Express the plane z=x in cylindrical and spherical coordinates. a) cykindrical z=rcos⁡(θ) b) spherical coordinates θ=arcsin⁡(cot⁡(ϕ))

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Express the plane z=x in cylindrical and spherical coordinates.  a) cykindrical  z=rcos⁡(θ)  b) spherical coordinates  θ=arcsin⁡(cot⁡(ϕ))

Clara Reese · Accepted Answer

Step 1  The plane z=x goes through the line intersection of the planes  x=0 and z=0 and makes a π4 angle with those planes.  So, a point (x,y,z)=(x,y,x)=(x,y) is on the plane.  a) Cylindrical coordinates:  Let x=rcos⁡θ, y=rsin⁡θ with r≥0 and 0≤θ≤2π  Then. z=x  ⇒z=rcos⁡θ  So, a point on the plane takes the form (x,y,z)=(rcos⁡θ, rsin⁡θ, rcos⁡θ)  Step 2  b) Spherical coordinates:  Let x=ρsin⁡ϕcos⁡θ, y=ρsin⁡ϕθ, z=ρcos⁡ϕ with ρ≥0, 0≤θ≤2π and 0≤ϕ≤π  Then, z=x  ⇒ρcos⁡ϕ=ρsin⁡ϕcos⁡θ  ⇒ρcos⁡ϕρsin⁡ϕ=cos⁡θ  ⇒cot⁡ϕ=cos⁡θ  ⇒θ=arccos⁡(cot⁡ϕ)  So, a point on the plane takes the form  (x,y,z)=(ρsin⁡ϕcos⁡θ, ρsin⁡ϕsin⁡θ, ρcos⁡ϕ)  =(ρsin⁡ϕcos⁡(arccos⁡(cot⁡ϕ)), ρsin⁡ϕsin⁡(arccos⁡(cot⁡ϕ)), ρcos⁡ϕ)  =(ρsin⁡ϕcos⁡(arccos⁡(cot⁡ϕ)), ρsin⁡ϕsin⁡(arcsin⁡(1−cot2⁡ϕ)), ρcos⁡ϕ)  =(ρsin⁡ϕcot⁡ϕ, ρsin⁡ϕ1−cot2⁡ϕ, ρcos⁡ϕ)  =(ρcos⁡ϕ, ρsin⁡ϕ1−cot2⁡ϕ, ρcos⁡ϕ)

Express the plane z=x in cylindrical and spherical coordinates. a) cykindrical z=r\cos(\theta) b) spherical coordinates \theta=\arcsin(\cot(\phi))

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