Find $\iint \frac{1}{\sqrt{x^2+y^2}}\;\mathrm{d}y\;\mathrm{d}x$ Given that $x^2+y^2\leq a^2$, evaluate 
$$
  \iint\limits_D \frac{1}{\sqrt{x^2+y^2}} \;\mathrm{d}A
$$

Now my initial idea is to use polar coordinates, but I do not get the right answer. 

This is what I have done: 

$$\int\limits_0^{2 \pi }\int\limits_0^a\frac{r}{a}\;\mathrm{d}r\;\mathrm{d}\theta=\pi  a$$

Now the answer I got I'm pretty sure is right, which makes me conclude that my setup of the initial integral is wrong since the answer should be $2\pi  a$ according to my textbook.

Could someone point out what I am missing?
$$\int\limits_0^{2 \pi }\int\limits_0^a\frac{r}{r}\;\mathrm{d}r\;\mathrm{d}\theta=2\pi  a$$
The integral is The integral of (1/r)r dr d theta and the limits are for from 0 to a and for theta from 0 to 2 pi giving the expected answer