use polar coordinates to evaluate $ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $ $$ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $$ converting to polar coordinates :

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2(cos^2 + sin^2)}} dydx $$
using trig substitution $cos^2 + sin^2 = 1$

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2}} dydx = \int_0^1 \int_0^{\sqrt{1-r^2}} e^r drdx  = \int_0^1  e^{\sqrt{1-r^2} } - 1  \space drdx  = e^1 - 1 - e^1 = -1  $$ Is this logic correct ?
If you draw that region in the plane you'll see that its the part of the unit disk contained in the first quadrant i.e calling the domain $D$ then $D = \{(r, \theta): 0 \leq r \leq ?, 0 \leq \theta \leq ?\}$. Fill those in. 
Now for the integrand, recall that you just change $x,y$ to their polar form and tac on the $r$ from the jacobian which accounts for the relationship between areas when making the coordinate change i.e,


$$\textbf{Area}= \int_{\theta=?}^{\theta=?} \int_{r=?}^{r=?} e^r \cdot r \ dr \ d\theta$$
What you have is a mess.

you can't say $\cos^2 + \sin^2 = 1$  you need some object.

When you convert to polar coordinates, your coordinate system will be in terms of $r, \theta.\ \ \ dx\ dy$ will be replaced

Start with.

$x = r \cos\theta\\y=r\sin\theta$

your limits.

$y \le \sqrt {1-x^2}\\
r\sin\theta \le \sqrt {1-r\cos\theta}\\
r^2\sin^2 \theta + r^2\cos^2 \theta le 1$
r le 1$

$y\ge 0\\
r\sin\theta \ge 0\\
\sin\theta \ge 0\\
0\le \theta \le \pi $

$x&gt;0\\
\cos\theta \ge 0\\
0\le \theta \le \frac {\pi}{2} $

The jacobian...
I am not sure I am ready educate you on this.  Look it up.
$dy\ dx = r\ dr\ d\theta$

$\int_0^{\frac {\pi}{2}}\int_0^1 re^r \ dr\ d\theta$

$\frac {\pi}{2} (r-1) e^r|_0^1\\
\frac {\pi}{2}$