VECTOR CALCULUS
total derivative, the tangent to the curver(λ) at any point is given by
dr
dλ=∂r
∂udu
dλ+∂r
∂vdv
dλ. (10.21)
The two curvesu=constantandv= constant passing through any pointPonSare calledcoordinate curves.Forthecurveu= constant, for example, we
havedu/dλ= 0, and so from (10.21) its tangent vector is in the direction∂r/∂v.
Similarly, the tangent vector to the curvev= constant is in the direction∂r/∂u.
If the surface is smooth then at any pointP onSthe vectors∂r/∂uand∂r/∂vare linearly independent and define thetangent planeTat the pointP(see
figure 10.4). A vector normal to the surface atPis given by
n=∂r
∂u×∂r
∂v. (10.22)
In the neighbourhood ofP, an infinitesimal vector displacementdris writtendr=∂r
∂udu+∂r
∂vdv.Theelement of areaatP, an infinitesimal parallelogram whose sides are the
coordinate curves, has magnitude
dS=∣
∣
∣
∣∂r
∂udu×∂r
∂vdv∣
∣
∣
∣=∣
∣
∣
∣∂r
∂u×∂r
∂v∣
∣
∣
∣du dv=|n|du dv. (10.23)Thus the total area of the surface is
A=∫∫R∣
∣
∣
∣∂r
∂u×∂r
∂v∣
∣
∣
∣du dv=∫∫R|n|du dv, (10.24)whereRis the region in theuv-plane corresponding to the range of parameter
values that define the surface.
Find the element of area on the surface of a sphere of radiusa, and hence calculate the
total surface area of the sphere.We can represent a pointron the surface of the sphere in terms of the two parametersθ
andφ:
r(θ, φ)=asinθcosφi+asinθsinφj+acosθk,whereθandφare the polar and azimuthal angles respectively. At any pointP,vectors
tangent to the coordinate curvesθ=constantandφ= constant are
∂r
∂θ=acosθcosφi+acosθsinφj−asinθk,∂r
∂φ=−asinθsinφi+asinθcosφj.