Spherical Coordinates — Question 6

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Question 6

Describe in spherical inequalities the solid inside x2+y2+z2≤16x^2+y^2+z^2\le 16, above the cone z=x2+y2z=\sqrt{x^2+y^2}, and in the half-space y≥0y\ge 0.

Tasks

  1. Determine bounds for ρ\rho, θ\theta, and ϕ\phi.

  2. Explain each angular bound geometrically.

  3. State which boundary pieces meet at the origin.

Original worksheet page 1: question and worked solution for 1-13-006
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Question 6 – Solution

Strategy. Translate the sphere to a radial bound, the half-space to azimuth, and compare zz with the cylindrical radius to bound inclination.

See the diagram in the original worksheet below.

Step 1: Radial bound The sphere gives 0≤ρ≤40\le\rho\le 4.

Step 2: Azimuth Since y=ρsin⁡ϕsin⁡θy=\rho\sin\phi\sin\theta and sin⁡ϕ≥0\sin\phi\ge 0, y≥0y\ge 0 is represented without overlap by 0≤θ≤π0\le\theta\le\pi.

Step 3: Inclination Above the cone means ρcos⁡ϕ≥ρsin⁡ϕ.\rho\cos\phi\ge\rho\sin\phi. For ρ>0\rho>0 this gives 0≤ϕ≤π/40\le\phi\le\pi/4. Consequently 0≤ρ≤4,0≤θ≤π,0≤ϕ≤π/4.\boxed{0\le\rho\le 4,\qquad 0\le\theta\le\pi,\qquad 0\le\phi\le\pi/4}. The sphere is the outer boundary; the cone and the two half-plane faces θ=0,π\theta=0,\pi all meet at the origin. Angular multiplicity there does not duplicate physical volume.

Original worksheet page 2: question and worked solution for 1-13-006

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