Double Integral

[Juggler123]

I need to find the area of the hyperbolic paraboloid z=xy contained within the cylinder x^2+y^2=1. I know I need to take a double integral but am having real difficulty finding the correct limits, so far I've got that;

\int dx\int dy

With the x limits being 1 and -1 and the upper y limit to be sqrt(1-x^2) I'm having trouble finding the lower y limit. Although to be honest I'm not completely sure about the other three limits! Sorry about my awful attempt at Latex-ing I don't know how to do it so couldn't write the limits of the integrals on the integral. Any help would be great! Thanks.

[MikeyW]

the problem is symmetric by pi/2 so I'd just stick to the (+,+) quadrant and your lower limit is y=0. Then your x limits would be 1 and 0.

The area in the entire domain is then four times the area in a single quadrant.

[HallsofIvy]

I have no idea what you mean by
so far I've got that;

\int dx\int dy

Shouldn't there be some function to be integrated in that? And it probably is NOT
\int dx\int dy
but rather
\int f(x,y) dxdy
Even ignoring the "f(x,y)" the two separate integrals implies that the two coordinates can be separated- which is not the case here- at least not in Cartesian coordinates.

The surface area of z= f(x,y) is given by
\int\int \sqrt{1+ \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2} dA
where dA is the differential of area in whatever coordinate system you are using, in the xy-plane. Because of the circular symmetry I would recommend changing to polar coordinates- where the two coordinate variables can be separated.