# Change of Variables in Double Volume Integral

• I
• pherytic
In summary, Greiner's Classical Electromagnetism book (page 126) has a derivation that involves a change of variables to \mathbf{z} and states that it is possible to move the 1/|\mathbf{r} - \mathbf{z}| term out of the \mathbf{r}' integral. This may seem incorrect, but it is justified because the integration with respect to \mathbf{r}' is done with \mathbf{z} held constant, and the domain of integration changes accordingly. This concept is helpful in understanding the substitution of independent variables in a mathematical derivation.

#### pherytic

In Greiner's Classical Electromagnetism book (page 126) he has a derivation equivalent to the following.
$$\int_V d^3r^{'} \nabla \int_V d^3r^{''}\frac {f(\bf r^{''})}{|\bf r + \bf r^{'}- \bf r^{''}|}$$

$$\bf z = \bf r^{''} - \bf r^{'}$$

$$\int_V d^3r^{'} \nabla \int_V d^3z \frac {f(\bf z + \bf r^{'})}{|\bf r - \bf z|}$$$$\nabla\int_V d^3z \frac{1}{|\bf r - \bf z|} \int_V d^3r^{'} {f(\bf z + \bf r^{'})}$$

The book says that after the change of variables to z, it is okay to just move the 1/|r-z| term out of the r' integral. But this doesn't make sense to me, given z is still a function of r'. Am I missing something or is this mathematically incorrect?

The integration with respect to $\mathbf{r}'$ is with $\mathbf{z}$ held constant. You have replaced the independent variables $\mathbf{r}'$ and $\mathbf{r}''$ with the independent variables $\mathbf{r}'$ and $\mathbf{z}$ by replacing $\mathbf{r}''$ with $\mathbf{r}' + \mathbf{z}$. It then makes sense to do the integral over $\mathbf{r}'$ first, precisely because you can regard $\|\mathbf{r} - \mathbf{z}\|$ as constant while doing so. The domain of integration will have changed because we require both $\mathbf{r}' \in V$ and $\mathbf{r}' + \mathbf{z} \in V$.

pherytic
Thank you, that really helped. If the two original variables are independent wrt each other, that has to still be true after substitution.