# Confusion about Change in Integration Variable [closed]

I'm working through example 3.2 of Zangwill's Modern Electrodynamics and have come across a change in integration variables that I just can't seem to get. The example has two different change distributions related by $$\rho_b(\lambda \mathbf{r}) = \rho_a(\mathbf{r})$$ and asks for the relationship between the electric potentials. According to the book, $$\int d^3 r'= \frac{1}{\lambda^2} \int d^3 (\lambda r').$$ When I try to work through the Jacobian, I get $$d^3 r' = \begin{vmatrix} \frac{1}{\lambda} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} d^3(\lambda r') = \frac{1}{\lambda} d^3(\lambda r'),$$ where I'm assuming changing from $$r'$$ to $$\lambda r'$$ is like going from $$\vec{x}$$ to $$\vec{y}$$ where $$y_1 = \lambda x_1, \hskip{1cm} y_2 = x_2\hskip{0.5cm} \text{and} \hskip{0.5cm} y_3 = x_3 .$$ What am I missing here?

## closed as off-topic by Aaron Stevens, stafusa, Kyle Kanos, Jon Custer, user191954 Feb 16 at 9:34

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• The change $\mathbf{r}\rightarrow\lambda\mathbf{r}$ means that you should make the change $x\rightarrow\lambda x$, $y\rightarrow\lambda y$ and $z\rightarrow\lambda z$. – TheAverageHijano Feb 15 at 8:33

If $$\boldsymbol r = \begin{vmatrix} x \\ y \\ z \\ \end{vmatrix} = \begin{vmatrix} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \\ \end{vmatrix}$$ the associated volume element is $$d^3 \boldsymbol r = \begin{vmatrix} r \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ r \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ r \cos \theta & - r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = r^2 \sin \theta dr d \theta d \phi$$ Similarly, if $$\lambda \boldsymbol r = \begin{vmatrix} \lambda x \\ \lambda y \\ \lambda z \\ \end{vmatrix} = \begin{vmatrix} \lambda r \sin \theta \cos \phi \\ \lambda r \sin \theta \sin \phi \\ \lambda r \cos \theta \\ \end{vmatrix}$$ the associated volume element is $$d^3 ( \lambda \boldsymbol r ) = \begin{vmatrix} \lambda r \sin \theta \cos \phi & \lambda r \cos \theta \cos \phi & - \lambda r \sin \theta \sin \phi \\ \lambda r \sin \theta \sin \phi & \lambda r \cos \theta \sin \phi & \lambda r \sin \theta \cos \phi \\ \lambda r \cos \theta & - \lambda r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = ( \lambda r)^2 \sin \theta dr d \theta d \phi = \lambda^2 ( r^2 \sin \theta dr d \theta d \phi )$$ So, comparing the two expressions above for $$d^3 \boldsymbol r$$ and $$d^3 ( \lambda \boldsymbol r )$$, you get $$d^3 ( \lambda \boldsymbol r ) = \lambda^2 d^3 \boldsymbol r$$ or $$d^3 r = \frac{1}{\lambda^2} d^3 ( \lambda \boldsymbol r )$$

I think you forget pieces of the integral that gives the potential :

$${{d}^{3}}(\lambda \overrightarrow{r'})={{\lambda }^{3}}{{d}^{3}}(\overrightarrow{r'})$$ as indicated in the comment.

But : $$\int{{{d}^{3}}(\overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \overrightarrow{r}-\overrightarrow{r'} \right|}}=\frac{1}{{{\lambda }^{2}}}\int{{{d}^{3}}(\lambda \overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \lambda \overrightarrow{r}-\lambda \overrightarrow{r'} \right|}}$$