# Spherical coordinates surface integral

1. Aug 7, 2009

### daudaudaudau

Hi.

I have this integral
$$\int_0^{2\pi}\int_0^\pi \mathbf A\cdot\hat r d\theta d\phi$$
where $\hat r$ is the position unit vector in spherical coordinates and $\mathbf A$ is a constant vector. Is it possible to evaluate this integral without calculating the dot product explicitly, i.e. without knowing that that $\hat r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ ?

Thanks.

2. Aug 7, 2009

### GPPaille

I guess that the answer is 0:

$$\int_0^{2\pi}\int_0^{\pi}{A\cdot\widehat{r}d\theta d\phi} = A \cdot \int_0^{2\pi}\int_0^{\pi}{\widehat{r}d\theta d\phi}$$

But you are now integrating normal vectors over the sphere, which is perfectly symmetric. Every vector can be summed with its opposite to give 0.