Flow Integrals and Circulation // Big Idea, Formula & Examples // Vector Calculus

[@tomscholten4819]

At 4:33, the last integral has boundaries a to b, but shouldnt that be C. Atleast accoording to the book it is C

[@DrTrefor]

Quite right, thank you!

[@tomscholten4819]

@@DrTrefor does C contain the x, y and z coordinates of the begin and end point? Great videos btw, they really help me studying😄

[@soccerbels7947]

@@tomscholten4819 c is just the boundary, it will eventually be a to b

[@soccerbels7947]

Boundaries *

[@hubenbu]

awesome!

[@ishaantrivedi9968]

this man's a legend.

[@copernicus6420]

I agree

[@adamsusel3801]

This man saving my life right on time. Doing vector calc in calc 3 right now. Could not have been better timed to post!

[@DrTrefor]

Nice! Good luck with calc 3:)

[@peterfriedman4912]

I'm not sure why Mr. Bazett's videos don't have more likes. All these (Multivariable Calculus) videos are fantastic! Thank you so much!

[@CaptchaSamurai]

The joy of feeling vector calculus is immense. Thank you for this concise, amazing series! ε>

[@kemalkayraergin5655]

thank you for making things simpler and putting it in some order to study us easily, appriciated it!

[@juncarlin2865]

I have to admit that this video is the best one I've ever watched about flow integrals.

[@qamarmoavia4031]

Thanks a lot :) . Dr. Trefor you always change implicit into explicit . LOVE YOU

[@fuzzyllamaa]

Best math videos on youtube !!!

[@leadtoexemplify]

I like the energy and the passion with which you are explaining them. thanks

[@abstractnonsense3253]

This is great content! If you continue like this you'll have all the essential university math on your channel in about a year.

[@DrTrefor]

That's the plan!

[@abstractnonsense3253]

@@DrTrefor Great! I've recommended your videos to a couple of student of mine. So you plan on having probability, statistics, complex analysis and expanding the differential equations?

[@xandiczr12444]

Those videos are just priceless! Thanks for the time!

[@DrTrefor]

Glad you like them!

[@harshaniperera6855]

Thank you sir !I am following your videos since 2 years .Now I am studying physics as my major subject. Still helpful to fulfil the foundation of every theory in physics and mathematics . Live long with good health!!!!!

[@cosmefulanito6386]

Greetings from Argentina!!! keep on the good work, thank you

[@mbelekamohelo3159]

We appreciate the visualizations, they jus blend perfect with the concept.

[@leadtoexemplify]

God and Knowledge is always with those selfless individuals who spread knowledge. God bless you

[@continnum_radhe-radhe]

Thank you so much sir for making these wonderful awesome videos 🙏🙏🙏

[@AdityaRaj-lj5wf]

what an exceptional explanation.

[@mania-jd3my]

the 10 people who disliked this video need to be arrested.

[@AnilSharma-qf5lh]

Great video as usual..

[@DrTrefor]

Thank you!

[@hectorgalva7495]

Awesome doctor, I like a lot your videos, your are very intuitive and I can understand better the wonderful of math.

[@julioreyram]

Nice examples. As always, good concept discussion and illustration. Thanks! :)

[@DJ-yj1vg]

Great series. 👍

[@federicovelasco4561]

Thanks for the video. Please review at 7:17, I think the equation should be corrected by M(dy) + N(dx). I think the order of the differentials is reversed.

[@momen8839]

this is the best channel for math .. I have some questions * Is ds =|dr| ? & ds=|dr/dt|*dt=|V|*dt & dr=dr/dt=V*dt ?

[@DrTrefor]

The main formula is that dr/ds=T, the unit tangent vector, while dr/dt=v the velocity vector

[@momen8839]

@@DrTrefor so dr/ds =vector on its magnitude so ds=|dr|

[@roshanpradeep8532]

6:03 the flow is negative 2pi, if I take r = sin(t) i + cos(t) j

[@Alannnn14]

Excellent 👌👍.

