Ian Tan we want to calculate upper half of sphere, so z>0. When he parametrized z=rcos(fi) you can ask your self: for which values cos is greater then 0? When fi goes from 0 to pi/2.
It's the factor you have to introduce when switching to spherical coordinates. Technically it should be ro^2*sin phi dro*dphi*dtheta, but since ro=1 in this case, he was able to omit it. Whenever you switch from one kind of coordinate system, such as from rectangular to polar, or cylindrical, or spherical, or your own made up system, you need to introduce multipliers. Sometimes they're just constants, but sometimes they also involve variables or functions.
im pretty sure you are allowed to change the surface when you are using stokes theorem as long as you keep the same boundary. if you had changed it to a disc at the base of the surface, the normal becomes and the integral is just the area of the unit disc which is pi(1)^2 which is equals to pi
by Stokes' theorem you can take any other (as long as requirements for the theorem are met) surface which has the same boundary as the original one. When taking that "another" surface as a unit disk centered at (0,0,0), the surface integral happens to be the area of that disk. pi * radius^2 = pi
@@tomctutor Yes, but after taking the dot product, you get 1, so all that is left is the integral and dS. This is the area. You also miss read their response on the last part... He put pi; the 2 is an exponent.
@8:18 i see how you get your normal vector by using special simmetries of the sphere, but I was looking for a more general way to find a normal vector for a surface. In my mind n is something orthogonal to the tangent space, so i should probably do the computations to find a vector which gives 0 when multiplied for a vector tangent to the surface. Nonetheless i find this process a bit difficult in higher dimensions, because i can't visualize a tangent vector fast enough. Anyway, very nice video :))))
He made the F dot dr integral too difficult. The integral from 0 to 2 pi is the same for cosine squared and sin squared. add in sin squared and then divide by two. cos squared plus sin squared = one. THe integral from 0 to 2 pi divided by two is pi.
So if we have had a complete sphere there would not be a line integral to do because we would have need to be equal to zero? And it doesn't matter the height of the upper part of the surface because we the same contour we will need to obtain the same line integral and the same result?
why parameterise at 11:40?? at that point, he has ∫∫z dS BUT that = ∫∫z dA/ n.k = ∫∫z dA/z because n.k = z, so ∫∫z dA and then go ∫∫ r dr dtheta and that gives the answer in 2 more lines.
when he went to spherical coordinates, he made the jacobian sin(phi) only. is this because even though the jacobian should be rho^2 sin(phi), rho is 1 and 1^2 is 1 so he just left it off?
Is he missing another surface? namely, the surface at the bottom of the hemisphere, which is the circle on the xy plane. we would have to compute the surface integral of this too right (of course, this will equal to zero)?
Sorry if Im too late with this reply, but no, the bottom would not equal 0. If you do it correctly, it should be -π since now the divergence theorem applies and the div(curl(F))=0. For the Kelvin-Stokes theorem, the surface doesnt have to be closed. It just has to be a piecewise-smooth surface that fills the region made by the curve
does anyone know why dS was equal to that? surely we dont multiply by the jacobian since were not transforming the region of integration... simply just parametrizing..
eurekkkkaaaaa!!!!!!!!! I have found a video with no racist comments conspiracy theories, blames jews for everything, atheist comments, et., calculus, you have gain a new friend for eternity.
Sorry if I'm too late but since the path is entirely contained within the xy plane, z=0. This also means dz=0 since the derivative of 0 is still 0. So F= and dr=. F•dr=0*dx+x*dy+y*0=xdy Hope that helps and again, sorry if im too late to help you for your final or something like that
@@tsshamoo2376 dont be sorry tsshamoo its explained in the video. these people should know better - instead of paying attention they just want the answer spoon fed to them. tik tok brain
+Cody Pulliam Hey genius, he was integrating over the unit sphere where rho would be one. Also when using Stoke's theorem, at least in Calc 3, you can only use up to two substitution variables.
+Assata Amaechi Osborne For a given φ, you can think of sin φ = r, where r is the radius of the horizontal circular cross section of the sphere, at height z = cos φ. Now the parameterizations of x and y are exactly the same as for the line integral.
apparently, it takes 10 sec or less for students at MIT to solve the problem at 2:00.
Lmao
its quite simple
Uh uh, he be skipping steps, I'm not advanced enough for those. Where PatrickJMT at lol
facts
What step did he skip? I would be happy to fill in the details, if you have a specific question.
when he does curl F he doesnt show how he did it haha and that is hard lol
spring1 could you explain the limits for fi? I don’t get why is it in between 0 and pi/2
Ian Tan
we want to calculate upper half of sphere, so z>0. When he parametrized z=rcos(fi) you can ask your self: for which values cos is greater then 0? When fi goes from 0 to pi/2.
