Finding the Area of a Hyperbolic Paraboloid

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SUMMARY

The area of the hyperbolic paraboloid defined by the equation z=xy, constrained within the cylinder x^2+y^2=1, can be calculated using a double integral. The correct limits for integration in the first quadrant are x from 0 to 1 and y from 0 to sqrt(1-x^2). The surface area formula requires the function to be integrated, which is given by the expression sqrt{1+(\partial f/\partial x)^2+(\partial f/\partial y)^2}. It is advisable to convert to polar coordinates for simplification due to the circular symmetry of the problem.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates
  • Knowledge of partial derivatives
  • Ability to interpret surface area formulas
NEXT STEPS
  • Study the application of double integrals in calculating areas
  • Learn how to convert Cartesian coordinates to polar coordinates
  • Review the derivation and application of the surface area formula for functions of two variables
  • Practice solving integrals involving circular symmetry
USEFUL FOR

Mathematicians, engineering students, and anyone involved in multivariable calculus or geometric analysis will benefit from this discussion.

Juggler123
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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.
 
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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.
 
I have no idea what you mean by
Juggler123 said:
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.
 

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