[@umlindeni9750]

Once again 😎💪

[@hubenbu]

First time to see the integral finger with a ring. haha Calculus: An Intuitive and Physical Approach by Morris Kline appears to be a great read.

[@continnum_radhe-radhe]

Sir ,please make videos on gamma and beta function .... 🙏🙏🙏

[@MsAlarman]

Excellent

[@eee_bangla]

nice explanation

[@abdelstar3999]

really love you, thak you so much. very useful

[@DrTrefor]

You are most welcome!

[@autonomesinklusionsreferat1251]

Wonderful presentation

[@DrTrefor]

Thanks a lot!

[@ileanadominguez6055]

Thank you very much :)

[@alevelsos]

7:53 shouldn't the derivative for M (-sint) be cos t?? Why is -sint written again?

[@alevelsos]

I think i actually got it, for the next time im here: M and N come from the equation F(x,y) refering to r(t) for example at 7:53 F(x,y) = -yi+ xj and the parametrized equation r(t) = cos(t)i + sin(t) so M=-yi = -sin(t) and N= +xj = cos(t) the dx comes from the i component in the parametrised equation. So dx = (cos(t))' = (-sin(t)) and the dy comes from the j component in the parameterized equation. So dy = (sin(t))' = cos(t)

[@MrJdcirbo]

Anyone else notice that the integral of that simple closed curve through -yi+xj was twice the area of the circle? Remember that for later... 😉

[@chandhiranphd]

Nice 😇

[@남양홍씨35대손]

awesome

[@Festus2022]

Great video. It answered some critical questions for me. What are the units of flow/circulation?

[@carultch]

Depends on what the vector field represents, and what the units of its coordinate directions are (usually distance units, but not necessarily). The most common introductory example you see as an application of this, is work, where the field has force units, and the input space is position in actual space, with distance units. In SI units, this would mean the units are a product of Newtons and meters, and since the result is a scalar, the units would be Joules. You could have a more abstract idea of space, for the space of the vector field. Such as phase space. There are applications of integrals like this in control system theory, where the spatial units of the field being integrated, have nothing to distance units like meters.

[@Festus2022]

Thanks for the reply!@@carultch

[@michaelfoster7038]

Champion

[@zsoca31]

Why is this any different than line integrals? Thank you!

[@carultch]

It's a special case of a line integral, with a closed path given, instead of just any path.

[@momen8839]

Nice .what is book that you explain from it?

[@DrTrefor]

I mainly use Thomas' Calculus, but this is all pretty standard fare

[@dominicellis1867]

Ok what is the difference between circulation and curl?

[@DrTrefor]

Circulation is a global property, a property that happens along a whole curve. Curl is something that exists at every point in a vector field.

[@dominicellis1867]

so the curl exists in the absence of a specific curve because it is of a field while the circulation only has meaning in tandem with a curve in a field? what would it physically mean for a field to have no curl but a positive circulation along a curve or to have a positive curl but no circulation along a curve?

[@carultch]

@@dominicellis1867 It just means the vector field has no curl locally, but has a circulation globally around that particular path. With special cases involving singularities, it's possible to have a net circulation, with zero curl "everywhere". One such example is . This function has zero curl everywhere except at the origin, where the function has a singularity and infinite curl. The result of this singularity, causes all closed loop integrals enclosing it, to have a net line integral of 2*pi in the CCW direction.

[@juanpablosm_767]

Is it possible to spot the curve (obtain the equation that describes C) by using the value of the line integral if the vector field is known? (Not necessarily a conservative field) For example: Trying to obtain which curve (or curves) uses a defined amount of energy from point A to B if moving through a fluid at constant speed (thus constant Friction force magnitude)

[@carultch]

