You can estimate the size of any diagram by realizing that $f\left(r\right)$ is of order 1 out to a distance of about the diameter of a molecule, and$f\approx 0$ beyond that. Hence, a three-dimensional integral of a product of f's will generally give a result that is of the order of the volume of a molecule. Estimate the sizes of all the diagrams shown explicitly in equation$8.20$ and explain why it was necessary to rewrite the series in exponential form.

We need to find the necessary to rewrite the series in exponential form.

In this problem we need to write size of the diagrams in equation , the diagrams are shown in the following figure, for simplicity I will indicate each equation by its number in the figure. The integral for these diagrams are:

$$\begin{gathered} \left(1\right)=\frac{1}{2}\frac{N\left(N-1\right)}{V^2}\int d^3{r}_1{d}^3{r}_2{f}_{12} \\ \left(2\right)=\frac{1}{2}\frac{N\left(N-1\right)\left(N-2\right)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23} \\ \left(3\right)=\frac{1}{8}\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)}{V^4}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{12}{f}_{34} \\ \left(4\right)=\frac{1}{6}\frac{N\left(N-1\right)\left(N-2\right)}{V^3}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{f}_{12}{f}_{23}{f}_{31} \\ \left(5\right)=\frac{1}{6}\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)}{V^4}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{14}{f}_{24}{f}_{34} \\ \left(6\right)=\frac{1}{4}\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)\left(N-4\right)}{V^5}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{f}_{12}{f}_{34}{f}_{45} \\ \left(7\right)=\frac{1}{8}\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)\left(N-4\right)\left(N\right)}{V^6} \\ \times \int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{d}^3{r}_5{d}^3{r}_6{f}_{12}{f}_{34}{f}_{56} \\ \left(8\right)=\frac{1}{6}\frac{N\left(N-1\right)\left(N-2\right)\left(N-3\right)}{V^4}\int d^3{r}_1{d}^3{r}_2{d}^3{r}_3{d}^3{r}_4{f}_{13}{f}_{12}{f}_{24} \end{gathered}$$

according to the above equation, we can write (realizing that $f\left(r\right)$is in order of $1$out of the distance about the diameter of the molecule and $f\approx 0$beyond that):

$\left(1\right)\approx \frac{Nv}{V}$

what I have done here is just approximate $N\approx N-1$, and since we have only one factor of we have one factor of $f$where is the sized volume of the molecule and the integration over the renaming factor of $r$gives the volume so the volume in the denominator reduced by one order. For the second diagram we get:

$\left(2\right)\approx \frac{N^3{v}^2}{V^2}$

and so on, we can write:

$\left(3\right)\approx \frac{N^4{v}^3}{V^3}$

For the fourth diagram, we need to change the coordinates of the diagram as follow:

$$\begin{gathered} {\overrightarrow{r}}_a={\overrightarrow{r}}_1-\overrightarrow{r} \\ {\overrightarrow{r}}_b={\overrightarrow{r}}_1-{\overrightarrow{r}}_3 \end{gathered}$$

so we can write:

$$\begin{gathered} \left(4\right)=\frac{1}{6}\frac{N^3}{V^2}\int d^3{r}_a{d}^3{r}_{b}f\left(r_a\right)f\left(r_b\right)f\left({\overrightarrow{r}}_a-{\overrightarrow{r}}_b\right) \\ \left(4\right)\approx \frac{N^3{v}^2}{V^2} \end{gathered}$$

Following the same method above, we can write the size for the rest of the diagram as:

$$\begin{gathered} \left(5\right)\approx \frac{N^4{v}^3}{V^3}\left(6\right)\approx \frac{N^5{v}^3}{V^3} \\ \left(7\right)\approx \frac{N^6{v}^2}{V^2}\left(8\right)\approx \frac{N^4{v}^3}{V^3} \end{gathered}$$