One other way to see why the area should be ld is through Pappus-Guldinus theorem. It states that (hyper)area and (hyper)volume of a (hyper) solid of revolution is equal to the product of (hyper) length or (hyper)area of what's being rotated times the distance traveled by its center of mass. So for a line segment its length is d and it's center of mass is in the middle, so the distance it travels is precisely l on the diagram
Also, you might not know this, but in hyperbolic polar coordinates, (x^2 - y^2 = r^2, x = r*cosh(θ), y = r*sinh(θ)), the 2-form differential for area is still rdrdθ.
sir I have seen your previous video and I have a question that why we use dl mean think as a fragment of a cone in case of surface area and use this equation and in case of volume we think as a small cylinder with width dx. why we don't think a fragment of cone in the case of volume?? or vice versa. please annswer me
For the people saying the smaller shape needs to be proven to be a sector, do know that the curved shape has uniform width. This means that if it was a full circle the inner shape would be concentric to the outer shape.
x = r cos t y = r sin t dx = cos t dr - r sin t dt dy = sin t dr + r cot t dr dx^dy = ( cos t dr - r sin t dt) ^ (sin t dr + r cos t dt) = r dr ^ dt dr ^ dt = - dt ^ dr dr ^ dr = - dr ^ dr = 0 dt ^ dt = 0
dx dy = d(r cos θ) d(r sin θ) = (cos θ dr - r sin θ dθ) (sin θ dr + r cos θ dθ) = sin θ cos θ dr dr + r cos² θ dr dθ - r sin² θ dθ dr - r² sin θ cos θ dθ = sin θ cos θ 0 + r cos² θ dr dθ + r sin² θ dr dθ - r² sin θ cos θ 0 = 0 + r (cos² θ + sin² θ) dr dθ + 0 = r 1 dr dθ = r dr dθ.
@@mrk3661 : Yes! And going along with that, dr dr = 0 (because if you swap the order, dr dr = −dr dr, so it must be 0). This is using the so-called wedge product, so it would be more proper to write dr ∧ dθ, dx ∧ dy, etc; but then it takes more space. There's also the complication that because we don't use orientation in area integrals, the area element is the absolute value |dx ∧ dy|, so the _really_ proper thing to write is |dx ∧ dy| = |r| |dr ∧ dθ|, which simplifies to r |dr ∧ dθ| because r ≥ 0. But now we're being very pedantic and the notation is getting complicated. So there's a sense in which dr dθ = −dθ dr, which you use when calculating the area element; what's really happening there is that dr ∧ dθ = −dθ ∧ dr. But there's also a sense in which dr dθ = dθ dr, so that the order of the variables doesn't matter when you set up the double integral; and what's really happening there is that |dr ∧ dθ| = |dθ ∧ dr| (because the absolute value kills the minus sign). This discrepancy used to confuse me until I understood what was going on.
Last time i checked dx,dy,dr,dtheta werent taught in geometry also genuine question, is this your first time doing math? even if you were solving the navier-stokes equation you still might do something like 5+2 in the middle, would you then go "HoW iS tHat FlUid MeChAnIcS tHaTs JuSt BaSiC aRiThMeTiC"? obviously this is just a small part of the process of solving double integrals and yes it is calculus, if you seriously think anything without a limit,integral,sum,or differential ceases to be calculus then im pretty sure you didnt understand the concept of calclulus at all
I think Area of curve rectangle is just an area of trapezoid
Is someone agree with me
Hey I just noticed that, too! Very cool!!
Area of a rectangle is a special case of a trapezoid A = ((b+b)/2)*a .
In summary:
1. r dr dtheta is polar coord integration
2. r dr r is “national talk like a pirate day”
OMG! ... Nobody had ever taught me this proof. Thanks a lot Steve Sir.
Hey bprp! I noticed a typo in your thumbnail. You put down dxdx in the thumbnail when you should have dydx or dxdy.
I noticed that too, very confused lol
agreed
I just fixed. Thank you for pointing it out!
I am glad to see a fellow with such mathematical insight start to make videos. Great stuff.
Thank you!
Could you make video for a geometrical proof for spherical dv element?
Big circular sector minus small circular sector.
Circular sector parameterized by angle theta in radians: Area = theta / (2 pi) * area of the circle.