Unfortunately no. There are an infinite number of paths that could have the same line integral. If you can lock-down the shape of the path in advance, so that there is simply one parameter of the shape to solve for, it is possible to generalize it, so that its line integral is equal to a desired value. Example: Consider the vector field . Suppose we are interested in a circular closed path, whose line integral equals 18*pi. We can use Green's theorem for a circle of an unknown radius R. Start by calculating the curl: curl = 2 Green's theorem tells us that the closed loop line integral, equals the area integral of the curl. Since the curl is constant, this integral reduces to simple multiplication. So this means the line integral around a CCW path, equals 2*pi*R^2. Set this result equal to 18*pi, and solve for R. 2*pi*R^2 = 18*pi R^2 = 9 R = 3 So this tells us that we'd like a circle of radius 3, to be the path to get a closed line integral of 18*pi

[@juanpablosm_767]

@@carultch Thanks. So, basically it is impossible without having somewhat of an imaginary sketch of the path or "environmental" constrictions? (As you said, e.g: it having to take a certain shape or not passing through certain surfaces or regions )

[@carultch]

@@juanpablosm_767Exactly. You need some way to lock-down the shape of the path, to just one degree of freedom, in order to solve for the specific path with the line integral equal to a value of interest. Just as there are numerous paths you could take to the top of a hill. No matter what path you take, the net work you do against gravity is still the same, since gravity is a conservative field, and work done against it or by it, only depends on initial and final locations. For my example of the non-conservative field with uniform curl, any closed shape that has an area of 9*pi, is a valid path that has a line integral that equals 18*pi. For instance, a square with side lengths of 3*sqrt(pi) would also have the same line integral. Or an ellipse with a semiminor axis of 1, and a semimajor axis of 9.

[@ΚωνσταντίνοςΛαζαρίδης-ξ9ι]

Thanks!

[@hectorgalva7495]

Hi Doctor, I have a question. I've seen in a lot of books some double, triple and multiple integrals (four, five,... and some on) that have in the middle of their symbols, a little circle like you show in this video. However, I mean, what does it represent when a double or triple integral has a circle in its symbols? I wait anxiously by your answer, greetings from Dominican Republic. 😊😊😊😊😊

[@DrTrefor]

That just indicated it is over a closed object. So a single integral with a circle is a line integral along a closed loop. Double would be closed surface.

[@hectorgalva7495]

@@DrTrefor WOW!! 😮😮It keep the same ideas then. Thank a lot by your answer doctor. 😊😊

[@yenmanagandlaaishwarya3220]

@Dr. Trefor Bazett After calculating the flow integral,we will be able to get +ve as well as -ve .so from the flow integral can we conclude whether it is circulating clockwise or anticlockwise? If yes,please tell how can we conclude?

[@carultch]

Yes. Positive means it circulates anticlockwise, while negative indicates it circulates clockwise.

[@duckymomo7935]

Flow integrals? Curl?

[@DrTrefor]

Curl up next!

[@tanish6035]

Sir, if field is force field then will f.tds along a close path or curve give work?? Plz reply..🙏🙏

[@carultch]

Yes. An application of this concept, is work. The full definition of work, is a line integral of a force field from original position to the destination. In the special case of a straight line path and a uniform force field, the integral reduces to a dot product. In the special case of a straight line path aligned with the uniform force field, the dot product reduces to just a product of magnitudes.

[@kiyoshimiyagawa1728]

It seems like the second vector field is just the cross product of the first one because 2D cross-products are unary operators and can take only one (2D) vector. But with the sign of minus

[@carultch]

2-D cross products aren't unary. They take in two vectors, and produce a third vector that is perpendicular to both original vectors. Given 2D vectors in only the x-y plane, this means the cross product is perpendicular to the plane of the vectors, and is exclusively in the z-direction. A 2-D cross product is a scalar, rather than a vector, if a 3rd dimension isn't applicable in the space you are considering. Torque and rotational quantities that are defined with a cross product would still exist in Flatland, but you'd describe their directions as either positive rotation or negative rotation, and wouldn't need to specify anything more than just a sign.

