Surface Area

If a curve is not parametrized, the arc length of y=f(x) from x=a to x=b is given by Integral from a to b of Sqrt[1+[f'(x)]^2] dx. This follows from the parametric arc length formula: if y=f(x), parametrize by {t,f(t)}, then r'(t)={1,f'(t)} and ||r'(t)||=Sqrt[1+[f'(t)]^2].

To use an analogy, the surface area of a surface z=S(x,y) is given by

Double Integral over the region of Sqrt[1+(DxS)^2+(DyS)^2] dx dy, where Dx stands for partial with respect to x and Dy is partial with respect to y.
The quantity Sqrt[1+(DxS)^2+(DyS)^2] dx dy is often written d(sigma).

Here capital sigma is a piecewise smooth surface z=S(x,y) with (x,y) in D.

Again, let's look at a simple example to convince ourselves this is reasonable. This is a hemisphere or radius 1. The surface area of a sphere is 4Pi r^2, so the answer we should get here is 2 Pi.

In[39]:=

  s[x_,y_]:=Sqrt[1-x^2-y^2]

In[40]:=

  Plot3D[s[x,y],{x,-1,1},{y,-1,1}];

In[41]:=

  Dx=D[s[x,y],x]
  Dy=D[s[x,y],y]

Out[41]=

            x
  -(-----------------)
              2    2
    Sqrt[1 - x  - y ]

Out[42]=

            y
  -(-----------------)
              2    2
    Sqrt[1 - x  - y ]

Note carefully Mathematica's double integral command. The y integral is done first, so it can depend on x.

In[43]:=

  Integrate[Sqrt[1+Dx^2+Dy^2],
    {x,-1,1},{y,-Sqrt[1-x^2],Sqrt[1-x^2]}]

Out[43]=

  2 Pi

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