Codex @codex  0

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

Use cylindrical coordinates $(\rho ,\theta ,z)$. The right circular cone is described by $0\le z\le 1$ and $0\le \rho \le a(1-z)$, and the divergence of $\mathbf{F}=(0,0,z^2)$ is $2z$. Direct integration gives

To check the boundary flux integral independently, first consider the base $z=0$. Its outward unit normal vector is $-\mathbf{k}$, but $\mathbf{F}=0$ there, so its flux is zero. Parametrize the lateral surface byThe cross productpoints outward: its horizontal component points away from the axis of the cone. This gives the vector surface element without separately normalizing it. Consequently the lateral flux integral isAdding the zero base contribution agrees with the volume integral, verifying the divergence theorem. The apex has zero surface area; the vector field is smooth there, so no additional flux is hidden at that point.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary $S=\partial V$, and let $\mathbf{F}$ have continuous first partial derivatives on a neighborhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. The right side is the outward flux; reversing the orientation reverses its sign.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

The divergence is $\nabla \cdot F=-1+0+1=0$. Take the hemisphere's outward orientation, and close it with the unit disk $D$ at $z=0$, whose outward normal is $-e_z$. The divergence theorem then gives ${\int }_{S}F\cdot dS+{\int }_{D}F\cdot dS=0$. On the disk, $F\cdot (-e_z)=-x^2$, and polar coordinates giveConsequently the curved hemisphere has fluxThe reversed hemisphere orientation would reverse the sign.

Choose the plane normal to be $+\hat{y}$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the surface integral factors into two Gaussian integrals:The original question does not specify an orientation; choosing $-\hat{y}$ instead gives $-\pi$.

The printed first component $\sin y/y$ is undefined on the entire integration plane. Its removable continuous extension there is $1$, since ${\lim }_{y\to 0}\sin y/y=1$. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

Take the sphere's outward normal. Since $F=k(x,y,z)$ with constant $k$, its divergence is $3k$. The divergence theorem givesThe translated centre does not enter the answer, because the divergence is constant. Equivalently the constant field produced by translating coordinates has zero net flux through a closed sphere. The inward orientation would negate the result.

For a sufficiently smooth vector field on a volume $V$ and its closed boundary $S$, the divergence theorem stateswith the unit normal directed outwards. HereHence the net volume flux through every closed surface in the liquid is zero. This verifies incompressible flow directly. It does not require the velocity to vanish pointwise on an arbitrary closed surface; inward and outward contributions can cancel.

For an oriented surface $S^{\prime }$ with compatible boundary direction, the Stokes theorem statesLet $H$ denote the removed patch and $S^{\prime }=S\setminus H$ the remaining cylinder. Keep the outward orientation inherited from the closed cylinder. Everywhere on the removed curved patch, the normal curl component is $-\alpha R$, so its flux is $-\alpha RA$, where $A$ is its actual curved area. Part (d) then givesConsequently the circulation around a hole in an oriented surface isThe positive direction here is induced by the remaining outward-oriented surface; locally it runs clockwise around the hole when viewed from outside. If the edge is instead traversed positively around the removed patch, the direction reverses and the answer is $-\alpha RA$. The question does not specify an edge direction, so the sign must be accompanied by this convention. The result depends only on the removed curved area, not on its shape. If the patch has several boundary components, use the sum of their induced line integrals. Since $[\alpha ]=1/(\text{length}\cdot \text{time})$, the result has the units of a velocity line integral, length squared per time.

The integrand field is exactly the curl computed in part (a), so use its polynomial vector potential $\mathbf{G}=(z^3,x^3,y^3)/3$. The upper cap meets the plane $z=0$ in an ellipse. Since the parameters are only assumed nonzero, set $a_0=|a|$ and $b_0=|b|$ to use positive semiaxes. Its correctly oriented boundary isOn that boundary, $G_x=0$ and $dz=0$, so only $G_{y}dy=x^{3}dy/3$ contributes. The Stokes theorem therefore givesThe last integral follows by expanding ${\cos }^{4}t=(3+4\cos 2t+\cos 4t)/8$. ThusThe result is independent of $c$. A sign check uses the divergence theorem: closing the cap with its base disk adds outward normal $-\hat{\mathbf{z}}$ there. Since $\nabla \cdot \mathbf{F}=0$, the cap flux equals ${\int }_{\mathrm{disk}}{x}^{2}dxdy>0$, consistent with the boxed expression and its absolute values.

The PDF diagram specifies the prism asIts triangular horizontal cross-section has area four. Calculate the outward flux on each face using its outward normal vector.

On $x=0$, the normal is $-{\mathbf{e}}_x$ and $F_x=0$, so the flux is zero. On $y=0$, the normal is $-{\mathbf{e}}_y$, givingOn the sloping vertical face $y+2x=4$, the outward normal is proportional to $(2,1,0)$. Both $F_x$ and $F_y$ vanish there, while $F_z$ is tangent to the face, so the flux is zero.

On the bottom $z=0$, the normal is $-{\mathbf{e}}_z$ and $F_z=1$, so the flux is minus the base area, namely $-4$. On the top $z=1$, $F_z=0$, so the flux is zero. Thus all five face values and their sum are

The specified unit normal vector is ${\mathbf{e}}_z$, while $F_z=0$ everywhere. Thus the normal component of the field vanishes pointwise, giving flux

Here $y=0$, the unit normal vector is ${\mathbf{e}}_y$, and $\mathbf{F}\cdot {\mathbf{e}}_y=x^2$. The square has $-1/2\le x,z\le 1/2$, so its outward flux is

Take the outward orientation of the complete closed surface, including its flat base. The constant part of the vector field has zero divergence, and the radial part has divergence $3\mathbf{a}\cdot \mathbf{b}$. The divergence theorem gives its flux as that constant times the solid hemisphere's volume, $2\pi R^3/3$:The constant field may have nonzero flux through the curved part alone, but its curved and base contributions cancel on the closed surface.

Use the orientation induced by the boundary order $O\to A\to B\to O$. The triangle's vector area element integrates to $\frac{1}{2}\mathbf{a}\times \mathbf{b}$. Every point $\mathbf{r}$ in its plane is a linear combination of $\mathbf{a},\mathbf{b}$, so $\mathbf{r}\cdot (\mathbf{a}\times \mathbf{b})=0$. The radial term of the field therefore contributes no flux. The constant term givesReversing the orientation reverses this sign. If $\mathbf{a}$ and $\mathbf{b}$ are collinear, the triangle has zero area and the displayed value is zero by the degenerate-area convention.

Use the outward unit normal on the unit sphere and the convention that $\theta$ is measured from the positive vertical axis. Then $\mathbf{F}\cdot \mathbf{n}=\sin \theta$, while the spherical coordinates area element is $dS=\sin \theta d\theta d\varphi$. The flux is therefore