Answer to Question #335775 in Analytic Geometry for Dhanush

Question

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

Answer 1

1 ) $yz=2\\$

We know that

$y=r\sinθ\sinϕ \\ z=r\cos{\theta}$

$yz= r\sinθ\sinϕ\ r\cos{\theta} \\=r^2\sin{\theta}\cos{\theta}\cos{\phi} \\=\frac{1}{2}r^2\sin{2\theta}\cos{\phi}$

Now, 

$yz=2 \\ \Rightarrow \frac{1}{2}r^2\sin{2\theta}\cos{\phi}=2 \\ \Rightarrow r^2\sin{2\theta}\cos{\phi}=4, \ r>0,\ \theta\in[0,\pi], \ \phi\in[0,2\pi]$

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]