Double Integral $\int\limits_1^2\int\limits_{1/Y}^y\sqrt{\frac yx}e^{\sqrt{xy}}\,dx\,dy$

Question

I'm working in this integration. $$\int_1^2\int_{1/Y}^y\sqrt{\frac yx}e^{\sqrt{xy}}\,dx\,dy$$ I make this: $$u=\sqrt{xy},v=\sqrt{\frac yx},x=\frac uv,y=uv$$ For calculate the integration limits, i make this: $$\frac{1}{y}\leq x\leq y,\frac{1}{uv}\leq\frac{u}{v}\leq uv,1\leq u\leq uv$$ And then: $$1\leq uv\leq2,\frac{1}{u}\leq v\leq\frac{2}{u}$$ And the integral is: $$\int_1^2\int_{1/Y}^y\sqrt{\frac yx}e^{\sqrt{xy}}\,dx\,dy=\int_1^{uv}\int_{1/u}^{2/u}ue^{v}\,du\,dv$$ But, my result's too bad, please help me!

Answer 1

Seeing that $$\frac {de^{\sqrt{xy}}}{dx}=e^{\sqrt{xy}} \sqrt y \frac{1}{2 \sqrt{x}}=\frac {1}{2} \sqrt {\frac{y}{x}}e^{\sqrt{xy}}$$

So, if let $t=e^{\sqrt{xy}}$ $$\int_{1}^{2}\int_{\frac {1}{y}}^{y} \sqrt {\frac{y}{x}}e^{\sqrt{xy}}dxdy=\int_{1}^{2}\int_{e}^{e^y}2dtdy=2e^2-4e$$