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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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what do you mean "compute the surface"? Do you want its area? –  Robert Mastragostino Jul 23 '12 at 20:35
    
Do you know a formula for the surface area involving partial derivatives? If you do, the problem is straightforward (especially in cylindrical coordinates). –  Rick Decker Jul 23 '12 at 20:50
    
Yes I want the area, I just fixed it. Thank you for pointing it out. –  gifty Jul 23 '12 at 20:50
    
$z=xy$ is a surface, not a curve. The set $S=\{(x,y,z): x^2+y^2<1\text{ and } z=xy\}$ is the part of the surface $z=xy$ that is inside the cylinder $x^2+y^2\lt1$. Is that what you mean? –  joriki Jul 23 '12 at 20:53
    
@joriki yes I do –  gifty Jul 23 '12 at 20:58

1 Answer 1

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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Thank you for the hint. I've used that the variable $\theta$ runs from $0$ to $2\pi$ and $r$ runs from $0$ to $1$. –  gifty Jul 23 '12 at 21:13
    
@gifty. Right you are. –  Rick Decker Jul 23 '12 at 21:16

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