- #36

Delta2

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It is a well known result (for example see https://en.wikipedia.org/wiki/Surface_integral#Surface_integrals_of_scalar_fields) that the surface of a function ##z=f(x,y)## is given by the integral ##\iint_A\sqrt{{\frac{\partial f}{\partial x}}^2+{\frac{\partial f}{\partial y}}^2+1}dxdy## where ##A=[-1,1]\times[-1,1]## in our case.

for ##z=f(x,y)=x^2+y^2## we have ##\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial y}=2y## hence the surface is ##\iint_A\sqrt{4x^2+4y^2+1}dxdy##

for ##z=f(x,y)=x^2-y^2## we have ##\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial y}=-2y## hence again the surface is ##\iint_A\sqrt{4x^2+4y^2+1}dxdy##

Q.E.D