To state this theorem, we also need to define the curl
of a vector field
F(x)=A(x,y,z)i + B(x,y,z)j
+C(x,y,z)k.
We will assume that F has continuous partial derivatives. The
curl is then defined by
There is a useful physical interpretation for the curl. Suppose that a
fluid is rotating about a fixed axis with angular velocity
ω. Define ω to be the vector with magnitude ω
and with direction along the axis of rotation. The velocity of an
element of the fluid located at the position with radius vector
x is v(x) = ω×x. With a
little work, one can show that ω =
½∇×v. Thus one half of the curl of the
velocity vector v is the vector ω mentioned above.
Stokes' Theorem Let S be an orientable surface bounded by a simple closed positively oriented curve C. If F is a continuously differentiable vector-valued function defined in a region containing S, then
We will first compute the line integral over C. In the xy-plane, C is
parameterized via
x(t) = 3 cos(t) i + 3 sin(t) j, 0 ≤ t
≤ 2π,
and so we have:
dx = (- 3 sin(t) i + 3 cos(t) j)dtWe now turn to finding the surface integral. ∫∫S ∇×F·n dσ. The normal compatible with the orientation of C is n = x/|x| = x/3. Thus, on the surface of the hemisphere S, we have
F(x(t)) = 2·3 sin(t)i + 32 cos(t)j - 02k = 2·3 sin(t)i + 32 cos(t)j
F·dx = (- 18 sin2(t) + 27 cos2(t))dt
∫CF·dx = ∫02π(- 18 sin2(t) + 27 cos2(t))dt = 9π
∫∫S ∇×F·n dσ = ∫∫S k·n dσSince both terms in Stokes's Theorem have the same value, we have verified the theorem in this case.
∫∫S ∇×F·n dσ = ∫02π ∫0½π cos(θ) 32 sin(θ)dφdθ
∫∫S ∇×F·n dσ = 9π
Divergence Theorem Let V be region in 3D bounded by a closed, piecewise smooth, orientable surface S; let the outward-drawn normal be n. Then,
To do this we must compute both integrals in the Divergence
Theorem. We will first do the volume integral. It is easy to check
that ∇·F = 3+1+2=6. Hence, we have that
∫∫∫V∇·FdV =
∫∫∫V6dV = 6π·42·5 =
480π
The surface integral must be broken into three parts: one for the top
cap, a second for the curved sides, and a third for the bottom cap.
∫∫S F·ndσ =
∫∫top F·ndσ +
∫∫sides F·ndσ +
∫∫bottom F·ndσ
The outward normals for the top and bottom caps are k and
−k, respectively. For the top (z = 5), we are integrating
F(x,y,5)·k = 2·5 = 10, and for the
bottom (z = 0), F(x,y,0)·(−k) =
−2·0 = 0. Hence, we have
∫∫top F·ndσ =
∫∫top 10dσ = 10π42 = 160π
∫∫bottom F·ndσ =
∫∫bottom 0dσ = 0
The integral over the curved sides will require a little more
effort. The outward normal (see my notes, Surfaces,
pg. 5) and area element are, respectively,
n = cos(θ)i + sin(θ)j and dσ =
4dθdz.
In addition, on the curved sides
F(4cos(θ),4sin(θ),z) = 12cos(θ)i +
sin(θ)j + 2zk, so F·n
= 12cos2(θ) + 4sin2(θ).
The surface integral over the curved sides is then given by
∫∫sides F·ndσ =
∫05∫02π
(12cos2(θ) + 4sin2(θ))4dθdz =
5·4(12π+4π)= 320π.
Combining these three integrals, we obtain
∫∫S F·ndσ =
160π+320π + 0 = 480π,
which agrees with the result from the volume integral. Thus we have
verified the Divergence Theorem in this case.