Brother, you are not getting enough views for your videos,but that doesn't change the fact that you are one of the top teacher/explainer/TH-camr on earth. Thanks for video
SPOILER ALERT for last example: Integral of 1+2y dy is y+y², evaluate from x² to 1 we get 2 - x² - x⁴ dx, integrate we get 2x - x³/3 - x⁵/5, evaluate from 0 to 1 we get 2 - 1/3 - 1/5, arithmetic and voila 22/15
@@philkirshenbaum8867 plot the graph and you will see the lower curve is y=x^2 and upper y=1(as mentioned in the question C moves from 1,1 to 0,1 along a line and the line is y=1)
Your videos are beyond impressive. Not only do I love your concise explanations, but also your graphics. One of the best mathematics channels on youtube!
Holy crap, you managed to explain this in 6 min and I was able to understand it, I was a bit confused after an hour and a half lecture, but this just made sense. It clicked.
Positive orientation is the reason a hollow circle inside a solid circle, that you are calculating, has the lines going clockwise. You are not calculating the hollow area, but the solid and it is on the left as you walk around it. Very easy concept that some people have issues understanding.
Its amazing how easily you explain this stuff. You explain exactly whats needed to be explained in such a concise manner. Perfect for people with ADHD lol
Your video is really good and I do appreciate the time and effort you put into it but I think you should mention that all points inside the D area must be differentiable and that depends on P Q as well as R for stokes theorem. I know that it sound obvious but if 1 or some points are not differentiable then we have to seek for approach to solve the problem.
Watching this on the train ride to my calc 3 final as a refresher, thanks for the easy to follow video that taught me more than my professor has the entire semester!
If you consider the vector field to represent a force, then the line integral from a to b across a curve C (aka the circulation) can be thought of as the work needed to move a particle subjected to said force from a to b.
It depends on what the vector field represents. Suppose the vector field is a force field, which is the go to application that motivates the concept of vector line integrals. The vector field would be force at any given point in space, and have the units of Newtons. The change in radius vector, dr, would have the units of meters. Upon integrating them together, you'd get Joules of work done, as a body moves along the given path, while subject to that force field.
So the technique to know the lower limit and the upper limit is to know the direction of the arrow. The starting point is the lower limit and it can be a function or a point while the upper limit is the intersection or the coordinate or so that's how I see it. Awesome!
I get that for the last example you should set the bounds of x as 0 to 1 and the bound of y as x^2 to 1, but why do we set up the bound of x as 0 to 1? When I graphed it, the left bound is zero but the right bound is the line graph y=x^2. Please help me
I have a question, the area of square in example at 4:10 shouldn't be 1? If we use equation for the area of square namely a^2 we should get 1 because length of the side is 1, but using green's theorem we got 1/2, someone can tell me why?
You are not calculating the area inside the square, since this is not just a normal integral. Here the line integral shows how much of the vector field that is given is in the direction of the curve, which in this case is the square. You can tell that it's not the area of it by using a different vector field as well, which will yield different results.
@@ProfessorDaveExplains Thanks for the response, your video on double integrals helped so much and clarified it all for me, one of the best maths videos I have ever seen so thanks for your work!
I have a quick question: While I understand the outcome of Line integral is Area, what is the outcome of the circular integral (conservative) ? Since it equals the double integral, Is the RHS is Volume? If that is the case, shall I assume that one special case of line integral which involves conservative Field on a closed curve over a surface yields volume? Please answer
Because you aren't just calculating the area of that square. You are multiplying each area element, with the curl of a vector field, across that square. Just like if you are given the density of a sheet of a square meter of plywood as 7 kg/m^2, and you calculate it for a piece that is 1 meter by 1 meter. When you calculate the area, it is 1 square meter. But when finding the mass, you shouldn't expect to get 1 kg. You should expect to get 7 kg.
If it did, we'd have no reason to use Green's theorem. If dQ/dx equaled dP/dy, then the vector field is conservative, and the closed loop line integral is trivial, because line integrals on conservative vector fields are independent of path, and closed path integrals of conservative fields are zero.
