Translate the rectangular equation to spherical

[whynot314]

Translate the rectangular equation to spherical and cylindrical equations.

http://www.texify.com/img/%5CLARGE%5C%21x%5E2%2By%5E2%2B2y-3x%2Bz%5E2%3D25.gif

 

[SammyS]

Translate the rectangular equation to spherical and cylindrical equations.

http://www.texify.com/img/%5CLARGE%5C%21x%5E2%2By%5E2%2B2y-3x%2Bz%5E2%3D25.gif

Hello whynot314. Welcome to PF!

What have you tried?

Where are you stuck?

We can't help you until you show us what you tried. It in the rules .

 

[whynot314]

I first transformed x^2+y^2+z^2 into ρ^2

ρ^2+2ρsinϕsinθ-3ρsinϕcosθ=25

then factored out a ρ

ρ(ρ+2sinϕsinθ-3sinϕcosθ)=25

I want to solve for ρ,

I am confused as to, weather or not subtract 25 and set it equal to zero or keep it like this and have two equations where

ρ=25

and


ρ=25-2sinϕsinθ+3sinϕcosθ

 

[HallsofIvy]

Then what you need is a good class in arithmetic! If ab= c it does NOT follow that "a= c or b= c!

If that isn't enough, if xy= 10, it does NOT follow that either x= 10 (and y= 1) or y= 10 (and x= 1). There are an infinite number of soutions to xy= 10.

 

[SammyS]

I first transformed x^2+y^2+z^2 into ρ^2

ρ^2+2ρsinϕsinθ-3ρsinϕcosθ=25

then factored out a ρ

ρ(ρ+2sinϕsinθ-3sinϕcosθ)=25

I want to solve for ρ,

I am confused as to, weather or not subtract 25 and set it equal to zero or keep it like this and have two equations where

ρ=25

and

ρ=25-2sinϕsinθ+3sinϕcosθ

How can you say ρ = 25 ?

ρ² + 2ρsinϕsinθ-3ρsinϕcosθ = 25 is quadratic in ρ .

Complete the square or use the quadratic formula to solve for ρ.

For either, you may want to rewrite your equation as:

ρ² + 2ρsinϕ(sinθ-cosθ) = 25​

To complete the square, add

\sin^2\phi\left(\sin\theta-\cos\theta\right)^2​

to both sides.

To use the quadratic formula instead, subtract 25 from both sides of

ρ² + 2ρsinϕ(sinθ-cosθ) = 25 .​