I have seen a picture for $dV$ so that $dV = r^{2} \sin(\theta)\,dr\,d\theta\,d\phi$. But how can I deduce things like $dA$ and $dV$? In a simpler coordinate (not sure about the name), $dA = r \,dr\,d\phi$, again no idea how to derive it (just given to me to memorize it). So explain what is $dA$ in polar coordinates?
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$\begingroup$Completing rcollyer's answer
$$\iint_S f(x,y) dx dy = \iint_S f(x(u,v),y(u,v)) \det[J(u,v)] du dv$$
for 2-D polar coordinates $u=r$ and $v=\theta$
Find out $\det[J(r,\theta)] =r$
Therefore $dA= r dr d\theta$
So in general your area element is $dA=\det[J(r,\theta)] du dv$
Jacobians are used extensively in Statistical mechanics, but usually most physics problems come in flavors of one in three coordinate systems(cart, spherical, cylindrical) so we do not bother with the Jacobian when geometric intuition gets us the right answer for these three cases readily, which I have written in the comment.
EDIT:
The Jacobian is just a product of the scale factors when you change from one (orthogonal) coordinate system to another. Please note, that all this is a part of nonstandard analysis which exploits the notion of an infinitesimal. Start working from 2-D. Your area element would be a region bounded by the lines $x,x+\Delta x,y, y+\Delta y$ and your area element is $\Delta x \Delta y$. Finally take the limit in $\Delta x ,\Delta y\rightarrow 0$ and this is understood when you write $\int \int dx dy$. Repeat with polar coordinates. To locate a curve start with the constant coordinate $r$ which is a circle of radius $r$. Then draw a curve of constant $\theta$. This is a ray which goes from the origin to infinity making an angle $\theta$ with the positive x axis. Now draw another constant curve $r+\Delta r$ and $\theta + \Delta \theta$. Look at the intersection of these two. When you take the limits, this can be approximated as a rectangle, so, area is $\Delta r (r \Delta \theta)$. The factor $r$ which converts the infinitesimal change in $\theta$ ti the corresponding displacement $rd\theta$ is called the scale factor $h_\theta$. For an orthogonal system the jacobian is the product of scale factors. For spherical coordinate system it is easy to show that: $h_r=1, h_\theta=r, h_\phi=r \sin \theta $ so $J(r,\theta,\phi)=r^2\sin\theta$ For cylindrical $h_\rho=1,h_\phi=\rho,h_z=1$ etc..
Now regarding your question about the integral, consider a transformation of a cartesian coordinate $x$ to some new coordinate $\xi$, you know that
$$\int f(x) dx = \int f(x(\xi)) \frac{dx}{d\xi}d\xi$$
The jacobian is $J(\xi)=\frac{dx}{d\xi}$
For surface integrals,
$$\iint_S f(x_1,x_2) dx_1 dx_2 = \iint_S f(x_1(\xi_1,\xi_2),x_2(\xi_1,\xi_2)) \det[J(\xi_1,\xi_2)] d\xi_1 d\xi_2$$
For volume integrals
$$\iiint f(x_1,x_2,x_3) dx_1 dx_2 dx_3 = \iiint f(x_1(\xi_1,\xi_2,\xi_3),x_2(\xi_1,\xi_2,\xi_3),x_3(\xi_1,\xi_2,\xi_3)) |J(\xi_1,\xi_2,\xi_3)| d\xi_1 d\xi_2 d\xi_3$$
EDIT2As I mentioned in the earlier comment, when you have to find the surface element in a 3D coordinate system, you have to first know what your surface is. This would depend on what actually your finction $f(x,y)$ is. Mostly in physical problems, the surface is traced out by keeping one coordinate system fixed and varying the other two (examples- sphere's surface is fixed radii and varying azimuthal and polar angles, a plane's surface is a fixed polar angle and varying azimuthal angle and radial coordinate, etc...)
$\endgroup$ 2 $\begingroup$The way to calculate this without appeal to a geometric construction is to use the Jacobian determinant. The Jacobian matrix is constructed by taking the partial derivative of the new coordinates ($\epsilon_i$) with respect to the old coordinates ($\delta_j$)
$$J_{ij} = \frac{\partial \epsilon_i}{\partial \delta_j}.$$
$\endgroup$ $\begingroup$To calculate this without using the Jacobian determinant. You can use the exterior derivative operator d to figure out this problem where for a function f $df = f_xdx+f_ydy+f_zdz$ We can use the other definitions of polar coordinates to solve for dx and dy in terms of r and $\theta$. Note that in this derivation we are using exterior products (Think cross product) of differentials not the usual point wise product and note: $dx\wedge dx = 0 $ and $dx\wedge dy = -dy \wedge dx$$$ x = r\cos(\theta) \\ y = r\sin(\theta) \\ dx = \frac{\partial}{\partial r} r\cos(\theta)dr + \frac{\partial}{\partial \theta} r\cos(\theta)d\theta \\ dx = \cos(\theta)dr - r\sin(\theta)d\theta \\ dy = \frac{\partial}{\partial r} r\sin(\theta)dr + \frac{\partial}{\partial \theta} r\sin(\theta)d\theta \\ dy = \sin(\theta)dr + r\cos(\theta)d\theta \\ \text{Multiplication yields} \\ dxdy = (\cos(\theta)dr - r\sin(\theta)d\theta) * (\sin(\theta)dr + r\cos(\theta)d\theta) \\dxdy = \cos(\theta)dr*\sin(\theta)dr - r^2\sin(\theta)d\theta*\cos(\theta)d\theta-r\sin^2(\theta)d\theta dr+r\cos^2(\theta)drd\theta \\dxdy = 0 -0+r(\cos^2(\theta)drd\theta-\sin^2(\theta)d\theta dr) \\dxdy = r(\cos^2(\theta)drd\theta+\sin^2(\theta)dr d\theta) \\dxdy = rdrd\theta( \cos^2(\theta)+\sin^2(\theta))=rdrd\theta \\\text{More accurately} \\dx\wedge dy = rdr\wedge d\theta \\\text{You can then integrate the differential 2-form} \\ \int dx\wedge dy = \int rdr\wedge d\theta = \iint rdrd\theta $$
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