Computing the Flux Across a Surface // Vector Calculus

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In this example we do an example of a surface integral, specifically computing the flux of a vector field across a surface (a parabaloid). While the surface was given explicitly, based on the region (a circle) we first switch to a parameterized of the surface and then use the surface integral formula for parameterized surfaces.
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Опубликовано:

8 дек 2020

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Комментарии : 76

@aafeer2227 8 месяцев назад

If you make a PDF I will buy it, and probably I will not be the only one. This really associates the formal math and the intuition in a striking way. Well designed, brilliant accomplishment.

@ReidCipriani 8 месяцев назад

Took me a while to find an actual problem as opposed to just explaining the formula. Thank you for uploading this.

@ANJA-mj1to 8 месяцев назад

We see vector field from the volumetric flux as well. Now I have indirectly equations for how volumetrix flux of fluid flowing though an unit surface in unite time as well as velocity of the flow. Velocity of flow is now better define like distance between two points by fluid. In the hydrology from the analytical solution we can meassure the saturation and pressure profiles as well as the total volumetric flux and breakthrough time. Thank you for your matematical approach because now it guides how to verify the theory, the analytical solution with compared numerical in another studies.

@nephiduff2856 3 года назад

First! Thanks for posting these videos. Couldn't have come at a better time. I have a calc III final next week!

@DrTrefor 3 года назад

good luck on your final!

@rodionraskolnikov6989 6 месяцев назад

great content!

@BORA0078 3 года назад

Amazing thank you so much

@ahoora46 2 года назад

love it when he gets excited

@mnada72 3 года назад

If we speak about flux I have the tendency to think about time so that quantity/volume can be computed. Is there something of that sort ?

@bredsj 3 года назад

Thanks, you explain so well!

@DrTrefor 3 года назад

Thank you!

@j.o.5957 3 года назад

Nice, I used the implicit formula and got a lil bit worried when you began using the other one, but we both got 3pi/2 so we cool. Question to self: I solved it by first computing nabla_g, then cross product that with p_hat. I didn't have to change its value, just keep it either i_hat, j_hat or k_hat, which answers the question on the previous video: I don't need to compute a crazy variation of i_hat, since nabla_g already has the relevant direction of the vectors. Also, how would I do it using parameterization without look at what you did? Let me try... So essentially: parameterize it, inserting the value of z = (...) into the z_value, call it r(r,theta), then find F(r,theta) which is given. Then it's purely computational. Lastly do we compute that it's r*d(r)*d(theta)? I want to figure that out

@limitless3462 3 года назад

thanks sir

@latifmuhammad8874 10 месяцев назад

I also noticed that Gauss' Divergence thm also works here

@dalisabe62 4 месяца назад

you should explain why the vector field happened to be aligned with the unit normal vector to the surface. This is generally not true with random surface. Is it so aligned due to the geometry of the paraboloid? More justification is needed. One more thing: the copordinates used are cylindrical coordinates, not polar coordinates in this case because the tangential vectors as functions of the parameters are F(r, thita, z). Other than that the presentation is very well done.

@cdawson875 2 года назад

I am confused about this example. When converting to polar coordinates, would we not need to use r*dr*d(theta)?

@yacksterthree3285 2 года назад

I'm certain someone will correct me if I'm wrong, but I'd wager that it's included inherently in the parametrization step. And failing that apply Cunningham's law. That is, rdrd(theta) comes from |r_u x r_v| drdv = d(sigma), that is r = |r_u x r_v|. Thus starting from the more general flux equation Flux = Int Int_S F * n d(sigma), we also find that n = (r_u x r_v)/|r_u x r_v|, thus because we use the normal vector it's canceled out and we are left with (r_u x r_v).

@latifmuhammad8874 10 месяцев назад

It is. The parameters being crossed together functions like taking the Jacobian and,ergo, need not be included

@seyit3552 4 месяца назад

@moulayabdallahcheriflouaza1354 2 года назад

Hello sir If w permute rø *rr the result will be negative What is the difference and how we know wich one first Thanks a lot for your videos Love for Algeria

@anujmishra6834 3 года назад

sir, i compute it with implicit method , with P Vector as k^ , and substituting Z = 1-x^2 -y^2 and got Double integral ( x^2 +y^2 +1 )dA as final expession . now i had set the limits of dx and dy as -1 to 1 . Therefore getting 20/3 as the final answer . why is it diffent ? where had i done the mistake ? pls enlight

@danielsuarez701 2 года назад

your bounds of integration are wrong, you can set either the bounds x or y to be from -1 to 1. Then, assuming the bounds for the x are -1 to 1, the bounds for y are -sqrt(1-x^2) to sqrt(1-x^2).

