# Surface integral

#### bruteforce

I want to integrate something in spherical coordinates
I have $$da=R^{2}sin(g)dgdh \hat{r}$$ with g and h angles
and $$\hat{r}=sin(g)cos(h) \hat{i}+sin(g)sin(h) \hat{j}+cos(g) \hat{k}$$

But what is now $$da_{x}=dydz \hat{i}$$ in spherical coordinates?
So I have the expression in ordinary coordinates and need to find it in spherical coordinates

thanks

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#### HallsofIvy

Homework Helper
On a spherical surface of radius 1, $\vec{r}= sin(\phi)cos(\theta)\vec{i}+ sin\phi)sin(\theta)\vec{j}+ cos(\phi)\vec{k}$ as you say.

Differentiating with respect to each variable,
$\vec{r}_\theta= -sin(\phi)cos(\theta)\vec{i}+ sin(\phi)cos(\theta)\vex{j}$
$\vec{r}_\phi= cos(\phi)cos(\theta)\vec{i}+ cos(\phi)sin(\theta)\vec{j}- sin(\phi)\vec{k}$

The "fundamental vector product" of a surface is the cross product of those two derivative vectors:
$$\left|\begin{array}\vec{i} & \vec{j} & \vec{k} \\ -sin(\phi)sin(\theta) & sin(\phi)cos(\theta) & 0 \\ cos(\phi)cos(\theta) & cos(\phi)sin(\theta) & -sin(\phi)\end{array}\right|= -sin^2(\phi)cos(\theta)\vec{i}- sin^2(\phi)sin(\theta)\vec{j}- sin(\phi)cos(\phi)\vec{k}$$
has length $sin(\phi)$ so the differential of surface area is $sin(\phi)d\phi d\theta$.

(In general if a surface is given by $\vec{r}(u, v)$ with parameters u and v, then the differential of surface area is $\left|\vec{r}_u\times\vec{r}_v|dudv$. Thats's worth knowing! In fact, the "vector differential" $\vec{r}_u\times\vec{r}_v du dv$ is a vector having the differential of surface area as length, normal to the surface and can be used to integrate vector fields over the surface.)

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