Trodimensional Fourier transform

[LagrangeEuler]

\mathcal{F}\{f(r)\}=\int e^{i\vec{k}\cdot \vec{r}}f(r)d\vec{r}
in spherical polar coordinates
\mathcal{F}\{f(r)\}=\int^{\infty}_0r^2dr\int^{\pi}_0\sin\theta d\theta\int^{\pi}_0d\varphi e^{ikr\cos \theta}f(r)

Why could I take $e^{ikr\cos \theta}$ and to take that $\theta$ is angle which goes from zero to $\pi$. Thanks for the answer.

 

[mathman]

You need to take φ from 0 to 2π in order to integrate over all space. As for your question (as best as I can understand it), the dot product between two vectors is the product of magnitude of the vectors and cosθ, where θ has the range [0,π].

 

[LagrangeEuler]

\mathcal{F}\{f(r)\}=\int e^{i\vec{k}\cdot \vec{r}}f(r)d\vec{r}
in spherical polar coordinates
\mathcal{F}\{f(r)\}=\int^{\infty}_0r^2dr\int^{\pi}_0\sin\theta d\theta\int^{\pi}_0d\varphi e^{ikr\cos \theta}f(r)

Why could I take $e^{ikr\cos \theta}$ and to take that $\theta$ is angle which goes from zero to $\pi$. Thanks for the answer.

Could I use this also?
\mathcal{F}\{f(\vec{r})\}=\int^{\infty}_0r^2dr\int^{\pi}_0\sin\theta d\theta\int^{\pi}_0d\varphi e^{ikr\cos \theta}f(\vec{r})

I thought that I can use $e^{i\vec{k}\cdot \vec{r}}=e^{ikr\cos \theta}$, only when $f(r)$ is radial function.