Evaluate $$\iint_{D} \exp\left(\frac xy\right) \,dx\,dy$$ where $D$ is an area in $\mathbb{R}^2$ bounded by $x=0, \ y=1, \ y=\sqrt{x}$.
I graphed the area and found that it is $[0,1]$ and the upper function is $x=y^2$ and the lower function is $x=0$.
So the integral is $$\iint_{D} \exp\left( \frac xy \right) \, dx\,dy=\int_{0}^{1}dy\int_{0}^{y^2} dx \, \exp\left( \frac xy\right)=\int_{0}^{1}\left(ye^y-y\right)dy=\frac{1}{2}$$
Is this correct?
Ye it seems correct, I obtain
$$\int \int_{D} e^{\frac{x}{y}}dx dy=\int_{0}^{1}\int_{x=0}^{x=y^2}e^{\frac{x}{y}} dx dy =\int_{0}^{1}[ye^{\frac{x}{y}}]_{x=0}^{x=y^2}dy=\\=\int_{0}^{1} (ye^y-y)dy=\left[ye^y-e^y-\frac{y^2}2\right]_0^1=\frac12$$