# Thread: decripting the integral

1. ## decripting the integral

dont be scared look at it from mathematical point of view

i just want to understand how they transform the painting to the integral

cant understand what is ds
and why it equals rd(phy)

maybe the confused phu with teta(cause i dont see phi)
i still cant see what that means
?

2. Originally Posted by transgalactic
dont be scared look at it from mathematical point of view

i just want to understand how they transform the painting to the integral

cant understand what is ds
and why it equals rd(phy)

maybe the confused phu with teta(cause i dont see phi)
i still cant see what that means
?

My best guess is it's a length element along the wire that carries the current. That's usually what ds means in that integral. But without knowing what the actual problem is (and what is that bowl shape in the sketch?) I can't tell for certain.

-Dan

3. i got a bowl spinning arround the Z axes with omega angular velocity
the bowl is charged with sigma charge density
i need to find out what is the magnetic feild in every point in the "z" axes
??

does it change some how your answer?

you are saying ds is a wire without width??
so ds is just a peace of line (its no a circle?)

having no width is a problem because we need to integrate from the start of the bowl till the top
and not having any width element doesnt give me the dx element in the integral

4. Originally Posted by transgalactic
i got a bowl spinning arround the Z axes with omega angular velocity
the bowl is charged with sigma charge density
i need to find out what is the magnetic feild in every point in the "z" axes
??

does it change some how your answer?

you are saying ds is a wire without width??
so ds is just a peace of line (its no a circle?)

having no width is a problem because we need to integrate from the start of the bowl till the top
and not having any width element doesnt give me the dx element in the integral
I would approach this somewhat differently. If you are allowed to use this methond you (presumably) know the formula for the B field due to a ring of charge. The bowl is simply a set of rings and you can add the contributions up.

-Dan

5. yess that exactly what thought to do
the physics part here is is automatic just put in the formua

the hard part is geometry
i did a sctech

k is the constant part(tired to write the constants all the time )
the formula is dB=$\displaystyle \frac{kIsin(\apha)dl}{d^2}$
d=$\displaystyle \sqrt{r^2+z^2}$
alpha is the angle between dl and the direction of we point we want to find the magnetic feild it
my latex doesnt compile dB=\frac{kIsin(\theta)dl}{d^2}
d=\sqrt{r^2+z^2}

and i dont know how to find dI??
I=integral of Jda

they did some thing different
i dont understand that

so it make sense do do what the original photo did
but idont understand the geometry
of how they get ds
and actually see it in the picture
and how they integrate
the formula is for dl??? (thin wire) thats the problem
we want to integrate by strips

6. oohhh sorry its not a bowl its a half of a solid ball

7. they break the charge unit into small flat squares
so we need to integrate by dI and ds

but its not been done