I really love the scene when the guy went and then came back :D
It's the factor you have to introduce when switching to spherical coordinates. Technically it should be ro^2*sin phi dro*dphi*dtheta, but since ro=1 in this case, he was able to omit it. Whenever you switch from one kind of coordinate system, such as from rectangular to polar, or cylindrical, or spherical, or your own made up system, you need to introduce multipliers. Sometimes they're just constants, but sometimes they also involve variables or functions.
That’s wrong
Stop spreading misinformation
It is “total nonsense”
Odeeeee
im pretty sure you are allowed to change the surface when you are using stokes theorem as long as you keep the same boundary. if you had changed it to a disc at the base of the surface, the normal becomes and the integral is just the area of the unit disc which is pi(1)^2 which is equals to pi
he's not working out a area, there's a vector field there?
and pi/2 =/= pi
by Stokes' theorem you can take any other (as long as requirements for the theorem are met) surface which has the same boundary as the original one. When taking that "another" surface as a unit disk centered at (0,0,0), the surface integral happens to be the area of that disk. pi * radius^2 = pi
@@tomctutor Yes, but after taking the dot product, you get 1, so all that is left is the integral and dS. This is the area. You also miss read their response on the last part... He put pi; the 2 is an exponent.
This is correct and it's how I prefer to solve these since it involves less complicated parametrizations so the risk of screwing up isn't as high
I love how this guy does intergration and the cross product like its simple arithmetic
How is your life after 7 years
4:51 "You have to remember way back to 1801" Yep yep, it definitely feels like it's been 215 years since single variable calc. Haha
when he actually means MIT course 18.01 LOL
Liam Patterson pretty sure he was joking
Who's man is this? He is fantastic!
Thank you sir, for blessing me with your knowledge.
@MrVandeju
We have that z=0 and dz=0. So, in his calculation, he does 2z*dx + x*dy + y*dz. Thus we are left with x*dy.
I've been confused with Stokes' Theorem for the past few days, and you just made it clear to me. Thank you!
briliant insructor
if z=o then fig would be 2 dimensional and limit willl be 0-π/2 and 0-2π results whole circle.
5:51 "So that was the Line Integral, very straightforward thing!".... I laughed so hard!
love his voice..
This guy is awesome!
Very nice explanation about the verification of stoke theorem
13:26 "Let me just check I'm not doing anything silly..."
*fades to black*
Oh. God this theorem is so interesting, verifying stoke theorem gives some basic ideas that what Stokes Theorem is all about.
Just a heads up, this is not for calculus beginners. You'd be lost when he advances without writing down steps.
@8:18 i see how you get your normal vector by using special simmetries of the sphere, but I was looking for a more general way to find a normal vector for a surface. In my mind n is something orthogonal to the tangent space, so i should probably do the computations to find a vector which gives 0 when multiplied for a vector tangent to the surface. Nonetheless i find this process a bit difficult in higher dimensions, because i can't visualize a tangent vector fast enough. Anyway, very nice video :))))
I love this guy!
He made the F dot dr integral too difficult. The integral from 0 to 2 pi is the same for cosine squared and sin squared. add in sin squared and then divide by two. cos squared plus sin squared = one. THe integral from 0 to 2 pi divided by two is pi.
Excellent teacher. Please make complex analysis videos.
So if we have had a complete sphere there would not be a line integral to do because we would have need to be equal to zero? And it doesn't matter the height of the upper part of the surface because we the same contour we will need to obtain the same line integral and the same result?
what don't you use x=rcost and y=rsint?
i like your style of showing this section. it's so cool--fast and clear.
Hourt Bora He did.
Hourt Bora r is 1.
ok i see. thanks guys!
why parameterise at 11:40??
at that point, he has ∫∫z dS BUT that = ∫∫z dA/ n.k = ∫∫z dA/z because n.k = z, so ∫∫z dA
and then go ∫∫ r dr dtheta and that gives the answer in 2 more lines.
+MrGiratina999 What's n.k?
n.k = normal vector dotted with the unit vector in the k or z direction.
Equals Sign: 10/10 would watch again
You deserve subscription.
God damn I LOVE THIS GUY I WANT TO TAKE ALL HIS CLASSES.
oh my, thank you so much.
This is well explained and helped me study better for my upcoming exam.
the best teacher!
I love your explanations!!!
Thank you
He forgot to mention the radius when parametrizing, if the radius was 2 this process would be slightly wrong.
No, because it's the unit sphere, so radius is 1. So you can ignore it in this case.
Tschaegger79 In this case, but it would be better to have an all encompassing explanation than to just ignore it.
