Bravo! You have done what no one else on the internet (or my calc 3 professor) had been able to do: explain the many different cases you encounter when you do surface integrals.
I have a vector calculus exam tomorrow and I got completely confused about all the formulas and how and where to use them. After watching your videos I finally figured it all out in a couple hours!!! Your videos are very direct and specific! The explanations are great! Thank you so much for the videos and your help! You are a lifesaver for students
That works for me! I really appreciate how concise it is! Many awesome videos drill deep into these concepts, but what I really need is ready-to-use skills!
Ooo I'm here to help with my understanding of exterior algebra, linear algebra, and geometric algebra. Thanks Dave, I've been loving your back and forth with narcissist James Tour and you're helping me out with my other passions too. I love it.
Thank you for the video! I could actually calculate the comprehension problem in my head b/c it's quite intuitive. Basically the surface S is an inclined plane (z=1+x) that given the x limits goes from z=1 up to z=2. Since we want to integrate f=z on that plane, we can take the area of the plane (i.e. 2*√2), and multiply it by the average value of z (i.e. z=1.5), leading to the final answer of 3*√2. NOTE: The area of the plane is 2*√2 b/c the length of one side of the plane is the y range (equal to 2) and the other is √2 b/c that other side of the plane goes across 1 unit (x=0 -> x=1) as it goes up 1 unit too (z=1 -> z=2).
So I noticed that the sqrt(2) is missing it seems he used the formula 6:05 it seem we are given a scalar field or a function or f(x,y,z) so this is the formula for scalar field / function when the surface "z" is in terms of x and y parametric equation which is at 6:10, the one he used at 11:48 is the formula for the vector field F. And if we plug it in, we'll get 3*sqrt(2)
11:40 - I got 2 times square root 2 instead. The length of Z is square root 2 between X=0 and 1, length of Y is 2 between Y=0 and 2. Area = length of Z times length of Y. Not sure if I missed something?
I got the same result the first time since I followed the 6:00 formula literally, but I think I realise now where I got it wrong: So the derivatives within the big square root(the magnitude of rU and rV) equals 2, giving square root 2. There is no doubt about it. We then have to do two integrals of this square root 2, which is also multiplied by the function, which is just plainly, z, which in turn, can be substituted by a simple x+1. I think we both followed the same path so far. The mistake that I made that was probably similar to yours was that I plugged in the limit value of x(0,1) BEFORE I even integrated it, which in hindsight is a rookie mistake. Meaning I calculated (x+1)*sqrt(2), so I thought "okay I will just plug in 1 for x", thus gaining (1+1)*sqrt(2) aka 2sqrt(2). I integrate it by adding an extra X, which is 1, so it remains 2sqrt(2). But this is the wrong order. You're supposed to INTEGRATE the result first and THEN plug in the number. Meaning that (x+1)*sqrt(2) should be integrated. By integrating it, you get ((x^2)/2+x)*sqrt(2). Distribute the root 2 across both terms. Now plug in the 1. Now you get 0.5sqrt(2)+1sqrt(2), or 1.5sqrt(2). From then on just proceed to integrate in regards to y. The integral will add an extra y, you plug in 2, and thus you get 2*1.5sqrt(2) or 3sqrt(2). Bingo.
Thanks for your video, it helps me comprehend the main concepts of surface integrals! And one thing that confused me is that at 9:00, why there is a vector r instead of the unit normal vector n?
made it, it seems the problem uses the formula of 6:10 because we have a function or a scalar field f(x,y,z) an not a vector field F or so that's how I comprehend it
I had to laugh out loud. The "double integral of the magnitude of the cross product of the derivative of r with respect to u and the derivative of r with respect to v". It's nearly the entire semester in one problem...
