Spherical Coordinates — Question 4

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Question 4

Rewrite the sphere x2+y2+z2=6zx^2+y^2+z^2=6z in spherical coordinates and identify its geometry.

Tasks

  1. Obtain the nontrivial spherical equation.

  2. Find its Cartesian center and radius.

  3. Explain the role of ρ=0\rho=0 when simplifying.

Original worksheet page 1: question and worked solution for 1-13-004
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Question 4 – Solution

Strategy. Substitute x2+y2+z2=ρ2x^2+y^2+z^2=\rho^2 and z=ρcos⁡ϕz=\rho\cos\phi, while preserving any solution lost by division.

See the diagram in the original worksheet below.

Step 1: Convert ρ2=6ρcos⁡ϕ⇔ρ(ρ−6cos⁡ϕ)=0.\rho^2=6\rho\cos\phi\quad\Longleftrightarrow\quad \rho(\rho-6\cos\phi)=0. The usual surface equation is ρ=6cos⁡ϕ,0≤ϕ≤π2,\boxed{\rho=6\cos\phi},\qquad 0\le\phi\le\frac\pi 2, and it includes the origin at ϕ=π/2\phi=\pi/2.

Step 2: Identify Completing the square in Cartesian form gives x2+y2+(z−3)2=9.x^2+y^2+(z-3)^2=9. It is the sphere centered at (0,0,3)\boxed{(0,0,3)} with radius 3\boxed{3}.

Step 3: Origin check Division by ρ\rho would temporarily discard ρ=0\rho=0. The origin satisfies the original equation and is recovered by ρ=6cos⁡(π/2)=0\rho=6\cos(\pi/2)=0; thus no point is absent from the final parametrized surface.

Original worksheet page 2: question and worked solution for 1-13-004

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