Let $e_1=(1,0,2), e_2:=(0,1,0)$ be the vector in $\mathbb R^3$.
We define $D:=\{(x,y,z)\in\mathbb {R}^3: x+2z=0; x^2+y^2+z^2\leq 1 \}$. I would like to compute exactly the following integral:
$$ A:=\int_D \left(\frac{1}{\Bigr((x+1)^2+(y+1)^2+(z+2)^2 \Bigr)^{10}} - \frac{1}{\Bigr(1^2+1^2+2^2 \Bigr)^{10}} \right) dS $$
Here $\int_D f dS$ means the surface integral of $S$ on defined surface domain $D$.
I want to know that $A$ is positive or negative
I am tried to change variable $(x,y)\mapsto (x,y,-x/2)$ or use the spherical coordinates. But the computation is very complicated because the function and the domain are not symmetric together!
Can you suggest me how to compute exactly the integral?
Consider translating to a coordinate system centered at $(-1,-1,-2)$ and rotated such that the new $z$ unit vector is parallel to the old $(1,0,2)$ vector and $y$ remains unrotated. This gives us the new integral
$$I = \int_{D'}\frac{1}{(x^2+y^2+z^2)^{10}}-\frac{1}{6^{10}}dS$$
where
$$D' = \{(x,y,z)\in\Bbb{R}^3:z = \sqrt{5}; x^2+(y-1)^2+(z-\sqrt{5})^2\leq 1\}$$
which simplifies the integral even further to
$$\int\limits_{x^2+(y-1)^2\leq 1}\frac{dA}{(x^2+y^2+5)^{10}}-\frac{\pi}{6^{10}}$$
The integral becomes
$$\int_0^\pi\int_0^{2\sin\theta}\frac{r\:dr\:d\theta}{(r^2+5)^{10}} = \frac{1}{18\cdot5^9}\int_0^\pi1-\frac{1}{(\frac{4}{5}\sin^2\theta+1)^9}\:d\theta$$
Using the substitution $t=\tan\theta$ we can get that integral evaluates to
$$\frac{1}{18\cdot5^9}\left(\pi - 2\int_0^\infty\frac{(t^2+1)^8dt}{\left(\frac{9}{5}t^2+1\right)^9}\right) = \frac{\pi}{18\cdot5^9}\left(1-\frac{1487675\sqrt{5}}{14348907}\right)$$
which is less than $\frac{\pi}{6^{10}}$, therefore $A$ is negative.