Question

Divergence Theorem
see attachment

Answer 1

hold up let me try to attempt #16...but geez when I take the partial derivatives, I get screwed over on the j term

Answer 2

@Euler271 @ChristopherToni

Answer 3

are you taking grad(F)?

Answer 4

the gradient of*

Answer 5

wait wait... I'm using divergence Theorem...I'll get an attachment on sonosfnosdfh

Answer 6

what's the entire formula?

Answer 7

k hold on

Answer 8

that front page should have the divergence

Answer 9

oh crud I have to find the outward flux using the divergence theorem

Answer 10

across the boundary of region d

Answer 11

this is just an example

Answer 12

but what's strange is that for the previous problems I didn't have to use the n part...you just have to find whatever the equation is with the divergence theorem and then afterwards you integrate according to what the D is saying

Answer 13

@ChristopherToni

Answer 14

you can also do it the normal way I think. Like you have it.

\[\int\limits_{0}^{2\pi}\int\limits_{1}^{\sqrt2}\int\limits_{2}^{-1} \nabla F\ rdzdrd \theta\]

Answer 15

but the problem is that I have a z lurking in its a**. How do I get rid of it?

Answer 16

damn z. if that was gone it would make my life easier

Answer 17

O_O
so I can start to do antiderivatives all the way in terms of z rdr and dtheta?

Answer 18

i'm pretty sure you can. do you have the solution?

Answer 19

not yet this is what I have so far..the r in the denominator... might cause problems or not

Answer 20

you know what, im really not sure anymore. i would forget what i said about being able to mix polar and z.

try doing it the lame way with x, y and z

but you might just need to use the surface way.

Answer 21

there's just...blah I got stuck in the middle of all this...the only way I could get through is accidentally getting the partial derivative of y wrong for the j term.

Answer 22

So, for starters, if \(\large F= \left\langle \ln(x^2+y^2),\dfrac{2z}{x}\arctan\dfrac{y}{x}, z\sqrt{x^2+y^2}\right\rangle\) we have that
\[\large \begin{aligned}\mathrm{div}\,F &= \nabla\cdot F \\ &= \frac{2x}{x^2+y^2}+\frac{2z}{x}\frac{x^2}{x^2+y^2}\cdot \frac{1}{x} + \sqrt{x^2+y^2}\\ &= \frac{2x+2z}{x^2+y^2}+\sqrt{x^2+y^2} \end{aligned}\]
In cylindrical coordinates, this would simplify to \(\large \dfrac{2\cos\theta}{r} +\dfrac{2z}{r^2}+r\).

Also, in cylindrical coordinates, \(\large 1\leq x^2+y^2\leq 2\implies 1\leq r\leq \sqrt{2}\text{ and } 0\leq \theta\leq 2\pi\) (since there was no restriction on which octant we're working in), and \(\large -1\leq z\leq 2\).

Therefore, the triple integral you should get in cylindrical coordinates is
\[\large \iiint\limits_G\mathrm{div}\,F\,dV = \int_0^{2\pi}\int_1^{\sqrt{2}}\int_{-1}^2 2\cos\theta+\frac{2z}{r}+r^2\,dz\,dr\,d\theta\]
I assume you can take things from here?

Answer 23

ok let me try getting this on paint and come back

Answer 24

hey! Wouldn't there be a dzrdrdtheta?

Answer 25

yes but he included the r already. its fully simplified and ready to integrate lol

Answer 26

Yea, I got a little lazy at the end and simplified it all the way through...lol

Answer 27

fjdsksalsdajdskl crap I added an r one sec

Answer 28

ok I got two different answers thanks to the slip up ROFL

Answer 29

have no idea which is right the black or the blue

Answer 30

o-o

Answer 31

ughhhh there's an error oh mannn!

Answer 32

You're missing some numbers throughout the calculation. XD

We have that
\[\large \begin{aligned} &\phantom{=}\int_0^{2\pi}\int_1^{\sqrt{2}}\int_{-1}^2 2\cos\theta+\frac{2z}{r}+r^2\,dz\,dr\,d\theta \\ &= \int_0^{2\pi}\int_1^{\sqrt{2}}\left.\left[2z\cos\theta + \frac{z^2}{r}+r^2z\right]\right|_{-1}^2\,dr\,d\theta\\ &= \int_0^{2\pi}\int_1^{\sqrt{2}}\left[\left( 4\cos\theta+\frac{4}{r}+2r^2\right) -\left(-2\cos\theta + \frac{1}{r} - r^2\right)\right]\,dr\,d\theta\\ &= \int_0^{2\pi}\int_1^{\sqrt{2}} 6\cos\theta + \frac{3}{r}+3r^2\,dr\,d\theta \\ &= \int_0^{2\pi}\left.\left[6r\cos\theta +3\ln r + r^3\right]\right|_1^{\sqrt{2}}\,d\theta\\ &= \int_0^{2\pi}\left[\left( 6\sqrt{2}\cos\theta +3\ln\sqrt{2} +(\sqrt{2})^3\right)-\left(6\cos\theta +3\ln 1 + 1\right)\right]\,d\theta\\ &= \int_0^{2\pi} (6\sqrt{2}-6)\cos\theta +3\ln\sqrt{2}+2\sqrt{2}-1 \,d\theta \\ &= \left.\left[(6\sqrt{2}-6)\sin\theta +(3\ln\sqrt{2}+2\sqrt{2}-1)\theta \right]\right|_0^{2\pi}\\ &= 2\pi(3\ln\sqrt{2} +2\sqrt{2} -1)\\ &= \pi(6\ln\sqrt{2}+4\sqrt{2}-2)\\ &= \pi (\ln 8 +4\sqrt{2}-2)\end{aligned}\]

Answer 33

i get the same result. i derived 3/r instead of integrating it

Answer 34

I might have finally got it, but the damn distribute the whole thing with r first threw my a** off

Answer 35

because I'm used to doing dzrdrdtheta

Answer 36

ok so all I need is just one more and that's #22 with this super box thing and I'm done

Answer 37

Ah, I was taught to do it the other way: \(r\,dz\,dr\,d\theta\) and then simplify everything first.. XD

Answer 38

not in my class hell no... lol

Answer 39

#22 only and that's it YAY!

Answer 40

Ugh...I loathe flux... >_>

Answer 41

For 22, though, note that \(\nabla\cdot F = 1-2+1 = 0\), but I'm not sure what to do from here. XD

Answer 42

damn me neither

Answer 43

maybe that's it? lol

Answer 44

well we could dydx it? since y =0 y = 1 x = 0 x =1