Calculating pressure inside a solid sphere

Let's imagine dividing a planet into a nearly infinite number of very thin layers, each with a thickness $dr$ . Consider a layer at distance $r$ from the center. At the top of this layer, the weight per unit area that is compressing the material and the equal pressure opposing that weight would be given by $w_r$ and $P_r$ . At the bottom of the layer, however, the two (equal) values must be slightly greater, because there is some extra weight within the layer, $dw$ , which would be equal to the density of the material in the layer, $\rho$ , multiplied by the volume of the layer, which is $dr$ times the area of the layer times the local value of the planet's internal gravity, $g_r$ .

$dw = \rho A dr g_r.$ As pressure is force per unit area, $dP = \frac{dw}{A} = \rho dr g_r.$ The total pressure can be calcuated by adding the pressure by all the layers above this layer.
$\begin{aligned} P &= \int_r^R \rho \frac{GMr}{R^3} \, dr \\ &= \frac{\rho GM}{2R^3} (R^2-r^2). \\ \end{aligned}$ Substitution of $\rho = \dfrac{3M}{4\pi R^3}$ , we get, $P = \dfrac {3}{8} \dfrac{GM^2}{\pi R^4} \left( 1 - \dfrac{r^2}{R^2} \right).$

This gives us, $a+b = \boxed{11}.$