Answer to Question #335775 in Analytic Geometry for Dhanush

Question #335775

Express the following surfaces in spherical coordinates (i) yz=2. (ii) y^2+z^2-x^2=1

1
Expert's answer
2022-05-05T12:22:34-0400

1 ) yz=2yz=2\\

We know that

y=rsinθsinϕz=rcosθy=r\sinθ\sinϕ \\ z=r\cos{\theta}


yz=rsinθsinϕ rcosθ=r2sinθcosθcosϕ=12r2sin2θcosϕyz= r\sinθ\sinϕ\ r\cos{\theta} \\=r^2\sin{\theta}\cos{\theta}\cos{\phi} \\=\frac{1}{2}r^2\sin{2\theta}\cos{\phi}

Now, 

yz=212r2sin2θcosϕ=2r2sin2θcosϕ=4, r>0, θ[0,π], ϕ[0,2π]yz=2 \\ \Rightarrow \frac{1}{2}r^2\sin{2\theta}\cos{\phi}=2 \\ \Rightarrow r^2\sin{2\theta}\cos{\phi}=4, \ r>0,\ \theta\in[0,\pi], \ \phi\in[0,2\pi]



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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