# Angle between vectors in higher dimensional integral

If I want to calculate a 3D-integral that contains the product of two vectors I can write $\int f(\vec{x}) e^{-i\vec{k}\cdot\vec{x}}\mathrm{d}^3 x = \int f(\vec{x})e^{-ikr\cos{\theta}} r^2\sin{\theta}\:\mathrm{d}r\:\mathrm{d}\theta\:\mathrm{d}\phi$

From wikipedia I know the higher dimensional volume element $r^{n-1}\sin^{n-2}{\phi_1}\sin^{n-3}\phi_2\dots\sin\phi_{n-2}\:\mathrm{d}r\:\mathrm{d}\phi_1\dots$

Which of these angles does the the theta in the first integral correspond to and why?

• Why don’t you try working out the change of coordinates for yourself? – amd Jun 28 '18 at 17:26
• I think I understand now that it is just a matter of orienting the coordinate system in a way that simplifies the integration the most. If I take the angle betweeen the vectors to be theta in the 3D-case, I can eliminate the $\sin\theta$ factor with the substitution $t=\cos\theta$. If it were phi I would get a Bessel function from the phi integration. If the f(x) doesn't depend on any angles I would orient the coordinates so that theta coincides with $\phi_{n-2}$ in higher dimensions. Sorry, I should have thought about this more on my own. – blackholedynamite Jun 28 '18 at 22:42