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=f(x,y) is given by

Double Integral over the region of Sqrt[1+(Dxf)^2+(Dyf)^2] dx dy, where Dx stands for partial with respect to x and Dy is partial with respect to y.

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

In[38]:=

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

In[39]:=

  Plot3D[f[x,y],{x,-1,1},{y,-1,1}]

  Plot3D::gval: 

     Function value 0. + 1. I at grid point xi = 1, yi = 1

       is not a real number.

  Plot3D::gval: 

     Function value 0. + 0.857143 I at grid point xi = 1, yi = 2

       is not a real number.

  Plot3D::gval: 

     Function value 0. + 0.714286 I at grid point xi = 1, yi = 3

       is not a real number.

  General::stop: 

     Further output of Plot3D::gval

       will be suppressed during this calculation.

Out[40]=

  -SurfaceGraphics-

In[41]:=

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

Up to Some Multivariable Calculus Ideas Part II