(865, #15) Find, to four decimal places, the are of the part of the surface z = 1 + x² y² that lies above the disc x² + y² <= 1..

Solution: In examining this surface it is important to note that it is not only symmetric to the x and y axes, but also to the lines y = x and y = -x

This means we can restrict our integral to the (1/8)th of the circle in the first half of the first quadrant. Calling this region R we have the total surface area given by

Switching to polar coordinates we get

To evaluate this numerically, we use the mid-point rule and the Maple method first employed in problem 7. When m = n = 90 ( a computation at 8100 points!) we get the value 3.321262811, which, to 4 decimal places is 3.3213.

g := (r,t) -> 8*r*sqrt( 1+r^6*sin(2*t)^2):
m := 90: n := 90:
sum( sum( g( (i-0.5)/m, (j-0.5)*(Pi/4)*(1/n) )*(1/m)*(Pi/4)*(1/n), i = 1..m), j = 1..n);

3.321262811

by: gm