- #1

Beelzedad

- 24

- 3

Consider a continuous charge distribution in volume ##V'##. Draw a closed surface ##S## inside the volume ##V'##.

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Consider the following multiple integral:

##\displaystyle B= \iint_S \Biggl( \iiint_{V'} \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ \right] dV' \Biggl) dS \tag1##

where

##R=|\mathbf{r}-\mathbf{r'}|##

##\mathbf{r'}=(x',y',z')## is coordinates of source points

##\mathbf{r}=(x,y,z)## is coordinates of field points

##\cos(\hat{R},\hat{n})## is the angle between ##R## and normal to surface element

##\rho'## is the charge density and is continuous throughout the volume ##V'##

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When ##\mathbf{r} \in S##, the function is not integrable in domain ##V'##. So we use change of variables:

##\displaystyle B= \iint_S \Biggl( \iiint_{V'} \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ {r'}^2 \sin \theta' \right] d\theta' d\phi' dr' \Biggr) dS \tag2##

Note that in this equation, ##\theta'## and ##\phi'## are w.r.t point ##\mathbf{r} \in S ##

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Now by changing the order of integration:

##\displaystyle B = \iiint_{V'} \Biggl( \iint_S \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ {r'}^2 \sin \theta' \right] dS \Biggr) d\theta' d\phi' dr' \tag3##

##\displaystyle B= \iiint_{V'} \Biggl( \iint_S \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \right] dS \Biggr) \rho'\ {r'}^2 \sin \theta' d\theta' d\phi' dr' \tag4##

Here in this last equation, ##\theta'## and ##\phi'## are w.r.t. which point?

[1]: https://i.stack.imgur.com/mz1VV.png

___________________________________________________________________________

Consider the following multiple integral:

##\displaystyle B= \iint_S \Biggl( \iiint_{V'} \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ \right] dV' \Biggl) dS \tag1##

where

##R=|\mathbf{r}-\mathbf{r'}|##

##\mathbf{r'}=(x',y',z')## is coordinates of source points

##\mathbf{r}=(x,y,z)## is coordinates of field points

##\cos(\hat{R},\hat{n})## is the angle between ##R## and normal to surface element

##\rho'## is the charge density and is continuous throughout the volume ##V'##

__________________________________________________________________________

When ##\mathbf{r} \in S##, the function is not integrable in domain ##V'##. So we use change of variables:

##\displaystyle B= \iint_S \Biggl( \iiint_{V'} \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ {r'}^2 \sin \theta' \right] d\theta' d\phi' dr' \Biggr) dS \tag2##

Note that in this equation, ##\theta'## and ##\phi'## are w.r.t point ##\mathbf{r} \in S ##

__________________________________________________________________________

Now by changing the order of integration:

##\displaystyle B = \iiint_{V'} \Biggl( \iint_S \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ {r'}^2 \sin \theta' \right] dS \Biggr) d\theta' d\phi' dr' \tag3##

##\displaystyle B= \iiint_{V'} \Biggl( \iint_S \left[ \dfrac{\cos(\hat{R},\hat{n})}{R^2} \right] dS \Biggr) \rho'\ {r'}^2 \sin \theta' d\theta' d\phi' dr' \tag4##

**Now here is my question:**Here in this last equation, ##\theta'## and ##\phi'## are w.r.t. which point?

[1]: https://i.stack.imgur.com/mz1VV.png

Last edited: