$\iiint \frac{1}{x^2+y^2+(z-2)^2}dA$ where $A=\{x^2+y^2+z^2 \leq 1\}$ check my answer!

Question

I would like someone to review my solution please, the original question is to calculate

$\iiint \frac{1}{x^2+y^2+(z-2)^2}dA$ where $A=\{x^2+y^2+z^2 \leq 1\}$

What I did:

First I changed variables to polar coordinates in order to simplify the boundaries:

$x=r\sin\theta cos\phi$, $y=r\sin\theta \sin\phi$, $z=r\cos\theta$, $0\leq r \leq 1$, $0\leq \theta \leq \pi$, $0 \leq \phi \leq 2\pi$ and the jacobian is $r^2\sin\theta$.

to get the following result:

$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta d\phi$$

We can see that $\phi$ doesn't appear in any integral, so we can just multiply by $2\pi-0=2\pi$ and forget about it:

$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta d\phi =2\pi \int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta$$

now I used substitution, $\alpha=-cos\theta$, $d\alpha =sin\theta d\theta$, $-1 \leq \alpha \leq 1$

$$2\pi \int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta=2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2sin\theta}{r^2+2r\alpha+4}dr\frac{d\alpha}{sin\theta}$$

cancel the sine to get:

$$2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2}{r^2+2r\alpha+4}drd\alpha$$

this integral is very difficult (at least wasn't apparent to me) if we integrate by $r$. luckily, Fubini's theorem states that we can integrate by $\alpha$ first and the result won't change:

$$2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2}{r^2+2r\alpha+4}drd\alpha=2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2+2r\alpha+4}d\alpha dr $$

The antiderivative of $$\frac{1}{r^2+2r\alpha+4}$$ with respect to alpha is $$\frac{\ln(r^2+2r\alpha+4)}{2r}$$

since $\alpha$ goes from $-1$ to $1$: $$\frac{\ln(r^2+2r+4)}{2r}-\frac{\ln(r^2-2r+4)}{2r} = \frac{\ln(r+2)^2-\ln(r-2)^2}{2r}=\frac{\ln(\frac{r+2}{r-2})}{r}$$ so:

$$2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2+2r\alpha+4}d\alpha dr=2\pi \int_{0}^{1}r\ln(\frac{r+2}{r-2})dr$$

Now I integrated by parts $\int uv'=uv-\int u'v$ where $u=\ln(\frac{r+2}{r-2})$, $u'=\frac{-4}{r^2-4}$, $v'=r$, $v=\frac{r^2}{2}$:

$$\int r\ln(\frac{r+2}{r-2})dr=\frac{r^2ln(\frac{r+2}{r-2})}{2}-\int\frac{-2r^2}{r^2-4}dr =\frac{r^2ln(\frac{r+2}{r-2})}{2}+2(r+\ln(\frac{r-2}{r+2})+2)$$

when $r=1$:

$$\frac{1^2ln(-3)}{2}+2(1+\ln(\frac{-1}{3})+2)=\frac{1}{2}\ln(-3)-2\ln(-3)+6=-\frac{3}{2}\ln(3)-\frac{3\pi i}{2}+6$$

when $r=0$:

$$0+2(0+\ln(-1)+2)=2\ln(-1)+4=2\ln(1)+2\pi i+4=4+2\pi i$$

Subtract the 2 results to get: $$-\frac{3}{2}\ln(3)-\frac{3\pi i}{2}+6-4-2\pi i=2-\frac{3}{2}\ln(3)-\frac{7\pi i}{2}$$

Multiply this result by $2\pi$ to reach the final answer which is $$4\pi-3\pi \ln(3)-7\pi^2i$$

Is this indeed the correct answer? Is there a better way of solving this? this seems like a very difficult way to do the exercise.

Answer 1

\begin{align*} \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{\pi} \frac{r^2\, \sin\theta}{r^2-4\, r\, \cos\theta+4}\, d\theta \, dr\, d\phi &= 2\, \pi\, \int_{0}^{1} \, \frac{r}{4}\, \log(r^2-4\, r\, \cos\theta +4) \Big|_0^\pi \, dr\\ &= 2\, \pi\, \int_{0}^{1} \, \frac{r}{4}\, \left(\log{(r+2)^2}-\log{(r-2)^2}\right)\, dr\\ &= 2\, \pi\, \Big(\frac{r^2}{8} \, \log(r+2)^2 - \frac{r^2}{8} \, \log(r-2)^2 + r - \log(r + 2) + \log(r - 2) \Big) \Big\vert_0^1\\ &= \pi\, \left(2-\frac{3}{2}\log{3}\right) \approx 1.1060968643 \end{align*}