Evaluate a double integral. 
To find:

$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$

R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$

$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$
$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$

i.e.

$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$
$$I=\int_0^2\int_{x^2}^4x(1+y^2)^{\frac{-1}{2}}\,dy\,dx$$

Where to from here?


\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
   I&=&  \int_0^4\left(\int_0^{\sqrt{y}}\frac{x}{\sqrt{1+y^2}}dx\right)dy\\
   &=& \int_0^4\frac{1}{\sqrt{1+y^2}}\left(\int_0^{\sqrt{y}}xdx\right)dy \\
   &=& \frac{1}{2} \int_0^4\frac{y}{\sqrt{1+y^2}} dy\\
   &=&\frac{1}{2}\left[\sqrt{1+y^2}\right]^4_0\\
   &=&\frac{1}{2}\left(\sqrt{17}-1\right)
\end{eqnarray*}
