We present a more tedious analytic solution.
We consider a portion of the surface area
Input :=
f[x_] = Sqrt[1 - x^2];
Input :=
s[w_] = Integrate[Sqrt[1 + f'[x]^2],{x,1-w,1}]
Output =
Pi 1
-- - ArcTan[(1 - w) Sqrt[--------]]
2 2
2 w - w
Input :=
curve = ParametricPlot[{s[w],1-w},{w,.0000001,1},
AspectRatio->Automatic]
Output =
-Graphics-
Input :=
s[1-u]
Output =
Pi 1
-- - ArcTan[Sqrt[--------------------] u]
2 2
2 (1 - u) - (1 - u)
If we collect terms we get.
Input :=
express = Pi/2 -
ArcTan[u Sqrt[1/Expand[2(1-u) - (1 - u)^2]]]
Output =
Pi 1
-- - ArcTan[u Sqrt[------]]
2 2
1 - u
And if we draw a triangle with hypotenuse 1 and sides u and Sqrt[1 - u^2] then it can be shown that express = ArcCos[u].
But a plot will "prove" that s[1-u] IS ArcCos[u].
Input :=
border = Plot[s[1-u],{u,0.0000001,.99999999},
AspectRatio->Automatic]
Output =
-Graphics-
Input :=
arcCos = Plot[ArcCos[u],{u,0,1},
AspectRatio->Automatic]
General::spell1:
Possible spelling error: new symbol name "arcCos"
is similar to existing symbol "ArcCos".
;[o]
General::spell1:
Possible spelling error: new symbol name "arcCos"
is similar to existing symbol "ArcCos".
Output =
-Graphics-
We see they match perfectly!
Input :=
Show[border,arcCos]
Output =
-Graphics-
We compute the area of the flat cut out sheet.
Input :=
Integrate[Cos[x],{x,0, Pi/2}]
Output =
1
We compute the desired surface area using the element of surface area in a double integral.
Input :=
g[x_,y_] = Sqrt[1 - x^2]
Output =
2
Sqrt[1 - x ]
Input :=
SA[x_,y_] = Sqrt[D[g[x,y],x]^2 + D[g[x,y],y]^2 + 1]
Output =
2
x
Sqrt[1 + ------]
2
1 - x
Input :=
inner[x_,y_] = Integrate[SA[x,y],{x,y,1}]
Output =
Pi 1
-- - ArcTan[y Sqrt[------]]
2 2
1 - y
Input :=
NIntegrate[inner[x,y],{y,0,1}]
Output =
1.
We see that they are the same.