16.6 Surface Area

Example

Find the surface area of the part of the surface z  =  x y that lies inside the cylinder (x^2) +  (y^2)  =  1.

Solution

The surface area is given by the formula A  =  ∫(∫_D) (Sqrt[1  +  
                    ((Subscript[f, x])^2) +  ((Subscript[f, y])^2)]) d A, where f(x, y)  =  x y, and D is the region inside the unit circle, (x^2) +  (y^2) ≤  1. Since (Subscript[f, x])  =  y and (Subscript[f, y]) =  x, the integral is A  =  ∫(∫_D) (Sqrt[1  +  
                    (x^2) +  (y^2)]) d A which can be done most easily in polar coordinates: (r^2)  =  (x^2) +  (y^2),  
      d A  =  r dr dθ, and D is described by 0  ≤  r  ≤  1, 0 ≤  θ  ≤  
        2  π. Thus, the surface area is given by \(TraditionalForm\`A\  = \ ∫\_0\%\(2  π\)\ \(∫\_0\%1\@\( 1\  + r\^2\)\ r\ dr\ dθ\)\) 
(Substitute u  =  1 + (r^2), du  =  
        2 r dr ).

(∫_0^(2  π)) (∫_0^1)(Sqrt[ 1  + (r^2)]) r \[DifferentialD]r \[DifferentialD]θ

The surface looks like a Pringle potato chip!

x  =  r Cos[θ];  y  =  r Sin[θ];
graph  =  
      ParametricPlot3D[{x, y, x y}, {r, 0, 1}, {θ, 0, 2  π}, 
        PlotPoints -> {20, 50}, 
        ViewPoint -> {3.215,  -0.175,  1.040}]