Why is dxdydz=ρ^2sinφdρdθdφ? (using Jacobian)

[@bprpcalculusbasics]

The geometry way: kzbin.info/www/bejne/q33CqqRtaJh2irs

[@xinpingdonohoe3978]

It's easier to do it in two steps. First is (x,y,z)→(r,θ,z) with (x,y)→(r,θ). x=r cos(θ), y=r sin(θ), x²+y²=r² dxdy=r drdθ, dxdydz=r drdθdz Then do the exact same thing again for (r,θ,z)→(ρ,θ,φ) with (r,z)→(ρ,φ). z=ρ cos(φ), r=ρ sin(φ), z²+r²=ρ² dzdr=ρ dρdφ, dzdrdθ=ρ dρdφdθ Combine them to get dxdydz=rρ dρdφdθ r=ρ sin(φ) dxdydz=ρ²sin(φ) dρdφdθ

[@thenew3dworldfan]

Have you heard about the negation of the epsilon delta definition? The statement is: There exists some positive ϵ such that no matter how small a positive δ you choose, there is always some x within the δ-neighborhood of c (but not equal to c) for which the distance between 𝑓(𝑥) and L is at least 𝜖.

[@Arycke]

Hey, great video! Typo in the title though sir. You have dpdødp

[@mohannad_139]

@@Arycke Yes he wrote dρdθdρ while he is supposed to write dρdθdφ

[@bprpcalculusbasics]

Thanks! I just fixed it.

[@jacokriek]

You waited a semester too long to make the calc 3 videos😅

[@JayOnDaCob]

Nah like a year, this class was by far my worst class in uni so far

[@k0pstl939]

I had to take it twice

[@yacovaaa9042]

what about dxdydzdw?

[@bitoty9357]

why does the jacobian works?

[@Seb2006-y4x]

What kind of Greek letter is that the one representing sine angle?

[@erroraftererror8329]

@@Seb2006-y4xTheta or phi?

[@Seb2006-y4x]

@@erroraftererror8329 i think phi

[@cyrusyeung8096]

Can you make a video explaining the concept of Jacobian marix? In particular, why dxdydz = ||J(ρ,φ,θ)||dρdφdθ.

[@mohannad_139]

How is your comment 2 months ago?

[@kunwarbtw981]

yeah lol

[@be-mr3tj]

Well to answer your question (I'll simplify to 2D) shifting the coordinates to a new point (x+dx, y+dy) comes with a new area. For rectangular cartesian coordinates this is is a rectangle with area dxdy. But for polar and other coordinate systems, this new additional area may not be a rectangle but can be approximated as a trapezium. This is essentially a rectangle in polar coordinates, but the area depends on r. (At different points, a small change in r,theta has differently scaled areas) The jacobian accounts for this scaling and thus any stretching in area that may occur when changing coordinates. I recommend looking into this further geometrically.

[@saharhaimyaccov4977]

If use x=y=z

[@maxvangulik1988]

now do it in reverse dpdQd@=|J(x,y,z)|dxdydz p=sqrt(x^2+y^2+z^2) Q=arctan(sqrt(x^2+y^2)/z) @=arctan(y/x) dp/dx=x/p dp/dy=y/p dp/dz=z/p dQ/dx=xz/rp^2 dQ/dy=yz/rp^2 dQ/dz=-r/p^2 d@/dx=-y/r^2 d@/dy=x/r^2 d@/dz=0 J(x,y,z)=|x/p y/p z/p; xz/rp^2 yz/rp^2 -r/p^2; -y/r^2 x/r^2 0| =x/p•|yz/rp^2 -r/p^2; x/r^2 0| -y/p•|xz/rp^2 -r/p^2; -y/r^2 0| +z/p•|xz/rp^2 yz/rp^2; -y/r^2 x/r^2| =x^2/rp^3+y^2/rp^3+z^2/rp^3 =1/rp =1/sqrt(x^2+y^2)sqrt(x^2+y^2+z^2) dpdQd@=dxdydz/sqrt((x^2+y^2)(x^2+y^2+z^2)) figures, but it was fun to work out :)