Volume of leminiscat look a like

[morea]

Homework Statement

Ok, so I need to calculate the volume of a body bounded with $$(x^2+y^2+z^2)^2=a^3x$$ using spherical coordinates

My actual question is, how does one determen the value for $$\phi$$ and $$\theta$$ in this example?

 

[tiny-tim]

welcome to pf!

hi morea! welcome to pf!

well, the LHS is easy …

so how do you convert a³x to spherical coordinates?

 

[morea]

thanx for reply and welcome

so how do you convert a3x to spherical coordinates?

Well, it may be a little tricky, but I guess you use the same conection as on the left side, that is

$$x= r cos \theta sin \phi$$

 

[tiny-tim]

hi morea!

(have a theta: θ and a phi: φ )

you got to learn to be the boss …

you choose the coordinates …

in this case, won't it be easier if you make x = rcosθ, y = rsinθcosφ, z = rsinθsinφ ?

 

[morea]

Well, I haven't consider that. To be perfectly honest, I wasnt aware that I could do that. But it certenly makes life a lot easier.
However, although this makes things simpler, I still don't see how to determine values for $$\varphi$$ and $$\theta$$

I got to say that I am impress by your expeditious

 

[tiny-tim]

(what happened to θ and φ i gave you? )

well, φ obviously goes from 0 to 2π

and you know r⁴ = a³rcosθ, so cosθ = (r/a)³, sooo … ?

 

[morea]

so, θ = arccos(r/a)3, that is - $\frac{\pi}{2}$ < θ > $\frac{\pi}{2}$ Is that right??

what happened to θ and φ i gave you?

What do you mean, to type them in a smaler font?

 

[tiny-tim]

so, θ = arccos(r/a)3, that is - $\frac{\pi}{2}$ < θ > $\frac{\pi}{2}$ Is that right??

no …

that's all of θ …

you need limits for θ, for each value of r

What do you mean, to type them in a smaler font?

no, i meant type them instead of bothering with latex!

 

[morea]

I feel kinda sily, but I don't get it
Do I just leave θ = arccos(r/a)3, but how I am I going to express r then?

no, i meant type them instead of bothering with latex!

he, well, as you might guess, english is not my first language, so I get things a litle slower. Thanks for the tip.

 

[tiny-tim]

Do I just leave θ = arccos(r/a)3, but how I am I going to express r then?

you need to integrate over all the points inside cosθ = (r/a)³ …

ie for all points with r ≤ (cosθ)^1/3/a, which is the same as cosθ ≥ (r/a)³, or θ ≤ arccos(r/a)³

 

[morea]

Great
Thank you, it was very helpfull

cheers