differentiation an integration Let $x,y$ and $r,\theta$ in Cartesian and polar cordinate. So
$$x=r\cos\theta  ,  y=r\sin \theta$$
Therefor $dx=\cos \theta dr-r\sin\theta d\theta $ and $dy=\sin \theta dr+r\cos \theta d\theta$. Therefore $$dA=dxdy=r\cos2\theta drd\theta+\sin2\theta/2(dr)^2-r^2\sin(2\theta)/2(d\theta)^2$$
My question is: with this $dA$ (elemant of surface) is it possible to find area of a shape for example a circle? I mean if it is possible to integrate what is the bound of integral?
 A: Your formulation is actually wrong, but in a way that is sufficiently subtle that it is hard to explain without appealing to "this is small". The best way to understand this is in terms of the cross product:
The basic reason is that the area you are considering is the area of a parallelogram (that happens to be a square since the vectors are orthogonal). The area of a parallelogram with sides given by two vectors $a,b$ in $\mathbb{R}^2$ is
$$ a \times b \quad (= a_1 b_2-a_2 b_1), $$
the cross product of $a$ and $b$. Hence the standard area element is $dx \times dy = 1 \, dx \, dy$, (which is a entirely standard abuse of notation). What you are doing is changing basis of this space, and looking at a different parallelogram.
To see this, you can look at the simpler transformation $x=u$, $y=u+v$. Then you find $dx=du$ and $dy = du+dv$, so $dx \times dy = du \times (du+dv) = du \times du + du \times dv = du \times dv = 1 \, du \, dv$, because the cross product of parallel vectors is zero.
(Another question: how do you propose to compute $\int (dr)^2$?)
For your example, you have
$$ dx \times dy = (\cos{\theta} \, dr -r \sin{\theta} \, d\theta) \times (\sin{\theta} \, dr + r\cos{\theta} \, d\theta) \\
= (\text{stuff})(dr \times dr+ d\theta \times d\theta) -r \sin^{2}{\theta} \, d\theta \times dr + r \cos^2{\theta} dr \times d\theta \\
= r (\cos^2{\theta}+\sin^2{\theta}) \, dr \times d\theta \\
= r \, dr \,  d\theta,
$$
since $dr \times d\theta = -d\theta \times dr$.

(By the way, I did this with the cross product because you were more likely to be familiar with it, but as Alex M. remarks, the generalisation to more than 3 dimensions is done with the exterior product, which does similar things, like antisymmetry, to the cross product, but in more generality (for example, $a$, $a \wedge b$ and $a \wedge b \wedge c$ are different sorts of objects (a vector, bivector and trivector, respectively, whereas $a \times b$ is again a vector (well, nearly - it's actually a pseudovector))). For more information on this there are some quite basic books on differential forms (this one is free on the internet, I've read another one with the same title, and it looks like there are a few others about). A more usual treatment is covered in courses on vector calculus (which in the UK is done in first year at university), where the thing is run through in terms of some sort of argument involving small parallelograms and vectors directly, and any "Methods in Mathematical Physics" text will cover this approach.
Really the hard step to justify is the $dx \wedge dy \to \, dx \, dy$ one: basically, the double integral operator
$$ \iint_A dx \wedge dy $$
is turned into an iterated integral,
$$ \int_a^b \left( \int_{c(y)}^{d(y)} \, dx \right) \, dy. $$
A: You're calculation is off. What you need is the change of variables formula. 
dA = dxdy = |D( T (x, y ) )|drd$\theta$
where T (x, y) = (rcos$\theta$, rsin$\theta$)
Essentially, dA = volume of infinitesimal parallelepiped in (r, $\theta$) coordinates. Which, is given by the determinant of the change of basis matrix.
In order to find the area of a circle, you're limits of integration over dA are from $\theta$ = 0 to $\theta$ = 2 $\pi$ and r=0 to r=R. 
http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
A: The reason why your computation of $\mathbb{d}A$ is not correct is that $\mathbb{d}x \mathbb{d}y$ is not a product like the usual one, but one that obeys slightly different rules; it is called "exterior product", it is denoted by a wedge and it obeys the following rules: $\mathbb{d}x \wedge \mathbb{d}x = \mathbb{d}y \wedge \mathbb{d}y = 0$ and $\mathbb{d}x \wedge \mathbb{d}y = - \mathbb{d}y \wedge \mathbb{d}x$. Try now to redo $\mathbb{d}A$ taking into account these rules and you'll get $2r \space \mathbb{d}r \wedge \mathbb{d}\theta$, which (for complicated reasons) is $r \space \mathbb{d}r \mathbb{d}\theta$ (the wedge and the $2$ are gone). If you're studying pure mathematics, then you'll learn this in differential geometry (or, possibly, in algebra, when discussing the tensor product and related concepts).
