Area of a "curved rectangle" (bonus: why dxdy=rdrdθ)

[@anuragguptamr.i.i.t.2329]

OMG! ... Nobody had ever taught me this proof. Thanks a lot Steve Sir.

[@evacuate_earth]

I am glad to see a fellow with such mathematical insight start to make videos. Great stuff.

[@bprpcalculusbasics]

Thank you!

[@chrisglosser7318]

In summary: 1. r dr dtheta is polar coord integration 2. r dr r is “national talk like a pirate day”

[@vincent.0705]

Hey bprp! I noticed a typo in your thumbnail. You put down dxdx in the thumbnail when you should have dydx or dxdy.

[@christianherrera4729]

I noticed that too, very confused lol

[@cdkw2]

agreed

[@bprpcalculusbasics]

I just fixed. Thank you for pointing it out!

[@vladislavanikin3398]

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

[@thexoxob9448]

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.

[@lumina_]

recently watched a few vids on the gaussian integral, so the bonus fact is helpful thanks

[@caniraso]

Would you please explain Jacobian matrix and determinant

[@General12th]

Amazing drawings!

[@AlbertTheGamer-gk7sn]

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.

[@bprpcalculusbasics]

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.

[@Ntt903]

Could you make video for a geometrical proof for spherical dv element?

[@jackkalver4644]

Ironically, it was easier to derive arc length in polar coordinates than area in polar coordinates.

[@anshumansingh5728]

Nice video Sir😊

[@recreater8659]

((angle/360) * Pi * r^2) - ((angle/360) * Pi * (r-w)^2) in here w is the width of the curved triangle

[@christianmirmo4942]

I like the identities poster you have. Where'd you get it?

[@radiationgaming889]

he sells it on his store

[@l9day]

"rdrr, get it? Hardy har har"

[@chitlitlah]

Don't cheat to get into the gifted school.

[@unitatersyt453]

Jacobian

[@Jono4174]

Annular sector

[@aMartianSpy]

r1, r2, d2

[@bprpcalculusbasics]

?

[@aMartianSpy]

r2d2 nvm ;)

[@DEYGAMEDU]

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

[@TranquilSeaOfMath]

This seems like a good problem to apply Cavalieri’s Principle.

[@vano__]

Ohh that makes a lot of sense!

[@anuragguptamr.i.i.t.2329]

Please correct the dx.dx in your thumbnail image.

[@bprpcalculusbasics]

Corrected. Thanks!

[@technicallightingfriend4247]

Gabriel horn

[@krzysztofs.8409]

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

[@Anmol_Sinha]

How is dr*dt = -dt*dr?

[@krzysztofs.8409]

@@Anmol_Sinha en.wikipedia.org/wiki/Differential_form

[@Anmol_Sinha]

@@krzysztofs.8409 thank you!

[@krzysztofs.8409]

@@Anmol_Sinha You're welcome :)

[@robertpearce8394]

Neat

[@IainDavies-z2l]

It's not a rectangle if it is curved.

[@abdeljeddi9188]

Is not true

[@zachansen8293]

how is this calculus? This is just a wedge minus a wedge. This is super basic geometry.

[@m3nny125]

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

[@tobybartels8426]

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]

drdØ = -dØdr ????

[@tobybartels8426]

@@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.

[@arhamdugar396]

It's not a rectangle if it's curved 😭