@momen8839 3 года назад

In physics we define flux ,e.g:electric flux as number of electric field lines .my question is how this definition is right,however we can draw infinite number of vector by control scale?

@DrTrefor 3 года назад

Ya if you prescribed a fixed density for drawing lines, then this becomes a fine definition, but then you can’t touch the slider any more;)

@Kappa647 3 года назад

Why is the outward normal vector the rr x rθ and not the oposite? I cant quite get it !!

@DrTrefor 3 года назад

We don't know in general which is which, the only thing to do is actually plug in values at one point to test whether it aligns with out interpretation of inward or outward

@Kappa647 3 года назад

@@DrTrefor ooooh that makes sense. Thanks !!

@aanandhatamil-4577 Год назад

Can we take both rr×rtheta and rtheta×rr as normal,and the answer is negative,why it is opposite and how could we know which should be chosen

@gumwrapper8826 Год назад

Is there a flux version of stokes theorem out there that could be used to solve this problem? I just completed my Calc III class and was wondering why Div Thm and Green's Thm have both circulation and flux forms but I haven't been able to find any information on if stoke's can be used for flux

@mrjazz2570 2 месяца назад

Stokes cannot be used as there is no curl when finding flux.

@MishaShvartsman Год назад

Great video, maybe dot product is easier to compute in explicit (x,y) coordinates to get the double integral of x^2 + y^2 + 1 over the disk x^2 + y^2 < 1 and then convert to polar. All the videos are great and I am going to steal some of them for my students as well

@ilikefootballandbaddecisions Год назад

What level do you teach?

@MishaShvartsman Год назад

@@ilikefootballandbaddecisions college

@ilikefootballandbaddecisions Год назад

@@MishaShvartsman ok, so I am ahead in the curriculum like 6 years, while it took me doing my normal curriculum, then looking up every single thing I could find for each grade, so I’m in sixth grade now, I did everything I could find a year ago, and then I also did 7th, 8th, 10th, and 11th on my own, then I skipped to calculus ( A calc 1 type class is what you get to do for 12th in my school if you stay at the top of the chart for all of school.) I just need to find motivation to do the 9th grade geometry.

@MishaShvartsman Год назад

@@ilikefootballandbaddecisions sounds like you got pretty far already, you may find 9th grade geometry somewhat trivial at your level. I have not taught high school so, I am not sure what the best source for it is. But in terms of problems in geometry the best and toughest problems in geometry come from the math magazine Kvant that are available online (look up Serge Tabachnikov Kvant Selecta)

@samuelsaldivar4903 Год назад

goat

@brendonchen4965 2 года назад

Hi Dr.Trefor, At 3:29 you say that r^2 = x^2 + y^2 so we can parameterize z component (1-x^2-y^2) as 1-r^2. however x^2 - y^2 is not x^2 + y^2; so how can we make this substitution?

@DrTrefor 2 года назад

It is -x^2, a negative in front of both terms:)

@sureshpatel-sg2xw 3 года назад

sir i'm unable to understand that whether take X or X...(i used R to represent position vector and< . > to represent vector).. and thank you sir your video series helps a lot..!!!

@DrTrefor 3 года назад

It depends on the choice of normals of the surface, you choose the order so it aligns.

@sureshpatel-sg2xw 3 года назад

@@DrTrefor thank u sir ..

@aanandhatamil-4577 Год назад

@@DrTrefor how to choose whether we have to use rr×rtheta or rtheta×rr?

@chandankar5032 3 года назад

From where i can find such question for exercise ,i mean could you recommend any book ?

@vaiterius 3 года назад

James Stewart Calculus 8th edition, found on chapter 16

@Yeheyeijwveusjw 2 года назад

Yeah

@shoshoshapan4982 3 года назад

can i use r(phi,theta) where r of a shpere is constant instead of r(r,theta) as a parametic?!

@DrTrefor 3 года назад

Doesn't work out nice as the surface is a parabaloid not a sphere

@shoshoshapan4982 3 года назад

@@DrTrefor okay thanks ^-^

@latifmuhammad8874 10 месяцев назад

Polar is much much easier

@iDor2 5 месяцев назад

Does anybody know why the bound for r should be 0 to 1, because I would interpret that as a cilinder with radius 1? Also, do you not have to add the r to the jacobian when you already considered the jacobian matrix?

