$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }$ D is given by $x^2+y^2+z^2 \leq 1$

Question

$$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }$$ D is given by $x^2+y^2+z^2\leq1$

I try to use $ \left\{\begin{matrix} x=r\sin \phi \cos \theta \\ y=r\sin \phi \sin \theta \\ z=r\cos\phi \end{matrix}\right. $ while the Jacobian is $r^2 \sin \phi$ and $ \left\{\begin{matrix} 0\leq \theta \leq 2\pi\\ 0\leq \phi \leq \pi \\0\leq r \leq 1\end{matrix}\right. $ $$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }\Rightarrow\int _{0}^{2\pi}d\theta \int _{0}^{\pi}d\phi \int _{0}^{1}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}dr$$ but it seems tough to do next.

Answer 1

$$I=\int _{0}^{2\pi}d\theta \int _{0}^{\pi}d\phi \int _{0}^{1}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}dr=2\pi\int _{0}^{1}dr \int _{0}^{\pi}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}d\phi.$$ Let $\cos\phi=t \implies \sin\phi d\phi=-dt$, then $$I=2\pi \int_{0}^{1}dr \int_{-1}^{1}\frac{r^2 dt}{\sqrt{r^2-rt+1/4}} dt=2\pi\int_{0}^{1} 2r (r+1/2)-|r-1/2|] dr =2\pi[\int_{0}^{1/2}4r^2 dr +\int_{1/2}^{1}2rdr ]$$ $$=2\pi[1/6+(1-1/4)]=\frac{11}{6}\pi.$$