Surfaces of Revolution
In single variable calculus, we considered the volume of a solid obtained by revolving a region about a given axis.
Such solids are bounded by the surface obtained by
revolving y = f( x) about the x-axis. As a parametric surface,
this surface of revolution can be represented by
|
r( u,v) =
á u, f( u) sin(v) , f( u) cos( v)
ñ |
|
where ( u,v) is in [ a,b] ×[ 0,2p] .
EXAMPLE 6 Find the equation of the surface of revolution
obtained by revolving the curve y = 2xe-x for x in [ 0,3]
about the x-axis
Find the tangent vectors ru and rv. Is the
parametrization orthogonal?
Solution: To do so, we notice that f( u) = 2ue-u, so
that the parametrization is
|
r( u,v) =
á u,2ue-usin( v),2ue-ucos( v)
ñ |
|
The graph of r( u,v) is shown below for (u,v) in [ 0,3] ×[ 0,2p] .
Notice now that ru =
á 1,2( e-u-ue-u)sin( v) ,2( e-u-ue-u) cos( v)
ñ and
|
rv =
á 0,-2ue-ucos( v) ,2ue-usin( v)
ñ |
|
Moreover, it is rather easy to show that ru· rv = 0, which implies that the parameterization is orthogonal.
More generally, if f( u) ³ 0 for all u in [a,b] , then
|
r( u,v) =
á g( u) ,f( u)sin( v) ,f( u) cos( v)
ñ |
|
is the surface obtained by revolving the curve
|
r( u,0) =
á g( u) ,0,f(u)
ñ |
|
about the x-axis, where u is in [ a,b] and v is in [ 0,2p] .
EXAMPLE 7 What surface of revolution is obtained from
revolving
|
r( u,0) =
á cos( u) ,0,3+sin( u)
ñ |
|
about the x-axis? Find the tangent vectors ru and rv. Is the parameterization orthogonal?
Solution: Since r( u,0) , u in [0,2p] , is a circle of radius 1 centered at ( 0,0,3), the surface is a torus parameterized by
|
r( u,v) =
á cos( u) ,(3+sin( u) ) sin( v) ,( 3+sin(u) ) cos( v)
ñ |
|
The graph of r( u,v) is shown below for (u,v) in [ 0,2p] ×[ 0,2p] .
Notice now that ru =
á -sin( u) ,cos( u) sin( v) ,cos( u) cos(v)
ñ and
|
rv =
á 0,-( 3+sin( u) ) cos( v) ,( 3+sin( u) ) sin( v)
ñ |
|
It follows that ru · rv = 0, which implies that
the parameterization is orthogonal.