Michael Corral: Vector Calculus

4.4 Surface Integrals and the Divergence Theorem 159
4.4.3(a)). Andasthecirclerevolvesaroundthez-axis,thelinesegmentfromtheorigin
to the center of that circle sweeps out an angle v with the positive x-axis (see Figure
4.4.3(b)). Thus, the torus can be parametrized as:
x=(b+acosu)cosv, y=(b+acosu)sinv, z=asinu, 0≤u≤2π, 0≤v≤2π
So for the position vector
we see that
r(u,v)= x(u,v)i+y(u,v)j+z(u,v)k
= (b+acosu)cosvi+(b+acosu)sinvj+asinuk
∂r
=−asinu cosvi−asinu sinvj+acosuk
∂u
∂r
∂v =−(b+acosu)sinvi+(b+acosu)cosvj+0k,
and so computing the cross product gives
∂r
∂u ×∂r =−a(b+acosu)cosv cosui−a(b+acosu)sinv cosuj−a(b+acosu)sinuk,
∂v
which has magnitude
Thus, the surface area of T is
S=
∥ ∥∥∥∥∥ ∂r
∂u ×∂r
∂v∥ = a(b+acosu).
=
=
=
=
1dσ
Σ
∫ 2π ∫ 2π
0 0
∫ 2π ∫ 2π
0
∫ 2π
0
∫ 2π
0
= 4π 2 ab
0
∂r
∥∂u ×∂r
∂v∥ dudv
a(b+acosu)dudv
(
abu+a 2 sinu∣
2πabdv
∣ u=2π
u=0
)
dv
Since ∂r
∂u and∂r ∂v
aretangenttothesurfaceΣ(i.e. lieinthetangentplanetoΣateach
point onΣ), then their cross product ∂r
∂u ×∂r ∂v
is perpendicular to the tangent plane to