I got 8/15 for the answer every time for the checking comprehension. I can't figure out what I am doing wrong here. If you are still looking at these video's comments, could you please help me? So I started by setting up the Pyx and Qxy in the double integral. So I got 1+2y. Now I get my bounds as from 0 to x^2. And for x I get from 0 to 1. So that becomes: y+y^2 then from 0 to x^2 I get x^2 + x^4. Then integrating again I get (x^3)/3 + (x^5)/5. Take that from 0 to 1: 1/3 + 1/5. That gets 5/15 + 3/15 = 8/15.
Your bounds of integration for y should be from x^2 to 1 rather than 0 to x^2. If you draw a picture it is easier to visualize why. Anyways, the function stops at y=1 and it begins at the equation y=x^2 that is given
Use the same method he described in the video, the only tricky thing is setting up bounds,. You can integrate using an integral calculator to check that the answer's right.
@@stylingandfitnessingwithty2713 The x bounds are from 0 to 1 and y bounds are from x^2 (lower) to 1 (upper). The correct integrand (what we're integrating) is 1 + 2y. Drawing the region will help you see the bounds.
@@Rogue_Art This took me so long to figure out, but with your help I could eventually figure out, that I'm simply not able to read 2d coordinates anymore - I read (0,1) as (1,0). Those Details can really screw you up..
Isn't the integral same as taking the area of the square for the first problem he solves? Then the answer would be 1.. ? IF IT IS 1/2 then what does that mean in relation to the square?
I did the same mistake. 1) The curve runs counter clockwise so you must have a + sign. 2) You have wrong bounds for y. Draw the shape and it will be easy to see that the lower bound is: y = x^2 & the upper bound is: y = 1. Using the correct bounds you'll find: +22/15 (the correct answer)
@@MonkeyOnFleek choose lower bound for x as 0 and the upper bound for x as 1, and then choose the lower bound for y as x squared and choose the upper bound for y as 1
The r refers to a radius vector from the origin to a point along the path you are integrating. Think of dr as "dinky change in r", which means an infinitesimal change in r along the integral. The vector r will be a vector function of space, where its components are each functions of a parameter (t) that describe its path. Think of t as time, and the x-position/y-position/z-position along the way, will each be functions of time that trace out this path. To find dr, you differentiate r with respect to t, essentially finding the velocity along this space curve, to get dr/dt. then multiply what you get by dt, and dt becomes the new variable of integration, for carrying out the line integral. The vector for dr/dt becomes part of the integrand.
I get 22/15 from line integral but 8/15 from Green's theorem... I've set the limit of y to be from 0 to x2 and x from 0 to 1 : still I get 8/15. Can someone help with 22/15 using Green's theorem?
Every time I search a topic and see professor Dave come up I know it’s about to all make sense. Thank you so much for all of your help
Yeah dude I’m in frisking fifth grade and I half understand greens theorem now.
i definitely asked !
Brother, you are not getting enough views for your videos,but that doesn't change the fact that you are one of the top teacher/explainer/TH-camr on earth.
Thanks for video
SPOILER ALERT for last example:
Integral of 1+2y dy is y+y², evaluate from x² to 1 we get 2 - x² - x⁴ dx, integrate we get 2x - x³/3 - x⁵/5, evaluate from 0 to 1 we get 2 - 1/3 - 1/5, arithmetic and voila 22/15
I may be dumb, but how did you get 1 + 2y ?
never mind....duh....I got it!
Why from x^2 to 1 instead of 0 to x^2?
@@philkirshenbaum8867 plot the graph and you will see the lower curve is y=x^2 and upper y=1(as mentioned in the question C moves from 1,1 to 0,1 along a line and the line is y=1)
@Alej Random Thanks for the spoiler dude, I was looking for if someone had shared the answer! :D
how come you can explain something in 5 min, that my college professor couldn't in 2 hours of lecture???
@Elian Aydin no it doesn't
@@igorfritz2973 They are bots.