But I didn't understand why is that that when the function is described with the parameters x and y- you decide to give it a certain formula -pdz/dx-qdz/dy+R basing it on a particular excercise that you came up with and than applying it to a completley different question at 9:06? can someone please explain because it's unclear to me why should you use specifically that formula if it came from a completely different question?
yes. this isnt rigorous but here it is. assume surface can be paramaterised into r(u,v) then a small change to r caused by one lf the variables is (dr) as dr/du is ratio of change in r caused by u, to u itself so dr/du *(du) is the actual change to r caused by u. this is a vector qauntaty. imagine limit sum of a difference in r's value being a tangential vector to the surface. now do the same for achange jn v an you have two vectors relating to a small change along the surface. by taking a cross product between these you then get a vector with a magnitude of the area of a small ds
You can parametrize it however you want, as long as you fully describe the surface in question, with your parametric equations. However, there often are ways you can parametrize it that will significantly simplify the remaining work. For instance, in your example, I could parametrize it as: x = u*cos(3*v) y = u*sin(3*v) z = 3*v But I wouldn't really accomplish anything by making a more complicated definition of my parameter v. I'll get the same results as if I kept it simple, and stuck with z=v, and defined everything else in terms of that.
All the problems in my calculus book are 10x harder than this. Especially the first steps, which are just given in this video. I wish you would raise the challenge a bit in your video, because this way I'm missing the vital part.
hmmmm. it doesn't change the outcome of the answer but when you were taking the determinant for finding Ru x Rv you got -cosv as the j component but i do not understand why it is negative. I am fairly certain that it should be positive. however when taking the magnitude of it the negative goes away so it doesn't actually change the answer. idk correct me if I am wrong.
Thank you dave, my college seriusly screwed my major cuz here we split calc 3 into 2 semesters to go more in depth, but that comes with the consequence of every major learning about vector fields on semester 4, the thing is that my major has EM physics on semester 3 (everyone else has it on sem 4) WICH IS FILLED WITH VECTOR FIELDS AND LINE INTEGRALS
Can anybody explain to me how (sinV)^2 + (-cosV)^2 becomes 1? I know (sinV)^2 + (cosV)^2 equals 1 but is (sinV)^2 + (-cosV)^2 equal to (sinV)^2 + (cosV)^2 ? Please help me out.
When you square any negative number, the answer is always positive, since a negative number times a negative number equals a positive one. That said, if we evaluate, for example, (-2)², the answer will be 4, since (-2)(-2) = 4. Note how this is the same as evaluating 2², which also equals 4. Note how (-2)² equals (-2)(-2). This is not the same as -2², which equals -(2)(2), which is the same as -4. The same happens with cos(v) and (-cos(v)). Since we are evaluating (-cos(v))(-cos(v)), we get a positive answer. If we were evaluating -(cos(v))(cos(v)), the answer would be negative (be aware of the negative sign outside the parenthesis). Since we know (-cos(v))² is positive and has the same magnitude as (cos(v))², which is also positive, then they must evaluate to exactly the same. Since (-cos(v))² = (cos(v))², (sin(v))² + (-cos(v))² also equals one.
Its just wonderful, clearly explained just try to eliminate those words appearing on screen that hinders clear view of the writings you're presenting, otherwise its so lit,,
I just don't get what function to integrate. In my calculus book, they just take y and integrate it over the surface, but it's absolutely not clear to me what I should double integrate :((
if in a vector field you must, first of all, parameterize the surface you are interested in, i mean, the surface in which the vector field is flowing through, remember surface integral calculate the flux throughout a surface in R3. once you parameterize your surface [ r(u,v)=(u,v,g(u,v)) being u,v the two parametrization variables ] and then you calculate the cross product and there you got the normal vector. after that, the force field you are given must be changed in terms of the u and v variables. after that you are left with two vectors and calculate teh dot product of them and there it is what you need to double integrate
It sucks that you don't have a video explaining Bézier curves yet. (Those are constructed by iterating interpolations between straight paths to control points until just one control segment remains, if that makes even remote sense to you!) But the ultimate generalisation of geometric curves is also the industry standard: NURBS.
