I like to think of it as a chunk of an orange peel

Wake up babe, blackpenredpen just whipped out his Pokeball

Another way to interpret it is that in spherical coordinates, it involves a radius and a solid angle: dV = ρdρdΩ, and since dΩ = ρsin(φ)dφdθ, so dV = ρ^2sin(φ)dρdφdθ.

REALLY nice drawings; thank you! Please do the Jacobian method next; I would love to see you explain it.

throughout the entier last semester i was looking for this title and now you uploaded it after the exam

Can you make a video about multiple indefinite integrals (like the antiderivative of a function of multiple variables)? It seems that there are no theorems or even any mention of them online and it seem wierd how such a simple thought has no information online.

Please make a video on Jacobian

Your drawing skills are immaculate

It's like you're explaining the perspective but in your own way I found this video interesting because it's one of the applications of camera lens

Also, a similar approach is used in determining the surface area of a sphere (or anything in spherical coordinates) dS = = . Surface area of a unit sphere: ∬dS = ∬√((ρ^2*sin(φ))^2+(ρ*sin(φ)*ρ_φ)^2+(ρ*ρ_θ)^2)dφdθ where ρ_x is the partial derivative respect of x in the corresponding spherical ρ(φ, θ) function. In this case, ρ(φ, θ) = 1. Since the partial derivatives of the function = 0, then you will get this simple form ∬ρ^2*sin(φ)dφdθ, which is basically the differential of volume of a sphere without the dρ, due to a property that in an n-ball, the derivative of the volume respect to its radius is its surface area, so, dV = SdR, and solving this integral ∬{{0, 2π}, {0, π}}sin(φ)dφdθ = ∫{0, 2π}2R^2dθ = 4πR^2.

Thank you

Do the cylindrical coordinates differential please!

I HAVE AN EXAM FROM THESE THANK YOU BOSS

Are you dealing with measure and integration theory (especially Lebesgue integral) ?

Spherical coordinates are double polar coordinates; you can transform from rectangular coordinates to cylindrical coordinates and then from cylindrical coordinates to spherical coordinates. Each transformation puts an appropriate radius factor into the volume element, first r and then ρ. Then you write r using ρ and φ to get the final result. From the last video, we know that x = r cos θ and y = r sin θ give us dx dy = r dr dθ, so we also know that z = ρ cos φ and r = ρ sin φ give us dz dr = ρ dρ dφ in the same way. So dx dy dz = r dr dθ dz = r dz dr dθ = r ρ dρ dφ dθ. And since r = ρ sin φ, this is ρ² sin φ dρ dφ dθ.

Yo master can u help me solving this integral? ∫arcsin((x-2)/(x+2))

Yay!!!

Sir please i need 100 derivative pdf

what math is this?

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dear bprp, i searched for this video yesterday due to my final exam being today. However, you didnt upload the video untill today(AFTER MY FINAL EXAM) IT IS UNBELIEVABLE to do such a thing. REZALET...... NOT WATCHING YOU ANYMORE.