$$\int\!\!\!\int\!\!\!\int_{V}\frac{(b-x)\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z}{\left(\sqrt{(b-x)^2+y^2+z^2}\right)^3},(b>a>0)$$ $$V:x^2+y^2+z^2\leq a^2$$
Let $x=r\sin\varphi\cos\theta,y=r\sin\varphi\sin\theta,z=r\cos\varphi,J=r^2\sin\varphi.$but i still can't solve the problem.it's too complex to calculate.does anyone have a better way?
what i think:
$$\int\!\!\!\int\!\!\!\int_{V}\frac{(b-x)\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z}{\left(\sqrt{(b-x)^2+y^2+z^2}\right)^3}=\oint_{x^2+y^2+z^2= a^2} \frac{\:\mathrm{d}y\:\mathrm{d}z}{\sqrt{(b-x)^2+y^2+z^2}}$$ but it's still not easy to calculate
When integrating over the ball $V$ of radius $a$ we choose the order of integration such that the integral with respect to $x$ is the outermost one.
For given $x\in[{-a},a]$ the variables $(y,z)$ vary over a disk $D_x$ of radius $\sqrt{a^2-x^2}$ centered at $(0,0)$. Furthermore we note that the integrand as a function of $y$ and $z$ is rotationally invariant. We therefore let $\rho:=\sqrt{y^2+z^2}$ and then obtain the following inner integral (the angular integral is for free): $$I(x):=\int_{D_x}{b-x\over\sqrt{(b-x)^2+y^2+z^2}}\>{\rm d}(y,z)=2\pi\int_0^{\sqrt{a^2-x^2}}{b-x\over\sqrt{(b-x)^2+\rho^2}}\>\rho\>d\rho\ .$$ This gives $$I(x)=2\pi(b-x)\sqrt{(b-x)^2+\rho^2}\>\Biggr|_{\rho=0}^{\rho=\sqrt{a^2-x^2}}=2\pi(b-x)\left(\sqrt{b^2+a^2-2bx}-(b-x)\right)\ .$$ The final result is then $$=\int_{-a}^aI(x)\>dx\ ,$$ which I may leave to you.