Solving the multiple integral $\int \int_D \sqrt{x^2+y^2+3}dx dy$, D:{$1\le x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3}$} $\iint_D \sqrt{x^2+y^2+3}\, dx dy$, $D=\left\{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3} \right\}$.
So I've started by drawing two circles, One with $R_1 =1$, Other with $R_2=2$
Then i drawed the lines $y_1=\sqrt3x$ and $y_2 = \frac x{\sqrt 3}$.
Using the polar system i'll need to calculate:
$\int^{\pi/3}_{\pi/6}dθ\int_1^2\sqrt{r^2+3}*r*dr$
Using substiution on: $\int_1^2\sqrt{r^2+3}*r*dr$ i get: $\frac {7^{3/2}}{3}-\frac {4^{3/2}}{3}$.
Then i need to multiply by $\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$ 
And the final answer is: $\frac {7^{3/2}-4^{3/2}}{18}\pi$.
The answer should be: $\frac {5(7\sqrt 7 -8)}{18}$, I'm pretty close but it seems like i'm missing something, I tried solving this one a couple of times and still got to the same answer.
I need help, What am i missing?

I think you've done almost everything right, but $\;1\le r\le\frac6{\sqrt{10}}=\frac{3\sqrt2}{\sqrt 5}\;$ ! , so
$$\int\limits_1^2r\sqrt{r^2+3}\,dr=\left.\frac12\frac23(r^2+3)^{3/2}\right|_1^2=\frac13\left(7^{3/2}-8\right)$$
and together with
$$\int\limits_{\pi/6}^{\pi/3}d\theta=\frac{\pi}6\;,\;\;\text{we get:}$$
$$\frac{\pi}{18}(7\sqrt 7-8)$$
I can't understand where does that five you wrote there come from...
