showing orthonormality of functions Let $ F$ be an unit sphere with $ F:= \partial K_1(o) \subset \mathbb{R}^3 $
For all continous functions $f,g \in C(F)$ define
$ <f,g>_n := \int_F f(x)g(x) do(x) $
And let
$ F_1(x( \theta, \phi)):= \sqrt{ \frac{3}{4 \pi }} sin \theta cos \phi $
$ F_2(x( \theta, \phi)):= \sqrt{ \frac{3}{4 \pi }} sin \theta sin \phi $
$F_3(x( \theta, \phi)):= \sqrt{ \frac{3}{4 \pi }} sin \theta cos \theta cos \phi$
$x( \theta, \phi) $ is the presentation of $x$ in $F$ in polarcoordinates $ \theta \in (o, \pi), \phi \in (0, 2\pi) $
I need to show that those functions are pairwise orthnormal.
okay, let's put 
$<F_1,F_2> = \int_F\sqrt{ \frac{3}{4 \pi }} sin \theta cos \phi\sqrt{ \frac{3}{4 \pi }} sin \theta sin \phi do(x) $
Now I am confused, what to put for the bounds for the integral? 
 A: Note that You integrate over a manifold $F=\partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
$$\Psi:(0,\pi)\times(0,2\pi)\rightarrow F,(\theta,\varphi)\mapsto (\sin\theta\cos\varphi,\sin\theta\sin\varphi,cos\theta).$$
Now You compute
$$\frac{\partial\Psi}{\partial\theta}=\begin{pmatrix}\cos\theta\cos\varphi\\ \cos\theta\sin\varphi\\-\sin\theta\end{pmatrix},\frac{\partial\Psi}{\partial\varphi}=\begin{pmatrix}-\sin\theta\sin\varphi\\\sin\theta\cos\varphi\\0\end{pmatrix}$$
and $|\frac{\partial\Psi}{\partial\theta}|^2=1,|\frac{\partial\Psi}{\partial\varphi}|^2=\sin^2\theta,\langle\frac{\partial\Psi}{\partial\theta},\frac{\partial\Psi}{\partial\varphi}\rangle=0$ to get the metric tensor :
$$G=\begin{pmatrix}1&0\\0&\sin^2\theta\end{pmatrix}.$$
Now for example
$$\langle F_1,F_2\rangle=\int_{(0,\pi)\times(0,2\pi)}F_1(\theta,\varphi)F_2(\theta,\varphi)\sqrt(|\det G|)d(\theta,\varphi)=$$
$$\int_0^{2\pi}\int_0^{\pi}\sqrt{\frac{3}{4\pi}}\sin\theta\cos\varphi\sqrt{\frac{3}{4\pi}}\sin\theta\sin\varphi|\sin\theta|d\theta d\varphi=$$
$$=\frac{3}{4\pi}\int_0^{\pi}\sin^3\theta d\theta\underbrace{\int_0^{2\pi}\cos\varphi\sin\varphi d\varphi}_{=0}=0$$
and 
$$\langle F_1,F_1\rangle=\int_0^{2\pi}\int_0^{\pi}(\sqrt{\frac{3}{4\pi}}\sin\theta\cos\varphi)^2|\sin\theta|d\theta d\varphi=$$
$$\frac{3}{4\pi}\underbrace{\int_0^{\pi}\sin^3\theta d\theta}_{=\frac{4}{3}}\underbrace{\int_0^{2\pi}\cos^2\varphi d\varphi}_{=\pi}=1$$.
You proceed similar with the other products.
