Final Exam 2015

2.

X2[u_,v_]={( 2+v Cos[u/2]) Cos[u],( 2+v Cos[u/2]) Sin[u],v Sin[u/2]}

ParametricPlot3D[X2[u,v],{u,0,2 Pi},{v,-1,1}]

X2u[u_,v_]=D[X2[u,v],u]
X2v[u_,v_]=D[X2[u,v],v]

g11[u_,v_]=Simplify[X2u[u,v].X2u[u,v]]
F=Simplify[X2u[u,v].X2v[u,v]]
G=Simplify[X2v[u,v].X2v[u,v]]

Integrate[Sqrt[g11[u,v]],{u,0,2 Pi}]

Integrate[Sqrt[g11[u,v]],{v,-1,1}]

halfarea=NIntegrate[Sqrt[g11[u,v]],{u,0,2 Pi},{v,-1,1}]

halfarea (36/125)

This means you need at least 8 quarts or 2 gallons.

competitors

A first possible “easy” but wrong approach would be to assume the area is the same as a circular cylinder of radius 2 and height 2:

cylinderarea=N[2 Pi (2) (2)]

The total paint in this case would be

cylinderarea (36/125)

So you would still buy two gallons, so it turns out this is an adequate approximation for this application.  Nevertheless, the calculated area is a little less than the actual area---but only by about half a square foot.

A second possibility would be to again assume you have a metric rectangle of width 2 but length the length of half the boundary curve.

boundarylength=NIntegrate[Sqrt[g11[u,1]],{u,0,2 Pi}]

2 boundarylength

2 boundarylength (36/125)

This gives an overestimate, though for practical purposes it is still adequate.  You’re still going to buy two gallons of paint.