The better question: Given a point P and closed curve X(t) in 3Space, what is the solid angle at P of the generalized cone joining X to P? --rwg On Mon, Jan 26, 2015 at 9:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
The plane area inside (x(t),y(t)) = ½∫x dy - y dx. E.g., the usual picture of the area under f(x), a<x<b becomes four line integrals--the three axis-aligned segments plus the actual curve of f, traversed anticlockwise: Integrate[f[x],{x,a,b}]==(Integrate[a - 0, {t, f[a], 0}] + Integrate[t*0 - 0, {t, a, b}] + Integrate[b - 0, {t, 0, f[b]}]+Integrate[t*D[f[t], t] - f[t]*D[t,t], {t, b, a}])/2 == (b f[b] - a f[a] + Integrate[t f'[t] - f[t], {t, b, a}])/2 , which follows from integration by parts. E.g., if f:=Exp, In[55]:= % /. f -> Exp
Out[55]= -E^a + E^b == 1/2 ((-2 + a) E^a - a E^a - (-2 + b) E^b + b E^b)
In[56]:= Simplify@%
Out[56]= True
What about on a sphere? Suppose we have a a unit vector X(t):=(x(t),y(t),z(t)) describing a closed curve on a sphere. (X(0)=X(1), say.) What's the area? There ought to be some nice thing analogous to ½∫x dy - y dx. --rwg