[@kiyoshimiyagawa1728]

@@carultch By 2D-"cross product" I meant a function that takes a vector on 2D space and returns an orthogonal vector. On 2D-space, there are only two orthogonal vectors to a vector of the same size, and such a function takes the one whose tip is closer to the tip of the original vector if we go counterclockwise. The 3D-cross product is the usual cross-product. But in this case there are infinitely many directions of orthogonal vectors of the same size, so we should specify another vector to choose the orthogonal one to both of them. In this case, its size is not of the size of a particular of two vectors but the area between them. Then another choice of two vectors appears. In this case, we apply the right-hand rule. For 4D-"cross products" we have to specify three vectors to find the orthogonal one to all of them and so on. We can generalize it by the determinant of N×N matrix, where N is the number of dimensions of the space where the vectors exist, the first row is the basis vectors, and the other rows are the argument vectors

[@playitback-os7mh]

Top!

[@gregorsamsa9762]

So how come the work is zero if I need to paddle a lot to keep myself on the curve?

[@carultch]

Because his explanation is an oversimplification, that doesn't account for friction, or also the effort to constrain yourself to the path. Assume friction is negligible, and assume an ideal frictionless track keeps you on the path.

[@KM-om1hm]

Anyone please answer, is it vector line integral?

[@carultch]

Yes. It's a special case of a vector line integral around a closed, non-intersecting path.

[@peanutpotato2394]

Where did u get M and N from? It’s so unclear

[@carultch]

M and N are the component functions of the vector field. Trefor uses M, N, and P, to refer to the component functions of the 3-D vector field, deliberately avoiding O for obvious reasons. Jimmy Stewart's Calculus book that I have, uses P, Q, and R for this purpose.

[@TheFpsPlayer01]

gj doctor

[@larsnielsen1606]

The units (dimensional analysis) of the line integral when F is a force field (e.g., units of Newtons or kg•m/s/s) and when T•ds has units of distance (e.g., m) is N•m (kg•m^2/s/s)...making T itself dimensionless, I suppose. N•m and Joules (J) are the same thing and these are units of work. Adding up the bits (each with dimensions N•m) while taking a limit gives a "sum" that has the same units...great. That makes sense to me. It also makes sense using the chain rule to replace T•ds with dr/dt • dt (a vector with units of distance per time, such as m/s, multiplied by a time increment, with units s); N•m/s •s gives N•m. Great. However, I am struggling with the dimensional analysis of the line integral in this video, which you are calling "flow" or "circulation." I am accustomed to flow having units of, for example, gallons per second (along a path) or maybe gallons per square meter per second (volume going through a cross-section such as a pipe. You say that here F is not a "force" field but a "velocity" field. So, the units for F are m/s. Again, the units for dr/dt are m/s; the units for dt is s; therefore, the overall units for "flow" are square-meters per second. I am not familiar with these units as a "flow"; I don't see it analogous to volume per second (m^3 per second) because this is (m^2 per second). What am I missing?

[@DrTrefor]

Velocity times distance, so m^2/s in the plane. WHat you are imagining is more like "flux", it is the amount of stuff or volume that crosses a threshold. We aren't measuring that here. We are measuring the degree to which your path is aligned with the underlying velocity field.

[@larsnielsen1606]

@@DrTrefor Thank you for the reply. The degree to which something is similar to something else I usually think of in unitless terms, such as 98% (with no units). I see from Wikipedia that the dimensions for circulation are indeed L^2 T^-1, for which I clearly haven't yet developed much physical intuition. Of course, I get that dot products are greatest the smaller the angle between the vectors (regardless of their respective dimensions). =) Perhaps my misunderstanding is somewhere in the difference between the projection of V onto ds and the dot product V • ds itself (the projection of V onto ds times ds). Summing together infinitesimal projections of V onto ds on a closed loop would at least give units of velocity, the same as the field. I guess my concern is that if I get an answer like +25 m^2 s^-1, I can interpret the sign of the answer easily enough, but intuition regarding the magnitude of the scalar is harder for me. I'm still thinking on it, just wanted to say, thank you for the reply.

[@justsaadunoyeah1234]

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[@justsaadunoyeah1234]

LaTeX

[@justsaadunoyeah1234]

Future me don't roast mi for writing these stupid comments in this reply section