Question A.5.1: Verify Stokes’ theorem for the vector field A(x, y) = (2x - ...
Verify Stokes’ theorem for the vector field A(x, y) = (2x – y)\hat{x} – yz^{2}\hat{y} + y^{2}z\hat{z} where S is the half-surface of the sphere x^{2} + y^{2} + z^{2} = a^{2} and C is its bounding curve in the x-y plane (Fig. A.10).
First, we evaluate \int_{s} (∇×A) · \hat{n} ds. It is easy to check that
(∇ × A) = \hat{z}.
In this case, the open surface S is defined by x^{2} + y^{2} + z^{2} = a^{2} with z ≥ 0. The unit vector \hat{n} on S is given by
\hat{n} = \frac{x\hat{x} +y\hat{y} +z\hat{z}}{a}.
Therefore,
(∇ × A) ⋅ \hat{n} =\frac{z}{a}.
We want to evaluate \int_{s} (z/a) ds over the hemisphere. We use spherical coordinates in which
z = r cos(θ),
and
ds = r^{2} sin(θ) dθ dΦ.
On S we have r = a and, therefore,
\int_{s}\frac{z}{a} ds = \int_{0}^{2π} \int_{0}^{π/2} a^{2} cos(θ) sin(θ) dθ dΦ
= a^{2}π. (A.43)
Next, we evaluate \oint_{c} A· dl using cylindrical coordinates. We transform A(x, y) into cylindrical coordinates using Table A.1. This gives
Table A.1 Coordinate transformations for Cartesian coordinates
| Spherical | Cylindrical | Cartesian |
| = sin(θ) cos(Φ)\hat{r}_{s} +cos(θ)cos(Φ)\hat{θ} – sin(Φ)\hat{Φ} | = cos(Φ)\hat{r}_{c} – sin(Φ)\hat{Φ} | \hat{x} |
| = sin(θ) sin(Φ)\hat{r}_{s} +cos(θ)sin(Φ)\hat{θ} + cos(Φ)\hat{Φ} | = sin(Φ)\hat{r}_{c} + cos(Φ)\hat{Φ} | \hat{y} |
| = cos(θ)\hat{r} -sin(θ)\hat{θ} | = \hat{z} | \hat{z} |
| = r_{s} sin(θ)cos(Φ) | = r_{c}cos(Φ) | x |
| = r_{s} sin(θ)sin(Φ) | = r_{c}sin(Φ) | y |
| = r_{s}cos(θ) | = z | z |
| = A_{r_{s}}sin(θ)cos(Φ) +A_{θ}cos(θ)cos(Φ) -A_{Φ}sin(Φ) | = A_{r_{c}}cos(Φ) -A_{Φ}sin(Φ) | A_{x} |
| = A_{r_{s}}sin(θ)sin(Φ) +A_{θ}cos(θ)sin(Φ) + A_{Φ}cos(Φ) | = A_{r_{c}}sin(Φ) -A_{Φ}cos(Φ) | A_{y} |
| = A_{r_{s}}cos(θ) – A_{θ}sin(θ) | = A_{z} | A_{z} |
A(r,Φ) = (2r cos(Φ) – r sin(Φ))(cos(Φ)\hat{r} – sin(Φ)\hat{Φ})
– r sin(Φ)z^{2}(sin(Φ)\hat{r} + cos(Φ)\hat{Φ})+r^{2} sin^{2}(Φ)z\hat{z}.
On the curve C, r = a, z = 0, and dl = a dΦ\hat{Φ}. Therefore,
A(r, Φ) · dl = (-2a^{2} cos(Φ) sin(Φ)+a^{2} sin^{2}(Φ)) dΦ
and
\oint_{c} A · dl = \int_{0}^{2π}(-2a^{2} cos(Φ) sin(Φ)+a^{2} sin^{2}(Φ)) dΦ
= a^{2}π. (A.44)
Notice that Eq. (A.44) equals Eq. (A.43), which verifies Stokes’ theorem.