Recent content by morea

Great Thank you, it was very helpfull cheers

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

so, θ = arccos(r/a)3, that is - \frac{\pi}{2} < θ > \frac{\pi}{2} Is that right?? What do you mean, to type them in a smaler font?

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

thanx for reply and welcome 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

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?