That is what I was wondering
when he went to spherical coordinates, he made the jacobian sin(phi) only. is this because even though the jacobian should be rho^2 sin(phi), rho is 1 and 1^2 is 1 so he just left it off?
yep
It's a unit sphere so ro is just 1
Most of us on the youtube do are not from the class, if you could go a bit in the background that would be great; instead of testing it right away
Is he missing another surface? namely, the surface at the bottom of the hemisphere, which is the circle on the xy plane. we would have to compute the surface integral of this too right (of course, this will equal to zero)?
Sorry if Im too late with this reply, but no, the bottom would not equal 0. If you do it correctly, it should be -π since now the divergence theorem applies and the div(curl(F))=0. For the Kelvin-Stokes theorem, the surface doesnt have to be closed. It just has to be a piecewise-smooth surface that fills the region made by the curve
does anyone know why dS was equal to that? surely we dont multiply by the jacobian since were not transforming the region of integration... simply just parametrizing..
Thank you very much :)
Love you guys!! Thanks
eurekkkkaaaaa!!!!!!!!! I have found a video with no racist comments conspiracy theories, blames jews for everything, atheist comments, et., calculus, you have gain a new friend for eternity.
thanks!!
curl(F) is wrong, it should be not ; although you get the same result anyway
nope, it's correct
wow, i was wrong! Thank you very much (:
Thank you sir, you made my life easier !
Well, thanks for helping, and also thanks for reminding me I am UT-Dallas material and not MIT.
Why is MIT uploading 360p videos...
I'm sitting here watching cat videos all of a sudden UA-cam's like you know what here calculus.
I couldnot understand 13.26 how do you got the ds value ???can anyone help
Thanks man.
very nice!
very helpful keep the good work up
at 13:20 , would there be limit of z from -pi/2 to pi/2
My three year old son informed it is actually a train track.
he computed the curl wrong, its , not
is dS interchangeable with dA?
mind blown
how to verify for not unit sphere
@4:57 why does F*dr = xdy, there is no explanation
Sorry if I'm too late but since the path is entirely contained within the xy plane, z=0. This also means dz=0 since the derivative of 0 is still 0. So F= and dr=. F•dr=0*dx+x*dy+y*0=xdy
Hope that helps and again, sorry if im too late to help you for your final or something like that
Tssha MoO2 k
There is explanation you just don’t pay attention.
@@tsshamoo2376 dont be sorry tsshamoo its explained in the video. these people should know better - instead of paying attention they just want the answer spoon fed to them. tik tok brain
14:00 nice equal sign lol
You're suppose to know that already..
This whole time I thought it was “strokes’ theorem” and not “stokes’ theorem” 🤦🏻♂️🤦🏻♂️🤦🏻♂️
Also… my brain 💀💀💀💀
How are we left with x dy? I dont get it whatt?
He explains it. You can replay the video multiple times too it’s not hard to miss. Pay attention.
M-I-T-K-E-Y M-O-U-S-E
Where did come ds from?
I watched this at 2X playback speed. :-D
4 hours lecture = 17 minutes video. =='
great video but he rushed through the dot product way too fast.
Its dot product… you want him to help you how to add too? Dumb criticism
quit watching once he left out the rho's in parametrization
+Cody Pulliam Hey genius, he was integrating over the unit sphere where rho would be one. Also when using Stoke's theorem, at least in Calc 3, you can only use up to two substitution variables.
+Cody Pulliam fail.
Well maybe if you actually gave it 2 seconds of thought you'd understand why he "left out the rho's".... Good luck in your finals.....
why is z zero ?
the circle is on the z=0 plane
why is dS equals to that? anyone knows?
Mfanafuthi Khoza this is the same confusion I face...
Mfanafuthi Khoza does anyone know the answer to this?
i thought d (boldface S) = n dS = n * |N| * dphi dtheta ?!?!
yo isn’t ds rho^2 sin(phi) not sin(phi)?!!!???!!?!!??!!?!??!!?!?!?
rho is 1 stupid
That is the most Awkward Silence of ALL TIME 2:00 XDDDDDDD Nice Explanation !! Loved it
Can someone be kind enough to explain to me why, upon parametrization, x=cos θsin φ etc.?
Assata Amaechi Osborne Those are the basic spherical coordinates. Check wikipedia about it
+Assata Amaechi Osborne For a given φ, you can think of sin φ = r, where r is the radius of the horizontal circular cross section of the sphere, at height z = cos φ. Now the parameterizations of x and y are exactly the same as for the line integral.
Isn't it?
;-P
sir how i can download this lecture
sounds like the joker
it's not pheeeeeeeee, it's phi -_-
7:26 Just write out the determinant man!!!
Horse shit, a length is not equal to a surface area. Why do you omit the units of the integrals?
thank you