Whoever came up with using u and v as the two variables would have chosen better variables if they'd had my handwriting. What on Earth were these people thinking. ςigma and τau would've worked just as nicely. μu or νu would've sufficed as well. u and v. picard facepalm.
@@HarpSeal I have a 1960s calc book that uses ξ to represent little Δx* when explaining Riemann sums, and also uses ξ elsewhere. it's kind of strange to see.
In the problem the f will come out as (0,0,1+x), rx=(1,0,1), ry=(0,1,0), rx*ry=(-1,0,1). And the answer after solving the double integral is coming as 3 not 3(2)^1/2.
We did it guys. We've successfully removed the numbers from math.
Bravo! You have done what no one else on the internet (or my calc 3 professor) had been able to do: explain the many different cases you encounter when you do surface integrals.
In my opinion, this topic is the most difficult to grasp out of all the topics in Calculus. However, difficulty is subjective
you've surely not studied partial differential equations yet
@@MBKfreestyle 🤓
@@MBKfreestylehe said that difficulty is subjective, oH yOu ArE sMaRt
@@MBKfreestyle PROVING THIS GUY'S POINT, I'D ARGUE THIS IS MORE DIFFICULT THAN PARTIAL DIFFERENTIAL EQUATIONS
@@MBKfreestylewe had this after PDE's
I have a vector calculus exam tomorrow and I got completely confused about all the formulas and how and where to use them. After watching your videos I finally figured it all out in a couple hours!!! Your videos are very direct and specific! The explanations are great! Thank you so much for the videos and your help! You are a lifesaver for students
That works for me! I really appreciate how concise it is! Many awesome videos drill deep into these concepts, but what I really need is ready-to-use skills!
Ooo I'm here to help with my understanding of exterior algebra, linear algebra, and geometric algebra. Thanks Dave, I've been loving your back and forth with narcissist James Tour and you're helping me out with my other passions too. I love it.
Just in time for my Calc III final, thank you so much
same story a year later lmao. Hope your final went good lol cuz I am gonna get fked in 2 hours
@@ritishgupta5201 It went alright, best of luck on your final 🙏
@@ritishgupta5201 I have mine today lol, how did yours go?
@@ritishgupta5201 lol update cmon
@@devan6012 hope youll do well, if its done, how did it go?
Thank you for the video!
I could actually calculate the comprehension problem in my head b/c it's quite intuitive.
Basically the surface S is an inclined plane (z=1+x) that given the x limits goes from z=1 up to z=2.
Since we want to integrate f=z on that plane, we can take the area of the plane (i.e. 2*√2), and multiply it by the average value of z (i.e. z=1.5), leading to the final answer of 3*√2.
NOTE: The area of the plane is 2*√2 b/c the length of one side of the plane is the y range (equal to 2) and the other is √2 b/c that other side of the plane goes across 1 unit (x=0 -> x=1) as it goes up 1 unit too (z=1 -> z=2).
justpaulo
&Look at Nikola Tesla over here 🙄-
Good on you M8!
I've been looking for a derivation of dS. seems all the videos online like to skip this detail. thank you
11:48 @Professor Dave Explains I got 3 as the answer. I think I did something wrong. Used double integral (-Pdelz/delx -Qdelz/dely + R)dxdy
I got 3 too after i parameterized z=1+x and did all that stuff so idk what’s up
yeah isnt it 3?
yep got 3 too I got stuck at this maybe we did something quite wrong i don't know where is it
So I noticed that the sqrt(2) is missing it seems he used the formula 6:05 it seem we are given a scalar field or a function or f(x,y,z) so this is the formula for scalar field / function when the surface "z" is in terms of x and y parametric equation which is at 6:10, the one he used at 11:48 is the formula for the vector field F. And if we plug it in, we'll get 3*sqrt(2)
Ight update I got it guys
You just explained it better than my Calculus II did in 6 months. Thank you!
