Question

given that x=ρsinϕcosθ,y=ρsinϕsinθ,z=ρcosϕ prove (x_phi_^2)+(y_phi_^2)+(z_phi_^2)=ρ^2

Answer 1

Do each at a time. \[x^2 = (\rho*sin(\phi)*cos(\theta))^2=\rho^2*sin^2(\phi)*cos^2(\theta)\]
\[y^2 = (\rho*sin(\phi)*sin(\theta))^2=\rho^2*sin^2(\phi)*sin^2(\theta)\]
\[z^2 = (\rho*cos(\phi))^2=\rho^2*cos^2(\phi)\]
Then sum them up \[\rho^2*sin^2(\phi)*cos^2(\theta) +\rho^2*sin^2(\phi)*sin^2(\theta)+ \rho^2*cos^2(\phi)\]
Notice you can pull out a rho^2 and group things up.
\[\rho^2*[sin^2(\phi)*cos^2(\theta) +sin^2(\phi)*sin^2(\theta)+ cos^2(\phi)]\]

Answer 2

Then you can group the \[sin^2(\phi)\] and you get the identity \[cos^2(\theta) +sin^2(\theta) =1\]
\[\rho^2*[sin^2(\phi)*(cos^2(\theta) +sin^2(\theta))+ cos^2(\phi)]\]
simplify that you get another identity of one
\[\rho^2*[sin^2(\phi)+ cos^2(\phi)]\]

Answer 3

Wow, thank you. That helped me a lot. I kept trying to find a way to make partial derivatives to work. Glad to know I was just over thinking it. =]