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Question: How do you convert \[{x^2} + {y^2} = z\] into spherical and cylindrical form?...

How do you convert x2+y2=z{x^2} + {y^2} = z into spherical and cylindrical form?

Explanation

Solution

Here in this question we have to convert the given cartesian form of an equation to the spherical and cylindrical form. While converting the cartesian form to cylindrical form, we use x=rcos⁡θ, y=rsin⁡θ,z=zx = r\cos \theta ,\,y = r\sin \theta ,z = z and for the spherical form we use x=ρsin⁡ϕcos⁡θ, y=ρcos⁡ϕsin⁡θ,z=ρcos⁡ϕx = \rho \sin \phi \cos \theta ,\,y = \rho \cos \phi \sin \theta ,z = \rho \cos \phi .Then on simplification we obtain the answer.

Complete step by step answer:
Cartesian form: A function or relation is written using the (x,y)(x,y) or (x,y,z)(x,y,z) coordinates.
Cylindrical form: A function or relation is written using the (rcos⁡θ,rsin⁡θ,z)(r\cos \theta ,r\sin \theta ,z) coordinates.
Spherical form: A function or a relation is written using the (ρsin⁡ϕcos⁡θ,ρcos⁡φsin⁡θ,ρcos⁡ϕ)(\rho \sin \phi \cos \theta ,\rho \cos \varphi \sin \theta ,\rho \cos \phi ) coordinates.
Now we have to convert the cartesian form into spherical and cylindrical form.
When we are converting the cartesian form to cylindrical form. We consider
x=rcos⁡θ, y=rsin⁡θ,z=zx = r\cos \theta ,\,y = r\sin \theta ,z = z--------(1)
Now consider the given question.
x2+y2=z{x^2} + {y^2} = z--------(2)

On substituting the equation (1) to the equation (2) we have
⇒(rcos⁡θ)2+(rsin⁡θ)2=z\Rightarrow {(r\cos \theta )^2} + {(r\sin \theta )^2} = z
⇒r2cos⁡2θ+r2sin⁡2θ=z\Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = z
Take r2{r^2} as a common in the LHS we have
⇒r2(cos⁡2θ+sin⁡2θ)=z\Rightarrow {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = z
By the trigonometric identity we have cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1, so we have
⇒r2=z\Rightarrow {r^2} = z
Therefore the above function represents the cylindrical form.
Now we convert the cartesian form into spherical form.
When we are converting the cartesian form to spherical form. We consider
x=ρsin⁡ϕcos⁡θ, y=ρcos⁡ϕsin⁡θ,z=ρcos⁡ϕx = \rho \sin \phi \cos \theta ,\,y = \rho \cos \phi \sin \theta ,z = \rho \cos \phi --------(3)
Now consider the given question.
x2+y2=z{x^2} + {y^2} = z--------(2)
To the equation (2) add z2{z^2} on both sides
⇒x2+y2+z2=z+z2\Rightarrow {x^2} + {y^2} + {z^2} = z + {z^2} ---- (4)
As we know that the x2+y2+z2=ρ2{x^2} + {y^2} + {z^2} = {\rho ^2}--- (5)

On substituting the equation (5) and equation (3) to the equation (2) we get
⇒ρ2=ρcos⁡ϕ+ρ2cos⁡2ϕ\Rightarrow {\rho ^2} = \rho \cos \phi + {\rho ^2}{\cos ^2}\phi
Take ρ2cos⁡2ϕ{\rho ^2}{\cos ^2}\phi to LHS
⇒ρ2−ρ2cos⁡2ϕ=ρcos⁡ϕ\Rightarrow {\rho ^2} - {\rho ^2}{\cos ^2}\phi = \rho \cos \phi
Take ρ\rho as a common in the LHS
⇒ρ2(1−cos⁡2ϕ)=ρcos⁡ϕ\Rightarrow {\rho ^2}(1 - {\cos ^2}\phi ) = \rho \cos \phi
On cancelling the ρ\rho both in LHS and RHS we get
⇒ρ(1−cos⁡2ϕ)=cos⁡ϕ\Rightarrow \rho (1 - {\cos ^2}\phi ) = \cos \phi
By the trigonometric identity we have (1−cos⁡2ϕ)=sin⁡2ϕ(1 - {\cos ^2}\phi ) = {\sin ^2}\phi , on substituting we have
⇒ρsin⁡2ϕ=cos⁡ϕ\Rightarrow \rho {\sin ^2}\phi = \cos \phi
On dividing both sides by sin⁡2ϕ{\sin ^2}\phi
∴ρ=cot⁡ϕcsc⁡ϕ\therefore \rho = \cot \phi \csc \phi
Hence this is the spherical form.

Note: We can convert the equation from cartesian form to spherical form, cylindrical form and vice versa, the only thing we have to remember is the values. Since both cylindrical and spherical forms involve the trigonometric ratios we should know about the trigonometric identities and it is implemented whenever it is used.