Question

Determine the flux

Answer 1

This is the formula but I think u may have to switch the coordinates and i am not sure how

Answer 2

well the shape is a cone, so I think you need to parameterize it in cylindrical coordinates
gotta review parameterization of surfaces though

Answer 3

... a cone that opens along the y-axis

Answer 4

ohh so a horizontal cone?

Answer 5

oh my bad, it's an elliptic parabaloid

Answer 6

that still opens along y

Answer 7

so which corrdinates wld i use ?

Answer 8

I'm thinking cylindrical, but I'm not positive...
\[x^2+z^2=r^2\]\[y=y\] but I forgot how to parameterize the surface..

Answer 9

hah i am forgetting everything tooo or maybe i never knew how to do it in the first place

Answer 10

I think we can just say\[y=x^2+z^2\]means the surface is parameterized by\[x\hat i+(x^2+y^2)\hat j+z\hat k\]

Answer 11

perhaps we have to leave this cartesian after all? hm...
typo above
\[\vec r(x,z)=x\hat i+(x^2+z^2)\hat j+z\hat k\]

Answer 12

ohhhhhh someone emailed me their solution, ummm maybe its correct. Wait ill attach it

Answer 13

I think we don't have to change the coordinates, though it may help

Answer 14

its number 10 its the 3rd last page

Answer 15

pg #12

Answer 16

I'm pretty sure they at least got the bounds of r wrong
\[\frac12\le r\le1\]don't you think?

Answer 17

Ya lol like he told me he is getting around a C in this course so like in that case i doubt his answers r correct

Answer 18

it opens along the y-axis, so I don't think you can say x=rcost, y=rsint
I think if you want to do cylindrical coordinates you need it as
\[\vec r(x,z)=x\hat i+(x^2+z^2)\hat j+z\hat k\]so maybe\[x=r\cos\theta\]\[z=r\sin\theta\]

Answer 19

the region below the shape is a circle in the xz-plane

Answer 20

so the bounds on this guy are\[\frac12\le r\le 1\]\[0\le\theta\le2\pi\]

Answer 21

brb

Answer 22

K thanks That was amazing seriously

Answer 23

so first get the surface in the form g(x,y,z), i.e.
\[g(x,y,z)=x^2+z^2-y\]from here I do this differntly that you I think...\[\nabla g=\langle2x,-1,2z\rangle\] the integrand will be the dot product of this and the vector function
after that convert to polar coordinates to integrate over the region D

Answer 24

hahah yes it is but ill figure it out. thanks

Answer 25

they are the same because \[\nabla g\]points in the same direction as \(\vec r_x\times\vec r_z\)...
hey guess where I'm gonna refer you :)

Answer 26

lol paulas online notes hahahha

Answer 27

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx
surprise, surprise
actually example 1 is almost identical to your problem with a different vector function, so it should be helpful

Answer 28

k thanks thats helpful

Answer 29

idk like i am just feeling overwhelmed and then i get like a brain block

Answer 30

calc III does that
I keep having to relearn these things too, as you can see :P

Answer 31

Thanks for helping me I have been stuck on this question for a few hrs hahahahah since last nite

Answer 32

well I hope you get the right answer; I'm pretty sure the setup is correct
I should really be learning python right now, so I gotta bounce
you're welcome, as always :)

Answer 33

@TuringTest, I think the shape is a paraboloid but all your other steps are correct...
so since its oriented away from the y axis, it will be oriented to outside of the paraboloid.... , so we make our n(unit normal vector) positive.
now we know that to get the flux, given a vector field and an equation of the surface, we evaluate
\[\int\limits_{}^{}\int\limits_{}^{}-Pg_{x}+Q-Rg_{z}dA\]
the one under the double integral is D, or the region of the paraboloid we will evaluate I cannot input it, so I left it like that
where F=<P,Q,R>=<x+z,1,z>, y=g(x,y)=x^2+z^2 So
-Pg_x=-2x^2-2xz
Q=1
-Rg_z=-2z^2
now we only need to find our region...

now in this case we might wanna use polar coordinates as Turing excellently suggested, where x=rcos theta z=rsin theta
so here
1/4<=r<=1
0<=theta<=2 pi
so we'll have
\[\int\limits_{0}^{2pi}\int\limits_{1/4}^{1}(-2x^2-2xz+1-2z^2)rdrd theta\]
\[\int\limits_{0}^{2pi}\int\limits_{1/4}^{1}(-2r^2\cos^2 \theta-2r^2\cos \theta \sin \theta+1-2r^2\sin^2 \theta)rdrd \theta\]
\[\int\limits\limits_{0}^{2pi}\int\limits\limits_{1/4}^{1}(\cos^2 \theta+\sin^2 \theta)(-2r^3)-r^3(2\cos \theta \sin \theta)+r drd \theta\]
\[\int\limits\limits_{0}^{2pi}\int\limits\limits_{1/4}^{1}-2r^3-r^3\sin2 \theta+r drd \theta\]
\[\int\limits\limits_{0}^{2pi}\int\limits\limits_{1/4}^{1}-r^3(2+\sin2 \theta)+r drd \theta\]
can you do it at this point :D