@@NameisU I know, I looked it up
@@NameisU wdym, who's bot
@@rahmatnasrul7263 the message in question, probably got deleted.
Your videos are beyond impressive. Not only do I love your concise explanations, but also your graphics. One of the best mathematics channels on youtube!
Thank you Mr. Green for making integrals easier and time-saving!
Awesome explanation. Love your humor and attitude... Surprisingly makes me want to study math more
I really wish i had seen your videos when I was struggling with vector calculus in my university 12 years ago. Take love and keep going.
Thank you for being a Physicist
ahhhaaa right 😁😁😁 that was such a physicist "explanation" of Green's theorem
Although he is mostly an Organic Chemistry tutor, but yeah he can teach any subject, an all rounder of course ✌💯
You saved my collegiate career. God bless you.
Your 5 minute video is the best lacture i have seen
Simple but yet extremely useful explanation, thanks Dave
Revisiting some calc topics and glad I picked these since i learned from all ur physics topics
Holy crap, you managed to explain this in 6 min and I was able to understand it, I was a bit confused after an hour and a half lecture, but this just made sense. It clicked.
How is this guy a professor in more than 10 subjects and a god in debating 😂😂😂😂
Positive orientation is the reason a hollow circle inside a solid circle, that you are calculating, has the lines going clockwise. You are not calculating the hollow area, but the solid and it is on the left as you walk around it. Very easy concept that some people have issues understanding.
Its amazing how easily you explain this stuff.
You explain exactly whats needed to be explained in such a concise manner.
Perfect for people with ADHD lol
This is genius. Simple and straight forward. Thank you so much,
Your video is really good and I do appreciate the time and effort you put into it but I think you should mention that all points inside the D area must be differentiable and that depends on P Q as well as R for stokes theorem. I know that it sound obvious but if 1 or some points are not differentiable then we have to seek for approach to solve the problem.
2:30 where does that seemingly arbitrary yet very specific vector field come from? Can anyone tell me please?
i am pretty sure it is given to us with the problem
@@hiba2475 thank you, I realised my mistake later on
Was confused for a week, and now enlightened in only 7 min
I found out who you were through your flat earth videos, stayed for the ones that are saving my ass in high school. God bless.
@@armgord Some people in high school enroll in a local community college to take multivariable calc
not me tho LOL
Watching this on the train ride to my calc 3 final as a refresher, thanks for the easy to follow video that taught me more than my professor has the entire semester!
That’s on you
I'm watching ur complete maths series I will get back to this soon 😍
Professor dave is the only thing holding up my academic career right now
after searching video after video, now i finally found it useful
Thank you for good lecture!
Good video!.. thanks..but I wanted to know what's the practical applications of finding the line integral of a 'Vector field' over any curve?
It's mainly a physics thing you can use to find the work done by a force field on an object
If you consider the vector field to represent a force, then the line integral from a to b across a curve C (aka the circulation) can be thought of as the work needed to move a particle subjected to said force from a to b.
Yeah, Green's theorem!
superb teaching
What a star you are
The Best Explanation 👌 thanks Sir
Can we also apply this Theorem for complex numbers?
Brilliant explanations dr dave
at 4:10 what does the answer "1/2" represent
like does it represent the area of something??
It depends on what the vector field represents. Suppose the vector field is a force field, which is the go to application that motivates the concept of vector line integrals. The vector field would be force at any given point in space, and have the units of Newtons. The change in radius vector, dr, would have the units of meters. Upon integrating them together, you'd get Joules of work done, as a body moves along the given path, while subject to that force field.
Is implicit differentiation valid in the condition for Greens theorem?
This is so easy to understand now, thanks, professor !! :)
Makes it seem like secondary school math great explanation
So the technique to know the lower limit and the upper limit is to know the direction of the arrow. The starting point is the lower limit and it can be a function or a point while the upper limit is the intersection or the coordinate or so that's how I see it. Awesome!
Great explanation! 🎉
Professor Dave is the best!
