I want to find the area of circle specified as a function of angle $\theta$, where the circle can be described as $r(\theta)=a$ means for every angle $0\le\theta\le2\pi$ the distance from the origin equals $a$.
So I evaluated the area using the double integral as follows $$A=\int_{0}^{2\pi}\int_{0}^{a}r\,dr\,d\theta=\int_{0}^{2\pi}\int_{0}^{a}a\,dr\,d\theta\tag{Since $r = a$}$$ $$=\int_{0}^{2\pi}a[r]^{a}_{0}\,d\theta=\int_{0}^{2\pi}a^2\,d\theta=a^2[\theta]^{2\pi}_{0}=2\pi a^2$$
Well, I get the result $2\pi a^2$ as opposed to $\pi a^2$. Where did I go wrong?
In your first line, you can't replace $r$ simply by $a$: you are crossing over the entire surface of your disk(*) ($r$ going from $0$ to $a$).
As a result, you are dealing with $\int_0^a{r dr}=\frac{a^2}{2}$. That $2$ in the denominator will compensate the one you have in your final result.
(*) I don't know how to say this correctly in English, but the circle is just the line, while the disk is the surface. In fact, a circle does not even have a surface (it's zero).