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

We will use the Jacobian to see why dxdydz=ρ^2sinφdρdθdρ. We must know this when we convert a triple integral from the rectangular coordinate to the spherical coordinate.
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[xinpingdonohoe]

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

[jacokriek]

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

[Arycke]

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

[cyrusyeung]

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

[maxvangulik]

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


figures, but it was fun to work out :)