One other way to see why the area should be ld is through Pappus-Guldinus theorem. It states that (hyper)area and (hyper)volume of a (hyper) solid of revolution is equal to the product of (hyper) length or (hyper)area of what's being rotated times the distance traveled by its center of mass. So for a line segment its length is d and it's center of mass is in the middle, so the distance it travels is precisely l on the diagram
Would you please explain Jacobian matrix and determinant
recently watched a few vids on the gaussian integral, so the bonus fact is helpful thanks
I like the identities poster you have. Where'd you get it?
he sells it on his store
Also, you might not know this, but in hyperbolic polar coordinates, (x^2 - y^2 = r^2, x = r*cosh(θ), y = r*sinh(θ)), the 2-form differential for area is still rdrdθ.
Yea I saw this a while ago and thought it was very cool, too. Not that many people use or talk about this. Similar to the hyperbolic substitution.
Amazing drawings!
Nice video Sir😊
sir I have seen your previous video and I have a question that why we use dl mean think as a fragment of a cone in case of surface area and use this equation and in case of volume we think as a small cylinder with width dx. why we don't think a fragment of cone in the case of volume?? or vice versa. please annswer me
Ironically, it was easier to derive arc length in polar coordinates than area in polar coordinates.
For the people saying the smaller shape needs to be proven to be a sector, do know that the curved shape has uniform width. This means that if it was a full circle the inner shape would be concentric to the outer shape.
"rdrr, get it? Hardy har har"
Don't cheat to get into the gifted school.
((angle/360) * Pi * r^2) - ((angle/360) * Pi * (r-w)^2) in here w is the width of the curved triangle
Jacobian
Annular sector
This seems like a good problem to apply Cavalieri’s Principle.
r1, r2, d2
?
r2d2
nvm ;)
Please correct the dx.dx in your thumbnail image.
Corrected. Thanks!
Ohh that makes a lot of sense!
Neat
Gabriel horn
x = r cos t
y = r sin t
dx = cos t dr - r sin t dt
dy = sin t dr + r cot t dr
dx^dy = ( cos t dr - r sin t dt) ^ (sin t dr + r cos t dt) = r dr ^ dt
dr ^ dt = - dt ^ dr
dr ^ dr = - dr ^ dr = 0
dt ^ dt = 0
How is dr*dt = -dt*dr?
@@Anmol_Sinha en.wikipedia.org/wiki/Differential_form
@@krzysztofs.8409 thank you!
@@Anmol_Sinha You're welcome :)
It's not a rectangle if it's curved 😭
It's not a rectangle if it is curved.
Is not true
dx dy = d(r cos θ) d(r sin θ) = (cos θ dr - r sin θ dθ) (sin θ dr + r cos θ dθ) = sin θ cos θ dr dr + r cos² θ dr dθ - r sin² θ dθ dr - r² sin θ cos θ dθ = sin θ cos θ 0 + r cos² θ dr dθ + r sin² θ dr dθ - r² sin θ cos θ 0 = 0 + r (cos² θ + sin² θ) dr dθ + 0 = r 1 dr dθ = r dr dθ.
drdØ = -dØdr ????
@@mrk3661 : Yes! And going along with that, dr dr = 0 (because if you swap the order, dr dr = −dr dr, so it must be 0). This is using the so-called wedge product, so it would be more proper to write dr ∧ dθ, dx ∧ dy, etc; but then it takes more space.
There's also the complication that because we don't use orientation in area integrals, the area element is the absolute value |dx ∧ dy|, so the _really_ proper thing to write is |dx ∧ dy| = |r| |dr ∧ dθ|, which simplifies to r |dr ∧ dθ| because r ≥ 0. But now we're being very pedantic and the notation is getting complicated.
So there's a sense in which dr dθ = −dθ dr, which you use when calculating the area element; what's really happening there is that dr ∧ dθ = −dθ ∧ dr. But there's also a sense in which dr dθ = dθ dr, so that the order of the variables doesn't matter when you set up the double integral; and what's really happening there is that |dr ∧ dθ| = |dθ ∧ dr| (because the absolute value kills the minus sign). This discrepancy used to confuse me until I understood what was going on.
how is this calculus? This is just a wedge minus a wedge. This is super basic geometry.
Last time i checked dx,dy,dr,dtheta werent taught in geometry
also genuine question, is this your first time doing math? even if you were solving the navier-stokes equation you still might do something like 5+2 in the middle, would you then go "HoW iS tHat FlUid MeChAnIcS tHaTs JuSt BaSiC aRiThMeTiC"?
obviously this is just a small part of the process of solving double integrals and yes it is calculus, if you seriously think anything without a limit,integral,sum,or differential ceases to be calculus then im pretty sure you didnt understand the concept of calclulus at all