In the problem starting at around 9:30 was the r supposed to be an n (normal vector)?
yes. U R right!
11:40 - I got 2 times square root 2 instead.
The length of Z is square root 2 between X=0 and 1, length of Y is 2 between Y=0 and 2.
Area = length of Z times length of Y.
Not sure if I missed something?
no same I keep getting 2 sqrt(2) as well. Might be a mistake on the answer.
I got the same result the first time since I followed the 6:00 formula literally, but I think I realise now where I got it wrong:
So the derivatives within the big square root(the magnitude of rU and rV) equals 2, giving square root 2. There is no doubt about it. We then have to do two integrals of this square root 2, which is also multiplied by the function, which is just plainly, z, which in turn, can be substituted by a simple x+1. I think we both followed the same path so far.
The mistake that I made that was probably similar to yours was that I plugged in the limit value of x(0,1) BEFORE I even integrated it, which in hindsight is a rookie mistake. Meaning I calculated (x+1)*sqrt(2), so I thought "okay I will just plug in 1 for x", thus gaining (1+1)*sqrt(2) aka 2sqrt(2). I integrate it by adding an extra X, which is 1, so it remains 2sqrt(2). But this is the wrong order.
You're supposed to INTEGRATE the result first and THEN plug in the number. Meaning that (x+1)*sqrt(2) should be integrated. By integrating it, you get ((x^2)/2+x)*sqrt(2). Distribute the root 2 across both terms. Now plug in the 1. Now you get 0.5sqrt(2)+1sqrt(2), or 1.5sqrt(2). From then on just proceed to integrate in regards to y. The integral will add an extra y, you plug in 2, and thus you get 2*1.5sqrt(2) or 3sqrt(2). Bingo.
i have no idea what I’m watching and I love it
Watch it before going to bed and try and ask yoir subconscious mind to hepp you understand it
the room vibe
@@daniellabinjo6046 Even the devil couldn't teach me this crap
Thanks for your video, it helps me comprehend the main concepts of surface integrals! And one thing that confused me is that at 9:00, why there is a vector r instead of the unit normal vector n?
i have the same question
I love all ur lectures, they were cool and clear 🔥♥️💯👍
made it, it seems the problem uses the formula of 6:10 because we have a function or a scalar field f(x,y,z) an not a vector field F or so that's how I comprehend it
I had to laugh out loud. The "double integral of the magnitude of the cross product of the derivative of r with respect to u and the derivative of r with respect to v". It's nearly the entire semester in one problem...
This might be the best video I’ve ever seen on math, god bless you Dr Dave
Requires patient and time ,positive mindset to grasp well this concept.
correction: @ 5:32 its mag of Rx X Ry. not Ru X Rv.
This helps me with the book introduction to electrodynamics, by Griffiths. Also we can compute the integral for any kind of surface.
same, the special integrals werent explained very thoroughly in griffiths e&m
This is very well explained, thank you for this!
But I didn't understand why is that that when the function is described with the parameters x and y- you decide to give it a certain formula -pdz/dx-qdz/dy+R basing it on a particular excercise that you came up with and than applying it to a completley different question at 9:06?
can someone please explain because it's unclear to me why should you use specifically that formula if it came from a completely different question?
Thus was almost as hard as your last couple of Italian language videos...
Hi any proof on why dS=|ru x rv|dudv? 1:29
yes. this isnt rigorous but here it is.
assume surface can be paramaterised into r(u,v) then a small change to r caused by one lf the variables is (dr) as dr/du is ratio of change in r caused by u, to u itself so dr/du *(du) is the actual change to r caused by u. this is a vector qauntaty. imagine limit sum of a difference in r's value being a tangential vector to the surface. now do the same for achange jn v an you have two vectors relating to a small change along the surface. by taking a cross product between these you then get a vector with a magnitude of the area of a small ds
Can parametrization of x,y,z be whatever I want?
