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

Рет қаралды 18,061

2 ай бұрын

We will investigate the area of a circular curved rectangle. We have discussed this shape in the surface area of revolution: • Integral formulas for ... . We can also use this area of a curved rectangle to see why dxdy=rdrdθ when we do double integrals.Here's how we can use change of coordinate to evaluate the Gaussian integral of e^(-x^2) from 0 to inf: • How Gauss solved the i...
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Пікірлер: 47

@user-wj1qb3qu1y 2 ай бұрын

I think Area of curve rectangle is just an area of trapezoid Is someone agree with me

@bprpcalculusbasics 2 ай бұрын

Hey I just noticed that, too! Very cool!!

@user-ie5bh3tl8h Ай бұрын

Area of a rectangle is a special case of a trapezoid A = ((b+b)/2)*a .

@chrisglosser7318 2 ай бұрын

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

@vincent.0705 2 ай бұрын

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

@christianherrera4729 2 ай бұрын

I noticed that too, very confused lol

@cdkw2 2 ай бұрын

agreed

@bprpcalculusbasics 2 ай бұрын

I just fixed. Thank you for pointing it out!

@evacuate_earth Күн бұрын

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

@bprpcalculusbasics Күн бұрын

Thank you!

@anuragguptamr.i.i.t.2329 2 ай бұрын

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

@lumina_ 2 ай бұрын

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

@emanuellandeholm5657 2 ай бұрын

Big circular sector minus small circular sector. Circular sector parameterized by angle theta in radians: Area = theta / (2 pi) * area of the circle.

@caniraso 2 ай бұрын

Would you please explain Jacobian matrix and determinant

@General12th 2 ай бұрын

Amazing drawings!

@anshumansingh5728 2 ай бұрын

Nice video Sir😊

@Ntt903 2 ай бұрын

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

@DEYGAMEDU 2 ай бұрын

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

@AlbertTheGamer-gk7sn 2 ай бұрын

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 2 ай бұрын

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.

@jackkalver4644 2 ай бұрын

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

@christianmirmo4942 2 ай бұрын

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

@radiationgaming889 2 ай бұрын

he sells it on his store

@l9day 2 ай бұрын

"rdrr, get it? Hardy har har"

@chitlitlah 2 ай бұрын

Don't cheat to get into the gifted school.

@unitatersyt453 2 ай бұрын

Jacobian

@recreater8659 2 ай бұрын

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

@thexoxob9448 23 күн бұрын

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.

@Jono4174 2 ай бұрын

Annular sector

@vano__ 2 ай бұрын

Ohh that makes a lot of sense!

@TranquilSeaOfMath 2 ай бұрын

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

@robertpearce8394 2 ай бұрын

Neat

@aMartianSpy 2 ай бұрын

r1, r2, d2

@bprpcalculusbasics 2 ай бұрын

@aMartianSpy 2 ай бұрын

r2d2 nvm ;)

@anuragguptamr.i.i.t.2329 2 ай бұрын

Please correct the dx.dx in your thumbnail image.

@bprpcalculusbasics 2 ай бұрын

Corrected. Thanks!

@technicallightingfriend4247 2 ай бұрын

Gabriel horn

@krzysztofs.8409 2 ай бұрын

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

@tobybartels8426 2 ай бұрын

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 2 ай бұрын

drdØ = -dØdr ????

@tobybartels8426 2 ай бұрын

@@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.

@user-wo6qn3vf9n 2 ай бұрын

It's not a rectangle if it is curved.

@abdeljeddi9188 2 ай бұрын

Is not true

@arhamdugar396 2 ай бұрын

It's not a rectangle if it's curved 😭

@zachansen8293 Ай бұрын

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

@m3nny125 Ай бұрын

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