Multivariable Calculus Help?

The paraboloid surface peaks at (0,0,2) and you're looking at a part of it that spills down into the first quadrant. It intersects the plane x=1 along the parabolic curve 2z = 3 - y^2, and intersects the plane y=1 along the curve 2z = 3 - x^2. The lowest point on the specified surface is (1,1,1). The two "corners" on the xz-plane and yz-plane are (1,0,3/2) and (0,1,3/2) respectively.

The surface integral can be turned into an ordinary double integral of

(xy) sqrt[(dz/dx)^2 + (dz/dy)^2 + 1] dx dy,

where the derivatives are partials, so you have a double integral of

(xy) sqrt(y^2 + x^2 + 1) dx dy.

Also note that the projection of the surface onto the xy-plane is simply a unit square, so the limits on the double integral are y = 0 to 1, x = 0 to 1.

If I do the "x" integration first, let u = y^2 + x^2 + 1, (1/2)du = x dx,

and the integrand is now

(1/2) y sqrt(u) du.

After integration w/r/t u, we have

(1/3) y u^(3/2) = (1/3) y (x^2 + y^2 + 1)^(3/2),

and we can immediately apply the limits x = 1 and 0, giving

(1/3) y [(y^2 + 2)^(3/2) - (y^2 + 1)^(3/2)].

So now you want to integrate from y = 0 to 1, the integrand

(1/3) y [(y^2 + 2)^(3/2) - (y^2 + 1)^(3/2)] dy.

This will be easily accomplished by two separate integrands,

(1/3)y (y^2 + 2)^(3/2) dy - (1/3)y (y^2 + 1)^(3/2) dy,

which I have separated because we need two separate substitutions,

v = y^2 + 2 and w = y^2 + 1.

As before, we'll have (1/2)dv = y dy and (1/2)dw = y dy,

so now the integrands are

(1/3) v^(3/2) dv - (1/3) w^(3/2) dw.

After integrating, you have

(2/15) v^(5/2) - (2/15) w^(5/2)

= (2/15) [(y^2 + 2)^(5/2) - (y^2 + 1)^(5/2)].

Finally, plug in y = 1 and 0, and you get

(2/15)[3^(5/2) - 2^(5/2) - 2^(5/2) + 1^(5/2)]

= (2/15)[3^(5/2) - 2^(7/2) + 1]

= approximately 0.7033.

This could be right; although the actual area of the surface must be greater than that of the unit square, the function (xy) that we were integrating has a value <1 at every interior point of the surface. So I am tentatively satisfied with my answer.