Problem in understanding orthogonal curvilinear coordinate like spherical coordinate or cylindrical coordinate Can someone explain your understanding to the orthogonal curvilinear coordinate like spherical coordinate or cylindrical coordinate, which i think is quite strange. I can actually do computation to problems with curvilinear coordinate like calculating the surface integral or the surface area of a surface using some standard way but i couldn't understand the meaning behind them. For instance, for a vector expressed in the form of standard basis, we can directly sum them up to get the resultant force but when it is expressed in spherical coordinate like $F=r^2\vec r+\theta  \vec \theta$ in this case you cannot directly sum the force, so when calculating a net for over a surface, if the force density are provided in terms of spherical coordinate, we couldn't directly use surface integral to get the answer. Anythings related are great, even tiny little things would help to make the concept more clear.Thx
 A: I'm not sure if I've understood your problem. But I'll give my best to explain this. Also, since you just want to get a better intuition of what's going on I'll not dive deep on rigourous constructions. First, intuitively: what's one possible way to interpret curvilinear coordinates ? Well, you can think like this: when you use the standard cartesian coordinate system in $\mathbb{R}^3$ you're representing a point as the intersection of $3$ planes. In other words, the point $p = (a,b,c)$ is the intersection of the planes $x =a$, $y =b$ and $z=c$. In $\mathbb{R}^n$ it's similar: you can think of a point as the intersection of $n$-hyperplanes (in $\mathbb{R}^2$ hyperplanes are lines and this works also).
Why not making this more general ? Let's then try to locate a point by intersecting $n$ hypersurfaces which need not to be hyperplanes. For instance: polar coordinates in $\mathbb{R}^2$, you find each point $(r,\theta)$ by the intersection of a circle and a line. Of course, that's just one interpretation.
Other way to look at it is by look at the coordinate lines. In $\mathbb{R}^n$ if you use standard cartesian coordinates, the coordinate lines will be the coordinate axis. However, on curvilinear systems those lines have non zero curvature (justifying the name curvilinear coordinates).
But how you think of vectors in such systems and how the hell you give this a little mathematical substance ? Well, we want to think of vectors as tangent to the coordinate lines, so we can do the following: consider $\phi : U \subset \mathbb{R}^n \to \mathbb{R}^n$ a coordinate transformation. One simple example is that of polar coordinates on the plane, where you have $n =2$ and the subset you're working with is 
$$U = \left\{(r, \theta) \in \mathbb{R}^2 \mid r\geq 0, \ \ 0 \leq \theta \leq 2\pi\right\}$$ 
And the change of coordinates is 
$$\phi(r,\theta) = (r \cos\theta, r\sin\theta)$$
Fixing $r$ and varying $\theta$ you get the parametrization of one coordinate line. Fixing $\theta$ and varying $r$ you get the parametrization of another coordinate line. The tangent vectors to these curves will depend on each point and will give you one different basis for expressing any vector at each point.
Well, this is just a little of what can be said. There's much more, and there are many levels of rigour you can use. Indeed all of this can be made much more general and precise using the notion of charts and pushforward. The chart gives the transformation of the coordinates, mapping points to new points, and the pushforward gives the transformation of vectors at the point.
Well, what's most important to understand is: curvilinear coordinates are called this way because the coordinate lines have non zero curvature, you cand understand a point located by curvilinear coordinates as the point obtained by intersecting hypersurfaces, each coordinate line gives one of the basis vectors you'll use to express the others in this system.
I hope this gives you a better idea of what you are doing when you work with curvilinear coordinates. Also, if you have any doubt, ask on comment. Good luck!
