The area of the graph of a differentiable function f(x,y)is given by
where fx and fy are the partial derivatives of f with respect to xand to y.
For the positive curvature example f(x,y)=-(x2+y2),the circle x2+y2=1 fits in thesurface exactly at height -1. The area enclosed is
where D is the disc of radius 1. Using polar coordinates the integral becomes
which can be evaluated (use the substitution u=1+4r2) as .
The negative example is more complicated because first one mustfind the circle in the (x,y)-plane which gives a circle in thegraph of circumference .I had to do it by trial anderror. The circle of radius r in the (x,y)-plane can beparametrized as ,.In the graph this circle becomes ,.Thelength of a parametrized curve is given by the integral
In this case the length is
using the identity beforedifferentiating, and the identity after. For r=.715 one gets length 6.28 by numerical integrationwhich is close enough.
The area computation goes the same way as for positive curvature,since the quantity 1+fx2 + fy2 is the same for both functions.The area enclosed by the circle of circumference 6.28 is calculatedas before but using .715 instead of 1. It comes out to be.
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