1 ) yz=2
We know that
y=rsinθsinϕz=rcosθ
yz=rsinθsinϕ rcosθ=r2sinθcosθcosϕ=21r2sin2θcosϕ
Now,
yz=2⇒21r2sin2θcosϕ=2⇒r2sin2θcosϕ=4, r>0, θ∈[0,π], ϕ∈[0,2π]
2) y2+z2-x2=1
We know that
х=r sinθ cosϕ
y=r sinθ sinϕ
z=r cosθ
y2+z2-x2=r2 sin2θ sin2ϕ + r2 cos2θ - r2 sin2θ cos2ϕ = r2 sin2θ (sin2ϕ -cos2ϕ )+ r2 cos2θ= r2 sin2θ (1- 2cosϕ ) +r2 cos2θ=
=r2 sin2θ - 2r2 sin2θcosϕ +r2 cos2θ = r2 (sin2θ+ cos2θ)- 2r2 sin2θcosϕ = r2 - 2r2 sin2θcosϕ =1
r2 - 2r2 sin2θcosϕ =1 , r>0, θ∈[0,π], ϕ∈[0,2π]
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