Surface Integrals, Stokes & Gauss

This is \(\langle x, y, 0\rangle\) — radially outward, as a closed can would want. Its length is 3, so \(dS = 3\,d\theta\,dz\) and

which is circumference \(6\pi\) times height 4. ✓

Larger than the unit disk (\(\pi\approx3.14\)) as any tilted surface must be. The \(x^2 + y^2\) in the integrand and the circular region both pointed to polar.

Spherical parametrisation with \(\rho = 2\): \(z = 2\cos\phi\), \(dS = \rho^2\sin\phi\,d\phi\,d\theta = 4\sin\phi\,d\phi\,d\theta\), \(0\le\phi\le\pi/2\).

Sanity: the hemisphere has area \(2\pi(2)^2 = 8\pi\), so the average of \(z\) over it is 1 — half the radius. Compare: the average of \(z\) over the solid half-ball was \(3R/8 = 0.75\) (Exam 4). The shell's mass sits higher than the solid's. ✓

Graph formula with \(g = 4 - x^2 - y^2\), \(g_x = -2x\), \(g_y = -2y\), over the disk \(x^2 + y^2\le4\):

The \(4xy\) term integrates to zero over the disk by symmetry (odd in \(x\)). The rest in polar:

The upward normal \(\langle -g_x, -g_y, 1\rangle = \langle 2x, 2y, 1\rangle\) is automatic from the graph formula — no orientation check needed as long as “upward” is what was asked.

For \(a = 1\): \(12\pi/5\approx7.54\). Computing this as a surface integral would mean parametrising the sphere and integrating \(\cos^4\) and \(\sin^4\) terms — the divergence theorem turned it into a one-line spherical integral. The two \(\rho^2\) factors are different things: one is the integrand, one is the Jacobian.

Let \(E\) be the solid half-ball. Its closed boundary is the hemisphere \(S\) (outward = upward) plus the disk \(D\) (outward = \(-\mathbf{k}\)).

(using \(\int_0^{\pi/2}\sin\phi\,d\phi = 1\) and \(\int_0^{\pi/2}\cos^2\phi\sin\phi\,d\phi = \tfrac13\).) On the disk, \(\mathbf{F}\cdot(-\mathbf{k}) = -z^3 = 0\) since \(z = 0\). So

Direct check (optional): on the unit sphere \(\mathbf{n} = \mathbf{r}\), so \(\mathbf{F}\cdot\mathbf{n} = x^2 + y^2 + z^4 = \sin^2\phi + \cos^4\phi\), and \(\int_0^{2\pi}\!\int_0^{\pi/2}(\sin^2\phi + \cos^4\phi)\sin\phi\,d\phi\,d\theta = 2\pi(\tfrac23 + \tfrac15)\). ✓

The boundary is the unit circle in the plane \(z = 0\), counterclockwise from above (right-hand rule with upward normal). On it \(z = 0\):

Confirm: \(\nabla\times\mathbf{F} = \langle \partial_y x - \partial_z z,\ \partial_z y - \partial_x x,\ \partial_x z - \partial_y y\rangle = \langle -1, -1, -1\rangle\). Through the flat unit disk with \(\mathbf{n} = \mathbf{k}\): \(\iint(-1)\,dA = -\pi\). ✓ Same boundary, same flux of curl, whichever cap — the Stokes lab's Paraboloid versus Flat disk.

(a) The field \(\langle y+z,\ z+x,\ x+y\rangle\) is \(\nabla(xy + yz + zx)\) — its curl is zero (check any component: \(\partial_y(x+y) - \partial_z(z+x) = 1 - 1 = 0\)). A conservative field around a closed loop: \(0\). No parametrising needed.

(b) \(\nabla\times\mathbf{G} = \langle 1, 1, 1\rangle\). Take \(S\) to be the piece of the plane \(z = 2y + 1\) inside the cylinder, a graph over the unit disk with \(g_y = 2\), \(g_x = 0\), upward normal \(\langle -g_x, -g_y, 1\rangle = \langle 0, -2, 1\rangle\):

The tilted ellipse \(C\) never had to be parametrised. Note that the projection of \(S\) onto the \(xy\)-plane is the unit disk regardless of how steep the plane is — the graph formula absorbs the tilt into the normal vector.

(a) By symmetry \(\mathbf{E} = E(r)\,\hat{\mathbf{r}}\) (radial, away from the axis). For a coaxial cylinder of radius \(r\) and length \(L\), the flux through the flat ends is zero (\(\mathbf{E}\parallel\) ends) and through the side is \(E(r)\cdot2\pi rL\):

(b) \(\hat{\mathbf{r}}/r = \langle x, y, 0\rangle/(x^2+y^2)\), so \(\mathbf{E} = \dfrac{\lambda}{2\pi\varepsilon_0}\dfrac{\langle x, y, 0\rangle}{x^2+y^2}\):

(c) Away from the line there is no charge, \(\rho = 0\), so \(\nabla\cdot\mathbf{E} = 0\) is exactly what the differential form demands. All the divergence is concentrated on the line itself, where \(\rho\) is infinite (a line charge is a delta function in the plane) and where the formula for \(\mathbf{E}\) is singular. The integral form sees that concentrated source as the enclosed charge \(\lambda L\); the differential form sees it as a spike in \(\rho\). Same physics, two languages, connected by the divergence theorem.

(a) \(\Phi_B = B_0 t\cdot\pi a^2\), so \(\mathcal{E} = -\dfrac{d\Phi_B}{dt} = -\pi a^2B_0\). The minus sign (Lenz) says the induced current opposes the growth of the flux.

(b) By symmetry \(\mathbf{E}\) is tangential with magnitude \(E(r)\). Faraday on a circle of radius \(r\): \(E(r)\cdot2\pi r = -\pi r^2 B_0\), so

(c)

This \(\mathbf{E}\) has curl, so it is not conservative and has no potential — there is no “voltage” here in the electrostatic sense, and a voltmeter's reading around this loop depends on how its leads are routed. That is the physical content of Stokes' theorem in a transformer.