Spherical Coordinates — Question 2

PDF ↗

Question 2

Convert the spherical point (ρ,θ,ϕ)=(6,3π/4,π/3)(\rho,\theta,\phi)=(6,3\pi/4,\pi/3) to rectangular and cylindrical coordinates. Keep all values exact.

Original worksheet page 1: question and worked solution for 6-13-002
Show solutionHide solution

Question 2 – Solution

Strategy First compute the cylindrical radius r=ρsin⁡ϕr=\rho\sin\phi and height z=ρcos⁡ϕz=\rho\cos\phi.

See the diagram in the original worksheet below.

Cylindrical form r=6sin⁡π3=33,z=6cos⁡π3=3.r=6\sin\frac{\pi}{3}=3\sqrt 3,\qquad z=6\cos\frac{\pi}{3}=3. Hence (r,θ,z)=(33,3π/4,3)\boxed{(r,\theta,z)=(3\sqrt 3,3\pi/4,3)}.

Rectangular form x=33cos⁡3π4=−362,y=33sin⁡3π4=362.x=3\sqrt 3\cos\frac{3\pi}{4}=-\frac{3\sqrt 6}{2},\qquad y=3\sqrt 3\sin\frac{3\pi}{4}=\frac{3\sqrt 6}{2}. Therefore (x,y,z)=(−362,362,3).\boxed{(x,y,z)=\left(-\frac{3\sqrt 6}{2},\frac{3\sqrt 6}{2},3\right)}.

Original worksheet page 2: question and worked solution for 6-13-002

Original worksheet layout. Use Enlarge or open the PDF for a closer view.