( it doesnt have to be "x=ucosv, y=usinv, z=v" yes?)
No,it actually depends on your function. He used those functions as an example.
I know I'm a bit late,but for future generations it'll be useful:)
You can parametrize it however you want, as long as you fully describe the surface in question, with your parametric equations. However, there often are ways you can parametrize it that will significantly simplify the remaining work.
For instance, in your example, I could parametrize it as:
x = u*cos(3*v)
y = u*sin(3*v)
z = 3*v
But I wouldn't really accomplish anything by making a more complicated definition of my parameter v. I'll get the same results as if I kept it simple, and stuck with z=v, and defined everything else in terms of that.
any update about the comprehension? help
I thought: let's explore some higher mathematics.
After 2 mins: I think I should close it now.
Thank you sir for your dedication and for making this free! 🙏
All the problems in my calculus book are 10x harder than this. Especially the first steps, which are just given in this video. I wish you would raise the challenge a bit in your video, because this way I'm missing the vital part.
whoa! thanks professor finally understood surface integrals.Took some time but finally its in , thanks again
hmmmm. it doesn't change the outcome of the answer but when you were taking the determinant for finding Ru x Rv you got -cosv as the j component but i do not understand why it is negative. I am fairly certain that it should be positive. however when taking the magnitude of it the negative goes away so it doesn't actually change the answer. idk correct me if I am wrong.
what book did you use?
or what book do you recommend for study?
Thank you so much, sir! I thought I won't catch up in my vector calc class, but I'm getting ahead now!
Thanks so much for making this a lot clearer than it was in my head :)
Thank you dave, my college seriusly screwed my major cuz here we split calc 3 into 2 semesters to go more in depth, but that comes with the consequence of every major learning about vector fields on semester 4, the thing is that my major has EM physics on semester 3 (everyone else has it on sem 4) WICH IS FILLED WITH VECTOR FIELDS AND LINE INTEGRALS
We could simplify so that the derivative of z is tan of angle of normal vector z and cross product with line integral of z normal vector?
can anyone explain how to derive the formula for surface integral of vector field
Does the parameters stay constant for each example. If not how do you choose them?
Can anybody explain to me how (sinV)^2 + (-cosV)^2 becomes 1?
I know (sinV)^2 + (cosV)^2 equals 1 but is (sinV)^2 + (-cosV)^2 equal to (sinV)^2 + (cosV)^2 ? Please help me out.
When you square any negative number, the answer is always positive, since a negative number times a negative number equals a positive one.
That said, if we evaluate, for example, (-2)², the answer will be 4, since (-2)(-2) = 4. Note how this is the same as evaluating 2², which also equals 4.
Note how (-2)² equals (-2)(-2). This is not the same as -2², which equals -(2)(2), which is the same as -4.
The same happens with cos(v) and (-cos(v)).
Since we are evaluating (-cos(v))(-cos(v)), we get a positive answer. If we were evaluating -(cos(v))(cos(v)), the answer would be negative (be aware of the negative sign outside the parenthesis).
Since we know (-cos(v))² is positive and has the same magnitude as (cos(v))², which is also positive, then they must evaluate to exactly the same.
Since (-cos(v))² = (cos(v))², (sin(v))² + (-cos(v))² also equals one.
I need to find the check comprehension song so I can jam it on replay while I solve the task
i have it on a five hour loop in a video in my "just for fun" playlst
@@ProfessorDaveExplains Hahah thanks, it's the perfect vibe!
If you start out with a 2-D vector, will it still be a double integral? Or single?
Beautifully explained.
This is simple I just can't remember anything on how to parameterize surfaces. WHICH I'm going to you for help lol. Amazing as always
thank you sir.....
