User:Skeptical scientist/scratchwork

$$\frac{1}{z}=\frac{1}{x+iy}=\frac{1}{x+iy}\cdot\frac{x-iy}{x-iy}=\frac{x-iy}{x^2+y^2}$$ $$\iiint_C \frac{G\rho(\vec{r}-(R-d,0,0))}{||\vec{r}-(R-d,0,0)||^3} d^3V(\vec{r})$$ $$\int_0^\infty 1 m^2 \cdot \rho(r) \, dr = \int_0^\infty 1 m^2 \cdot \rho_0e^{-kgr} \, dr = \left [ -\frac{\rho_0 \cdot 1 m^2}{kg}e^{-kgr} \right ]_0^\infty=\frac{\rho_0 \cdot 1 m^2}{kg},$$ $$g\int_h^\infty \rho(r) \, dr = g\int_h^\infty \rho_0e^{-kgr} \, dr = g \left [ -\frac{\rho_0 }{kg}e^{-kgr} \right ]_h^\infty=\frac{\rho_0 e^{-kgh}}{k}=\frac{\rho(h)}{k}=P(h).$$

$$\int_0^{2\pi}\int_0^\infty \frac{Gmd}{h^2+r^2} \frac{h}{\sqrt{r^2+h^2}} r dr d\theta$$ $$=\pi \int_0^\infty \frac{Gmdh}{(h^2+r^2)^{3/2}} 2r dr$$ $$=\pi \int_{h^2}^\infty \frac{Gmdh}{u^{3/2}} du$$ $$=\pi \left[ -2\frac{Gmdh}{\sqrt{u}} \right]_{h^2}^\infty$$