MapleHyperboloidArea.mws

Surface Area Example

Arc Length and Surface Area Example 3

>    with(plots): 

Warning, the name changecoords has been redefined

Find the area of the surface formed by revolving the graph of the curve given parametrically about the y-axis.

x = 4tan(t),     y = 3sec(t)     t-interval:  [ 0 , Pi/3 ]

The surface is part of a hyperboloid.

>    implicitplot3d(-9*x^2+16*y^2-9*z^2=144,x=-4*sqrt(3)..4*sqrt(3),y=0..6,z=-4*sqrt(3)..4*sqrt(3),axes=boxed,scaling=constrained,numpoints=2500);

[Maple Plot]

>    Area:=8*Pi*Int(tan(t)*sqrt((4*(sec(t))^2)^2+(3*sec(t)*tan(t))^2),t=0..Pi/3);

Area := 8*Pi*Int(tan(t)*(16*sec(t)^4+9*sec(t)^2*tan(t)^2)^(1/2),t = 0 .. 1/3*Pi)

>    value(%);

8*Pi*(-2+9/10*ln(3)-9/20*ln(10*91^(1/2)+91)+9/20*ln(10*91^(1/2)-91)+91^(1/2))

>    evalf(%);

171.9507978

>