@Professor
@4.45 it should be not
I just hate how this didn't exist when I took this class at university
Nicely explained . Thanks
Thank You So Much Sir.
thanks helps lots
Hi I keep getting 3/2 as my answer for the question at the end. Could you please explain it to me
thanks Manav
THANK YOU SO MUCH! ❤
I'm so happy... Love you professor 👍❤
excellent explanation
Thank you so much Prof Dave!
This was the topic i didn't get to study for the exam. Oh well, its not that bad
Thank you verymuch for the informative presentation
thanks prof
Just starting out calculus 3. One week left for the final exam.
So easy, thanks!
Dr I got 3 as the answer have i done an error??
I also got 3, but I dont know what's wrong with my process....
I got -3 also
Me too
thank dude
Its just wonderful, clearly explained just try to eliminate those words appearing on screen that hinders clear view of the writings you're presenting, otherwise its so lit,,
Just hit the CC button, my friend. It's on your end.
@@ProfessorDaveExplains I got the answer as 3 for comprehension. Could you check?
Thank you master!
I just don't get what function to integrate. In my calculus book, they just take y and integrate it over the surface, but it's absolutely not clear to me what I should double integrate :((
if in a vector field you must, first of all, parameterize the surface you are interested in, i mean, the surface in which the vector field is flowing through, remember surface integral calculate the flux throughout a surface in R3. once you parameterize your surface [ r(u,v)=(u,v,g(u,v)) being u,v the two parametrization variables ] and then you calculate the cross product and there you got the normal vector. after that, the force field you are given must be changed in terms of the u and v variables. after that you are left with two vectors and calculate teh dot product of them and there it is what you need to double integrate
Are da of equal lenths
THANK YOU
Think u need to color code ur graphs
Thank you.
Are da of equal lengths
Last example that you have done, the answer is supposed to be 1/3
Helped me a lot, thank you!
thank you
idk but you got the unit normal vector (n) in new way. what i mean is that i get the n in different way n= grad.S/|grad.S|
dawg whaaat da haaaaaaaaleeeeeeee
8’59, flux F.ndS instead F.rdS...
I don't think you're right...
NICE VIDEO, LETS GOO
It sucks that you don't have a video explaining Bézier curves yet. (Those are constructed by iterating interpolations between straight paths to control points until just one control segment remains, if that makes even remote sense to you!)
But the ultimate generalisation of geometric curves is also the industry standard: NURBS.
Are you gonna do Theodore Roosevelt tomorrow? You said you do the president videos once per month.
I'm a little behind, it'll be next week.
Great
Thank you ! It helped me a lot
Hey salut Nic, bonne chance demain là
Nic Florant Ahahahah merci toi aussi big 😂😂😂 fuck Celse right
What Playlist is this
math
Whoever came up with using u and v as the two variables would have chosen better variables if they'd had my handwriting. What on Earth were these people thinking. ςigma and τau would've worked just as nicely. μu or νu would've sufficed as well. u and v. picard facepalm.
Let’s not forget the people who chose to use ξ in sets
@@HarpSeal I have a 1960s calc book that uses ξ to represent little Δx* when explaining Riemann sums, and also uses ξ elsewhere. it's kind of strange to see.
😬I'm definitely more confused than when I started.
T shirt so cool
Can please anyone help me with comprehension
this is late but z=x+1 is plane in 3d space that can be describe by = so you have something in the form of
@@fernandoportal5422 Can you elaborate?
T H A N K Y O U
nice
Your texts subtitle make me confuse to see and to understand
In the problem the f will come out as (0,0,1+x), rx=(1,0,1), ry=(0,1,0), rx*ry=(-1,0,1). And the answer after solving the double integral is coming as 3 not 3(2)^1/2.
final tmo, if i dont get an 80% i fail😿
i kid you not, thats exactly what I need too. 80% just to get a D. I’m mentally prepped to retake. Did you pass?
10/10
Thanku physics jesus
👌👌👌
💯💯💯
❤❤❤
what happened to those hairs.
i watched this and solved a question worth 15 marks in 2 minutes, my teacher gave me $200 for that
?