Surface area of a part of z=xy in a cylinder?

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Homework Statement


What is the surface area of the part of z=xy that lies in the cylinder given by x^2+y^2=1?


Homework Equations


SA=(double integral) sqrt(Fx^2+Fy^2+1)dA


The Attempt at a Solution


I've graphed it but I don't know what to integrate or the bounds.. :/ any help? thank you!
 
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You can write z= xy as the vector equation \vec{r}(x,y)= x\vec{i}+ y\vec{j}+z\vec{k}= x\vec{i}+ y\vec{j}+ xy\vec{k}. The two derivative vectors \vec{r}_x= \vec{i}+ y\vec{k} and \vec{r}_y= \vec{j}+ x\vec{k} are in the tangent plane at each point. Their cross product, \vec{r}_x\times\vec{r}_y= -y\vec{i}- x\vec{j}+\vec{k}, the "fundamental vector product" for the surface, is perpendicular to the surface and its length gives the "differential of surface area"- dS= \sqrt{y^2+ x^2+ 1}dydx.

Because the problem asks for the surface area of the part inside the cylinder x^2+ y<br /> ^2= 1, that circle is the boundary. You might want to put it in polar coordinates.
 
Awesome I finally got it! Thanks so much for your help! :)
 
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