In the first section the formula (19) for the distribution of electric potential along the radius of the ball, depending on the distribution of volume charge density. Once again, we recall it here: \[\varphi(r) = {1 \over \varepsilon_0} \left( \frac{1}{r} \int \limits_0^r r^2 \rho (r) \Bbb{d} r + \int \limits_r^R R\, \rho (r) \Bbb{d} r \right) \qquad (1)\] The objective of this section is to deduce an approximate formula for the potential difference between the layer surface of the bowl and layer on some known depth \(h\), and it is at least several orders of magnitude smaller than the full radius of the ball: \(R \gg h\). She is using the absolute difference of potentials at two points: \[\Delta \varphi(h) = \varphi(R) \varphi(R-h) \qquad (2)\] Therefore: \[\Delta \varphi(h) = {1 \over \varepsilon_0} \left( \frac{1}{R} \int \limits_0^R R^2 \rho (r) \Bbb{d} r - \frac{1}{R-h} \int \limits_0^{R-h} r^2 \rho (r) \Bbb{d} r + \int \limits_R^R R\, \rho (r) \Bbb{d} r \int \limits_{R-h}^R R\, \rho (r) \Bbb{d} r \right) \qquad (3)\] As you can see, third from the left integral is equal to zero, so the desired formula takes the following form: \[\Delta \varphi(h) = {1 \over \varepsilon_0} \left( \frac{1}{R} \int \limits_0^R R^2 \rho (r) \Bbb{d} r - \frac{1}{R-h} \int \limits_0^{R-h} r^2 \rho (r) \Bbb{d} r - \int \limits_{R-h}^R R\, \rho (r) \Bbb{d} r \right) \qquad (4)\] Due to the approximate expansion in Maclaurin series [1] we know that: \[\frac{1}{R-h} \approx {1 + \delta + \delta^2 \over R}, \quad \delta= \frac{h}{R} \qquad (5)\] In this decomposition, hereafter, we will apply the precision to the second term. Now the expression (4) we can write as \[\Delta \varphi(h) = {1 \over \varepsilon_0} \left( \frac{1}{R} \int \limits_{R-h}^R R^2 \rho (r) \Bbb{d} r \int \limits_{R-h}^R R\, \rho (r) \Bbb{d} r - {\delta + \delta^2 \over R} \int \limits_0^{R-h} r^2 \rho (r) \Bbb{d} r \right) \qquad (6)\] the Difference between the upper and lower boundaries of the first and the second integral is very small, and therefore this value can be found using numerical integration [2], in particular, the method of trapezoids. To calculate such integrals, we divide the interval (\(R..h\)) on two plots: (\(R..h/2\)) and (\(R-h/2..R-h\)). Then approximate the integral would be: \[\int \limits_{R-h}^R f(r) \Bbb{d} r \approx \frac{h}{2} \left({f(R) + f(R-h) \over 2} + f(R - h/2) \right) \qquad (7)\] by Converting the thus the first and the second integral in (6), summarizing and reducing the obtained values, we give this expression to the following form: \[\Delta \varphi(h) \approx {h \over \varepsilon_0} \left( {h \over 2} \rho (R) + {1 + h/R \over R^2} \int \limits_0^{R-h} r^2 \rho (r) \Bbb{d} r \right) \qquad (8)\] This is the desired approximate formula for finding the potential difference between the surface layer and ball layer at a depth \(h\), where \(R \gg h\). The accuracy of calculations exceeds \((h/R)^2\).

To do this, imagine the right integral from the expression (8) in this form: \[ \int \limits_0^{R-h} r^2 \rho (r) \Bbb{d} r = \int \limits_0^{R} r^2 \rho (r) \Bbb{d} r \int \limits_{R-h}^{R} r^2 \rho (r) \Bbb{d} r \qquad (9)\] In the resulting formula for the second integral are very close to the border and can be approximately found using the expression (7), just as we did before. After some transformations we get a more convenient option desired formula: \[\Delta \varphi(h) \approx {h \over \varepsilon_0} \left( {1 + h/R \over R^2} \int \limits_0^{R} r^2 \rho (r) \Bbb{d} r - {h \over 2} \rho (R) \right) \qquad (10)\] It is the upper bound of the first integral is sought is already full radius, which should greatly simplify the approximation of complex functions.