I get that for the last example you should set the bounds of x as 0 to 1 and the bound of y as x^2 to 1, but why do we set up the bound of x as 0 to 1? When I graphed it, the left bound is zero but the right bound is the line graph y=x^2. Please help me
You are a real genius
😜 My brain! Keep up the good work and I like the new hair.
I have a question, the area of square in example at 4:10 shouldn't be 1? If we use equation for the area of square namely a^2 we should get 1 because length of the side is 1, but using green's theorem we got 1/2, someone can tell me why?
You are not calculating the area inside the square, since this is not just a normal integral. Here the line integral shows how much of the vector field that is given is in the direction of the curve, which in this case is the square. You can tell that it's not the area of it by using a different vector field as well, which will yield different results.
@@vanadium1021 thanks for explanation;))
Sir can you also do a lesson on Laplace transform
In the triangle example, why are the y limits from 0-->y=x and not simply 0-->y=1 like in the first example of the square?
because the upper limit is the curve itself
@@ProfessorDaveExplains Thanks for the response, your video on double integrals helped so much and clarified it all for me, one of the best maths videos I have ever seen so thanks for your work!
Thanks you very much, Professor!
Thank you sir, you explain it excellently.
plz,could you explain how to get out vector field for the contour
I have a quick question: While I understand the outcome of Line integral is Area, what is the outcome of the circular integral (conservative) ? Since it equals the double integral, Is the RHS is Volume? If that is the case, shall I assume that one special case of line integral which involves conservative Field on a closed curve over a surface yields volume? Please answer
Your videos are amazing
thank you
Stoke's Theorem next!
close enough... welcome back mathematical Dominick Reyes
for the problem at the end, we take the inner integral to go from an "exact" extent and the inner one from the "furthest" extent
Awesome video, super explanation❤️👍tysm sir
why the limits of integration of x and y is from 0 to 1? can someone please explain. Thanks
How do you get the P and Q ?
But why did you switch the dxdy to dydx? And the bounds are set strange too
I think on 0.24 time you have a mistake in line integral. Shouldn't it be dl in the end?
no interpretation about conservation?
Great video
Where does (2x - x) comes from?
Professor
Pls explain it for 3d vector field and 3d curve
I have a problem pls help
If we have square with side = 1 then area is 1*1=1
But with green theorem it is 1/2
How it works? What is wrong with me?
Because you aren't just calculating the area of that square. You are multiplying each area element, with the curl of a vector field, across that square.
Just like if you are given the density of a sheet of a square meter of plywood as 7 kg/m^2, and you calculate it for a piece that is 1 meter by 1 meter. When you calculate the area, it is 1 square meter. But when finding the mass, you shouldn't expect to get 1 kg. You should expect to get 7 kg.
Thank you very much sir!!
Does Green's theorem imply that dQ/dx = dP/dy, because of Cauchy theorem on closed and analytic curves?
If it did, we'd have no reason to use Green's theorem. If dQ/dx equaled dP/dy, then the vector field is conservative, and the closed loop line integral is trivial, because line integrals on conservative vector fields are independent of path, and closed path integrals of conservative fields are zero.
I got 8/15 for the answer every time for the checking comprehension. I can't figure out what I am doing wrong here. If you are still looking at these video's comments, could you please help me?
So I started by setting up the Pyx and Qxy in the double integral. So I got 1+2y. Now I get my bounds as from 0 to x^2. And for x I get from 0 to 1. So that becomes: y+y^2 then from 0 to x^2 I get x^2 + x^4. Then integrating again I get
(x^3)/3 + (x^5)/5. Take that from 0 to 1: 1/3 + 1/5. That gets 5/15 + 3/15 = 8/15.
Your bounds of integration for y should be from x^2 to 1 rather than 0 to x^2. If you draw a picture it is easier to visualize why. Anyways, the function stops at y=1 and it begins at the equation y=x^2 that is given
@@henrymartens4079
Oh ok that makes sense. Thanks for taking the time to help! Have a good rest of your day :)
I feel, it will be better to visualize that y varies from 0 to 1 and X varies from 0 to √y.
