can be thought of as defining anelementofvolume dV in the shape of a parallelepiped
with vector sides A =hudu ˆe 1 ,B =hvdv ˆe 2 and C =hwdw ˆe 3. The volume of this
parallelepiped is given by
dV =|A·(B×C)|=|(hudu ˆe 1 )·((hvdv eˆ 2 )×(hwdw ˆe 3 )|=huhvhwdu dv dw
In rectangular coordinates (x, y, z )one finds hx= 1, h y= 1, and hz= 1 and the element
of volume is dV =dxdydz.
In cylindrical coordinates (r, θ, z )where x=rcos θ, y =rsin θ, z =zone finds
hr= 1, h θ=rand hz= 1 and the element of volume is dV =r dr dθ dz
In spherical coordinates (ρ, θ, φ )where x=ρsin θcos φ, y =ρsin θsin φ, z =ρcos θ, one
finds hρ= 1, h θ=ρ, h φ=ρsin θand the element of volume is dV =ρ^2 sin θ dρ dθ dφ
These elements of volume must be summed over appropriate regions of space in
order to calculate volume integrals of the form
∫∫∫
f(x, y, z )dxdydz,
∫∫∫
f(r, θ, z )r dr dθ dz,
∫∫∫
f(ρ, θ, φ )ρ^2 sin θ dρ dθ dφ
Surface Placed in a Scalar Field
If a surface is placed in a region of a scalar field f(x, y, z),one can divide the
surface into nsmall areas
∆S 1 , ∆S 2 ,... ,∆Sn.
For nlarge, define fi=fi(xi, y i, zi) as the value of the scalar field over the surface
element ∆Si as iranges from 1 to n. The summation of the elements fi∆Siover all i
as nincreases without bound defines the surface integral
∫∫
R
f(x, y, z)ˆendS =
∫∫
R
f(x, y, z )dS= limn→∞
∑n
i=1
fi(xi, y i, zi)∆ Si, (7 .97)
where the integration is determined by the way one represents the element of surface
area dS . The integral can be represented in different forms depending upon how the
given surface is specified.
Surface Placed in a Vector Field
For a surface Sin a region of a vector field F=F(x, y, z )the integral
∫∫
R
F·dS=
∫∫
R
F·ˆendS (7 .98)
