This question is a follow-up to a previous question.
I am dealing with the following type of integral:
$$ f(\vec x) = \int_{\mathbb R} \text{d}^3y\; (\vec x\cdot \vec y)^n H(A-|\vec y-\tfrac{1}{2}\vec x|)H(B-|\vec y+\tfrac{1}{2}\vec x|)\qquad A,B>0 $$
I know that the integration region is a sphere-sphere-intersection. For $n=1$ the integral should vanish due to symmetry arguments. But how would I even start to tackle this integral for $n=2$? (Let's settle on $n=2$ for this question)
Some of my thoughts:
- Maybe spherical coordinates are useful? In order to write $(\vec x\cdot\vec y)^2$ as $x^2y^2\cos^2\theta$ with $\theta=\angle(\vec x,\vec y)$
- When I see $H(a-x)$ simply as the condition $x\leq a$, I can deduce the following inequalities: $$ \begin{align} \frac{x}{2}\cos\theta-\frac{1}{2}\sqrt{4A+x^2(\cos^2\theta-1)} &\leq y \leq \frac{x}{2}\cos\theta+\frac{1}{2}\sqrt{4A+x^2(\cos^2\theta-1)}\\ -\frac{x}{2}\cos\theta-\frac{1}{2}\sqrt{4B+x^2(\cos^2\theta-1)} &\leq y \leq -\frac{x}{2}\cos\theta+\frac{1}{2}\sqrt{4B+x^2(\cos^2\theta-1)} \end{align} $$ They both give me a range for $y\equiv |\vec y|$.
- Maybe I should separate the $\theta$ integration from 0 to $\pi/2$ and then from $\pi/2$ to $\pi$. This way I’m covering both spheres separately.