@jeroeshak 4 месяца назад

If you are working with the parametric way you're not supposed to use jacobian R goes from 0-1 cuz you should compute the R of the volume's shadow "its like the biggest circle you can have"

@Pindakaasruiktgoor 6 месяцев назад

At 5:09 shouldn't the J-hat come out to a negative 2 r² sin (theta) ? Where did the fourth minus come from? Anyway god bless you for everything you do, I'm not only understanding vector calculus for the first time thanks to you but also having fun doing it!

@Pindakaasruiktgoor 6 месяцев назад

My bad, I used the wrong algebraic formula here, I didn't notice the typo in the video where you explained cross-products at first. All good now! Thank you for all you do!

@riveratsu2691 7 месяцев назад

Could you use divergence theorem on this and get the same answer?

@cjjk9142 6 месяцев назад

no but you can alter it so it becomes closed but its easier to just do it as done by op

@ziyancheng8122 Год назад

Actually if z is a function with respect to both x and y, we could put the partial derivatives -gx and -gy into the x and y components of the normal vector respectively with z component equal to 1(in condition of positive z direction). Therefore we could skip the annoying cross product calculation process. 😆

@latifmuhammad8874 10 месяцев назад

Yep

@wanmuhammadtaufiqabdullaha7744 3 года назад

Dr, for the value of n. It magnitued of ru x rv right ???

@DrTrefor 3 года назад

n is a vector so no it is just the cross product, not it’s magnitude

@wanmuhammadtaufiqabdullaha7744 3 года назад

@@DrTrefor dr, can i have ur phone nmber. I want to ask so many question about this topic. I am pure math student. Can dr ???

@KK-rg3nj 3 года назад

@@wanmuhammadtaufiqabdullaha7744 are you serious?

@aanandhatamil-4577 Год назад

@@DrTrefor then when we take rv×ru instead of ru×rv ,we get opposite value,how could we know which is correct to choose

@woodychelton5590 8 месяцев назад

this approach didn't work when i had F(x,y,z) = 3z k . Instead, online sources said I had to use the divergence theorem. Why does this approach not work for that problem?

@cjjk9142 6 месяцев назад

theres a hole in the bottom, its not a closed surface hence cant use Div thm

@eldi 3 года назад

Why didn't you include an r in r dr dtheta ?

@DrTrefor 3 года назад

That would be if I converted from Cartesian to polar. I didn’t do that. I parameterized it in polar from the beginning so no conversion and thus no r

@eldi 3 года назад

@@DrTrefor So how would it be different if you parametrized it afterwards? It seems for me like you were given a function and a vectorfield, and then parametrized it after it was given. My intuition would have said include r in dr dtheta. I'm just trying to understand:)

@eldi 3 года назад

@@DrTrefor Oh, so the difference is that if you didn't use polar from the start, you instead just kept x, y and z, and then when attempting to perform the integral - you then changed the variables to polar - then you would have to incluede an r?

@eldi 3 года назад

@@DrTrefor I believe I figured it out! Took a while and had to ask my teacher as well, but I attempted to compute this flux without parametrizing (just keeping my x, y and z) and then included the r, and just as you said, I got the same integral :))

@danteconciatori1094 3 года назад

no jercobian?

@stevenchen6752 Год назад

how come he didnt add the r to dr? i thought whenever using dr you have to add r

@richardvalentinonainggolan328 Год назад

if i don't get it wrong, you don't add r when you use it for find surface area of the parameterized surface. but, indeed, when we integrate anything from coordinate to polar, we need to change dA to rdrdtheta

@stevenchen6752 Год назад

@@richardvalentinonainggolan328 but this is for flux no?

@richardvalentinonainggolan328 Год назад

@@stevenchen6752 that is for parameterized surface and to find flux across the surface or sum of the curl in the surface, you need to integrate differential to area surface (2 option: parameterized or implicit) in your formula. flux across the surface= double integral of F(dot)n dS sum of the curl in the surface or circulation on its boundary (stokes theorem)= double integral of del(cross)F(dot)n dS both of them using dS (differential of surface area). sorry for my bad english :(

@cjjk9142 6 месяцев назад

year late unfortunately but if you use the rxr then you dont need the jacobian part