Integral of the function $\left[\frac{y}{x}\right]\sqrt \frac{y}{x}e^{\sqrt \frac{y}{x}}$ How do I solve this integral 

$$\iint_{D}\left[\frac{y}{x}\right]\sqrt \frac{y}{x}e^{\sqrt \frac{y}{x}}dA$$ where $D=\{(x,y): 1\le x\le 2\quad and\quad  x\le y\le 3x \}$

I have no problems dealing with integrals of the form $\int_{a}^b[x]$ by using definition. Just by watching the D you can get this, still how would the definition of ceiling/floor function applied to $y/x $ be?   
$$\iint_{D}\left[\frac{y}{x}\right]\sqrt \frac{y}{x}e^{\sqrt \frac{y}{x}}dA=\int_{1}^{2}\int_{x}^{3x}\left[\frac{y}{x}\right]\sqrt \frac{y}{x}e^{\sqrt \frac{y}{x}}dydx$$ 
Maybe some change $a=y/x$ before trying the use of the definition?                     
 A: $x = r \cos t\\
y = r \sin t\\
\frac yx = \tan t\\
dy dx = r dr dt$
limits:
$r \cos t < r \sin t < 3r \cos t\\
\cot t < 1 < 3\cot t\\
\frac 13> \tan t > 1\\
1< r \cos t< 2\\
\sec t<r < 2 \sec t$
$\int_{\tan^{-1} \frac13}^{\tan^{-1} 1}\int_{\sec t}^{2\sec t} \tan^{\frac32} t e^{\tan t} r dr dt\\
\int_{tan^{-1} \frac13}^{tan^{-1} 1} \frac 12 \tan^{\frac32} t e^{\tan t} r^2|_{\sec t}^{2\sec t} dt\\
\int_{tan^{-1} \frac13}^{tan^{-1} 1} \frac 12 \tan^{\frac32} t e^{\tan t} (3\sec^2t) dt\\
\int_{tan^{-1} \frac13}^{tan^{-1} 1} \frac 32 \tan^{\frac32} e^{\tan t} \sec^2t) dt\\
u = \tan t\\
du = \sec^2 t\\
\int_{\frac13}^{1} \frac 32 u^{\frac 32} e^u du\\
$
A: let $u=\frac{y}{x}$ and $v=y$ then
$$\frac{\partial u}{\partial x}=-\frac{y}{{{x}^{2}}}\,\,\,\,\,,\,\,\,\,\,\frac{\partial u}{\partial y}=\frac{1}{x}\,$$
$$\frac{\partial v}{\partial x}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\frac{\partial v}{\partial y}=1\,$$
so
$$\frac{\partial (u,v)}{\partial (x,y)}=-\frac{y}{{{x}^{2}}}=-\frac{{{u}^{2}}}{v}\,\,\,\,\to \,\,\,\,\,\left| \frac{\partial (x,y)}{\partial (u,v)} \right|=\frac{v}{{{u}^{2}}}$$
$$\iint_{D}\left[\frac{y}{x}\right]\sqrt \frac{y}{x}e^{\sqrt \frac{y}{x}}dA=\int_{1}^{6}{\int_{1}^{3}{[u]\sqrt{u}}}\,{{e}^{u}}\frac{v}{u^2}dudv$$
