Area of a Curved Surface Find the area of the part o the surface $z=xy$ that lies within the cylinder $x^2+y^2=1$. 
I'm not sure how to set up the surface integral to compute this.

The formula of the are of the surface given as a graph of the function $z=f(x,y)$ over the region $(x,y) \in D$ is
$$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA$$
In this case $D$ is the disk of radius $1$ with center at $(0,0)$:
$D=\{(x,y): x^2+y^2 \leq 1\}$
$$z=f(x,y)=xy$$
$$f_x=y, f_y=x$$
So, we have the following:
$$A(S)=\iint_D \sqrt{1+f_x^2+f_y^2}dA=\iint_D \sqrt{1+y^2+x^2}dA$$
Now we use polar coordinates:
$$x=r \cos{\theta}, y=r \sin{\theta}, 0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi$$
$$dA=dxdy=r dr d \theta$$
So we have the following:
$$A(S)=\iint_D \sqrt{1+y^2+x^2}dA=\int_0^{2 \pi} \int_0^1 \sqrt{1+r^2}rdr d \theta=\int_0^{2 \pi} \int_0^1 (1+r^2)^{1/2}rdr d \theta=\int_0^{2 \pi} \int_0^1 \frac{d}{dr} \left (\frac{1}{3}(1+r^2)^{3/2} \right ) dr d \theta =\int_0^{2 \pi} \left [ \frac{1}{3}(1+r^2)^{3/2} \right ] _0^1  d \theta=\int_0^{2 \pi} \left (\frac{1}{3} 2^{3/2}-\frac{1}{3} \right )  d \theta=\frac{1}{3}  ( 2^{3/2}-1 ) 2 \pi =\frac{2 \pi }{3}  ( 2 \sqrt{2}-1 )$$

