Finding da_x in Spherical Coordinates

[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

 

[HallsofIvy]

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.)