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Let $S\subset \mathbb{R}^3$ be the surface enclosed by the (infinitely long) open cylinder in $\mathbb{R}^2$ (given by the equation $x^2+y^2<1$) and the surface $z=xy$. Explicitly, $S=\{(x,y,z): x^2+y^2<1\text{ and } z=xy\}$. How do I compute the area of $S$, by using a double integral.

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    @joriki yes I do2012-07-23

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I presume you know the double integral for the surface area $S$ of a surface defined by $z=f(x, y)$:

$ S=\iint\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\ dx\,dy $

The two partial derivatives are easy enough. Once you have that, you'll find it easiest to work in cylindrical coordinates so $dx\,dy$ will become $r\,dr\,d\theta$, and the bounding region will be the circle $r=1$. You should be able to take it from there.

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    @gifty. Right you are.2012-07-23