13—Vector Calculus 2 330ˆr
θ
θˆ
Example
Verify Gauss’s Theorem for the solid hemisphere,r≤R,
0 ≤θ≤π/ 2 , 0 ≤φ≤ 2 π. Use the vector field
F~=rαrˆ^2 sinθ+θβrθˆ^2 cos^2 φ+φγrˆ sinθcos^2 φ (13.16)
Doing the surface integral on the hemisphereˆn=rˆ, and on the bottom flat disk you havenˆ=θˆ. The
surface integral is then assembled from two pieces,
∮F~.dA~=
∫
hemisphˆrαr^2 sinθ.rdAˆ +
∫
diskθβrθˆ^2 cos^2 φ.θdAˆ
=
∫π/ 20R^2 sinθdθ
∫ 2 π0dφαR^2 sinθ+
∫R
0rdr
∫ 2 π0dφβr(π/2)^2 cos^2 φ
=απ^2 R^4 /2 +βπ^3 R^3 / 12 (13.17)
Now do the volume integral of the divergence, using Eq. (9.16).
divF~=
1
r^2
∂
∂r
αr^4 sinθ+
1
rsinθ
∂
∂θ
βrθ^2 sinθcos^2 φ+
1
rsinθ
∂
∂φ
γrsinθcos^2 φ
= 4αrsinθ+βcos^2 φ[2θ+θ^2 cotθ] + 2γsinφcosφ
Theγterm in the volume integral is zero because the2 sinφcosφ= sin 2φfactor averages to zero
over allφ.
∫R
0drr^2
∫π/ 20sinθdθ
∫ 2 π0dφ
[
4 αrsinθ+βcos^2 φ[2θ+θ^2 cotθ]
]
= 4α.
R^4
4
. 2 π.π
4
+β.
R^3
3
.π.
∫π/ 20dθsinθ[2θ+θ^2 cotθ]
=απ^2 R^2 /2 +βπ^3 R^3 / 12
The last integration used parametric differentiation starting from
∫π/ 20 dθcoskθ, with differentiation
with respect tok.
13.4 Stokes’ Theorem
The expression for the curl in terms of integrals is Eq. (9.17),
curl~v= lim
V→ 01
V
∮
dA~×~v (13.18)
Use the same reasoning that leads from the definition of the divergence to Eqs. (13.14) and (13.15)
(see problem13.6), and this leads to the analog of Gauss’s theorem, but with cross products.
∮SdA~×~v=
∫
Vcurl~vdV (13.19)
This isn’t yet in a form that is all that convenient, and a special case is both easier to interpret and
more useful in applications. First apply it to a particular volume, one that is very thin and small. Take