@@henrymartens4079 Thankss.
well he certainly knows a lot about a lot of stuff. Thanks for your help.
thank you sir from Japan u help me a lot
i still dont get it
thank you so much!
Im having trouble solving the comprehension exercise any chance you can help?
Dude you're a genius. But wait, what happened to the hair?
give us back the haiiiirrrrrrrrrrrrrrr :OOOOO
where do u live ,sir ????
why we need to set 0 to y = x? Thanks
my king
how did he get 22/15 for the last problem.
UP UP UP
Use the same method he described in the video, the only tricky thing is setting up bounds,. You can integrate using an integral calculator to check that the answer's right.
@@TheEpicGod111 what were the correct bounds for y?
@@stylingandfitnessingwithty2713 The x bounds are from 0 to 1 and y bounds are from x^2 (lower) to 1 (upper). The correct integrand (what we're integrating) is 1 + 2y. Drawing the region will help you see the bounds.
@@Rogue_Art This took me so long to figure out, but with your help I could eventually figure out, that I'm simply not able to read 2d coordinates anymore - I read (0,1) as (1,0). Those Details can really screw you up..
Thanks
sir what is difference between green theorem and stroke theorem ? thank u sir amarjit india
I see. I no longer hate you, green
Isn't the integral same as taking the area of the square for the first problem he solves? Then the answer would be 1.. ? IF IT IS 1/2 then what does that mean in relation to the square?
No, the line integral is being solved with green's theorem, not the area
Relative to the last exercise; I guess the result is (-8/15)!? (not 22/15)
Please kindly check it out?
I did the same mistake.
1) The curve runs counter clockwise so you must have a + sign.
2) You have wrong bounds for y.
Draw the shape and it will be easy to see that the lower bound is: y = x^2 &
the upper bound is: y = 1.
Using the correct bounds you'll find: +22/15 (the correct answer)
@@Invalid571 but the upper bound should be y=x^2 and the lower should be 0
got -8/15 too
@@MonkeyOnFleek I got 22/15
@@MonkeyOnFleek choose lower bound for x as 0 and the upper bound for x as 1, and then choose the lower bound for y as x squared and choose the upper bound for y as 1
What exactly is dr in F.dr? Which vector is r?
The r refers to a radius vector from the origin to a point along the path you are integrating. Think of dr as "dinky change in r", which means an infinitesimal change in r along the integral.
The vector r will be a vector function of space, where its components are each functions of a parameter (t) that describe its path. Think of t as time, and the x-position/y-position/z-position along the way, will each be functions of time that trace out this path. To find dr, you differentiate r with respect to t, essentially finding the velocity along this space curve, to get dr/dt. then multiply what you get by dt, and dt becomes the new variable of integration, for carrying out the line integral. The vector for dr/dt becomes part of the integrand.
In the square example why is D 1/2? It looks like it should be 1 to me :(
Oh it's something to do with the vector field, okay... Hard to visualise.
Your explanations are EXCEPTIONAL. Thank you. You look good with shor hair also. :)
Nice video. S2
I get 22/15 from line integral but 8/15 from Green's theorem... I've set the limit of y to be from 0 to x2 and x from 0 to 1 : still I get 8/15.
Can someone help with 22/15 using Green's theorem?
Set the y limit from x^2 to 1
But what is the reason behind setting limit from x^2 to 0 ??
@@mehakali1465 need to draw the a) b) c), then u will see the bounded area for x-axis from 0 till y=x^2
studying this before a test, wish me luck
Wait whaaat, where is the long hair? ;(
im having troubles finding the limits for the last question
How do I contact you in person, pls?
darn i wish the practice problem at the end had a quick solution explanation
Did you get my email on a new theory of gravity?
Isnt it 8/15?
thats what i got
It is 22/15
@@himahelmy6206 hi bro/sis choose the bounds correctly draw a diagram and visualise that
@@himahelmy6206 upper bound for x=1, lower bound for x = 0, and upper bound for y =1 and the lower bound for y= x squared
Caitlyn
ily
gren theodor
wtf u must’ve known we were learning this