh! Do I take the gradient vector of the surface and divide it by its magnitude?

Answer 7

\[\triangledown S=\frac{2x i+2yj-2zk}{?}\]I'd get something like rho in spherical coordinates in the denominator when I took the magnitude, how do I deal with that?

Answer 8

(?)

Answer 9

@Kainui

Answer 10

Well the normal vector is just a radial vector. You can always convert to spherical coordinates, which is probably your best bet. Sorry I'm reading a book so I'm not really all here right now.

Answer 11

No problem, I've got three hours before my final and then I'll be in the immediate timeframe done with worrying over this, I just need to remember and figure out enough prior.

Answer 12

Well, it's close to a good conversion, but not quite because of the minus sign in there; I'm wondering if anybody else has a good idea. @zepdrix

Answer 13

Wait you said this was a cone?

Answer 14

Yeah, it's a cone when graphed, x^2+y^2=z^2; I think I figured out n, the denominator will be a constant when the constraints are involved

Answer 15

nevermind, I'm still not really sure about n..

Answer 16

Oh sorry my bad I misread it and though it as a sphere. You're better off in cylindrical coordinates.

Answer 17

I don't know if this picture helps, but essntially the downward component is constant while the other one changes direction as you change angle.

Answer 18

I'm just trying to figure out the meaningful math behind that; both the k and j component are constant, I read it as, but I don't know how to translate that into math, right now.

Answer 19

So the z component is constant, while the x and y components are not.

Answer 20

So find the slope of a line perpendicular to the surface of the cone, normalize it, and allow the component in the xy plane to be spread out with sintheta and cos theta.

Answer 21

So it'll look like this I think. I think I might have it pointing in the wrong direction, so add some negative signs in there or whatever to fix it \[\large \bar n = \frac{\sqrt{2}}{2}\cos \theta \ \hat i+\frac{\sqrt{2}}{2}\sin \theta \ \hat j+\frac{\sqrt{2}}{2} \hat k\]

Answer 22

(Lord help me jesus)

Answer 23

This exam is about to be a lot of not-good if I can't figure out a systematic way to find n prior.

Answer 24

Thank you for helping me so far.

Answer 25

You're thinking too had about the integrals when really you should be visualizing the objects and concepts. I mean, I can't really help you there, but I find it too difficult to try to do it how you're thinking. I mean, a cone is truly something independent of a coordinate system. You just have to choose whatever coordinate system has the most symmetry in common with your object of interest.

Answer 26

Yeah, I know that's not what you want to be reading lol sorry.

Answer 27

No problem, I understand that cylindrical coordinate systems would be fantastic for this, I just need to learn how to *SYSTEMATICALLY FIND N* regardless of what that is, lol.

I think I'm going to go sulk in a corner/use the bathroom for like 10 minutes.

Answer 28

You can take the cross product to find the normal vector, and then divide by the magnitude.

Answer 29

BAM

Answer 30

See, that makes sense to me, lol. What two vectors should I be taking the cross product of?

Answer 31

?

Answer 32

hey still stuck on this ?

Answer 33

Focusing purely on conceptuals now. If you want to take a shot, go for it, one hour and a half prior to my exam.

Answer 34

part a or part b ?

Answer 35

il give you the solution may be if it helps

Answer 36

Yeah, that would help, right now I'm trying my best not to do problems but run through a ton of worked problems and observe how they were done.

Answer 37

part b :

\[\vec{v} = \langle yz,-xz,z^2\rangle\]
\[\mathrm{div}(\vec{v}) = 0+0+2z = 2z\]

\[\begin{align}\iint_S \vec{v}\cdot \vec{n}dS &= \iiint_D 2z ~dV\\~\\
&= \int\limits_{0}^{2\pi}\int\limits_0^{\pi/4}\int\limits_{\sec\phi}^{2\sec\phi } 2\rho \cos\phi ~\rho^2\sin\phi d\rho d\phi d\theta\\~\\
&= \int\limits_{0}^{2\pi}\int\limits_0^{\pi/4}\int\limits_{\sec\phi}^{2\sec\phi } ~\rho^3 \sin(2\phi ) d\rho d\phi d\theta\\~\\
&= \frac{15\pi}{2}
\end{align} \]

Answer 38

see if that makes more or less sense
(dont wry about evaluating the integral for now, just see if you're comfortable with the setup)

Answer 39

part a : S1 = curved surface

\[f(x,y,z) = x^2+y^2-z^2\]
\[\hat{n}dS =-\dfrac{\langle 2x,2y,-2z \rangle }{\langle 2x,2y,-2z \rangle \cdot\langle 0,0,1\rangle } = \dfrac{\langle x,y,-z\rangle }{z} ~dxdy\]

\[\begin{align}\iint_S \vec{v}\cdot \vec{n}dS &= \iint_S \langle yz,-xz,z^2\rangle \cdot \dfrac{\langle x,y,-z\rangle }{z} ~dxdy\\~\\
&=\iint_S -z^2 ~dxdy\\~\\
&=-\iint_S x^2+y^2 ~dxdy\\~\\
&=-\int\limits_0^{2\pi}\int\limits_{1}^2 r^2~rdrd\theta \\~\\
&= - \frac{15\pi}{2}
\end{align} \]

Answer 40

part a : S2 = bottom plane ( z = 1)

\[\hat{n}dS = \langle 0,0,-1\rangle ~dxdy\]

\[\begin{align}\iint_S \vec{v}\cdot \vec{n}dS &= \iint_S \langle yz,-xz,z^2\rangle \cdot \langle 0,0,-1\rangle ~dxdy\\~\\
&=\iint_S -z^2 ~dxdy\\~\\
&=-\iint_S 1 ~dxdy\\~\\
&=-\pi
\end{align} \]

Answer 41

part a : S3 = top plane ( z = 2)

\[\hat{n}dS = \langle 0,0,1\rangle ~dxdy\]

\[\begin{align}\iint_S \vec{v}\cdot \vec{n}dS &= \iint_S \langle yz,-xz,z^2\rangle \cdot \langle 0,0,1\rangle ~dxdy\\~\\
&=\iint_S z^2 ~dxdy\\~\\
&=\iint_S 2^2 ~dxdy\\~\\
&=16\pi
\end{align} \]

Answer 42

total flux out of closed surface = S1 + S2 + S3 \[-\frac{15\pi}{2} -\pi + 16\pi = \frac{15\pi}{2}\]


which checks out with the part a so we are good!

Answer 43

Ignore messages in this thread. I'm using OpenStudy's built-in LaTeX editor for the moment for other purposes.

Answer 44

okie.. there is another freindly editor online which i dont like but i see many ppl using it here
http://www.codecogs.com/latex/eqneditor.php

Answer 45

Yeah, I'm aware of that one, I don't like it either, heh. Ended up just waiting until TeX studio and MiKTeX were up and running.