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Orient the triangular boundary from $P=(1,0,0)$ to $Q=(0,1,0)$ to $R=(0,0,1)$ and back to $P$. This orientation corresponds to the positive normal vector $(Q-P)\times (R-P)=(1,1,1)$. The cyclic coordinate rotation $(x,y,z)\mapsto (z,x,y)$ sends the three directed edges to one another and transforms the vector field by the same rotation. Since rotations preserve dot products, the three edge contributions to the line integral are equal.

On the first edge use $x=1-t$, $y=t$, $z=0$, $0\le t\le 1$. ThenIts integral is $2/3$, so the required circulation is

For a sufficiently smooth vector field on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the Stokes theorem statesTherefore the requested surface integral, with all normal components positive, is also $2$.

To verify it directly, parametrize the triangle by $\mathbf{r}(y,z)=(1-y-z,y,z)$ on $y\ge 0$, $z\ge 0$, $y+z\le 1$. Its two tangent vectors are ${\mathbf{r}}_y=(-1,1,0)$ and ${\mathbf{r}}_z=(-1,0,1)$, giving the vector area elementDifferentiation of the vector field gives $\nabla \times A=2(y+z,z+x,x+y)$. On the triangle $x+y+z=1$, so its dot product with $(1,1,1)$ is $4$. Consequentlyconfirming both the magnitude and the orientation sign.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

Take the unit normal to point outwards. The divergence of the field isThe divergence theorem turns the closed surface integral into a volume integral over $[0,1{]}^3$:For a direct check of the surface integral, on $x=1$ the outward component is $F_x=2y$, contributing ${\int }_0^1{\int }_0^{1}2ydydz=1$. On $x=0$ it vanishes. On both $y=0$ and $y=1$, $F_y=xy(y-1{)}^2=0$. On $z=1$, $F_z=0$, whereas on $z=0$ the field component is $F_z=-1$ and the outward normal is $-\hat{\mathbf{z}}$, giving a contribution $1$. The six face contributions are therefore $1,0,0,0,0,1$, again totaling 2. The positive bottom-face contribution is a consequence of its outward orientation.

The divergence of a cross product identity and the stated static field equations giveFor an outward-oriented closed surface, the divergence theorem therefore gives the flux of the specified Poynting vector normalization:No open-surface conclusion follows from this equation without including its closing pieces.

For the particular fields, direct calculation givesThe fields also have $\nabla \times \mathbf{E}=0$, $\nabla \cdot \mathbf{B}=0$, and $\nabla \cdot \mathbf{J}=0$, consistently with the required equations. Close the outward spherical patch by the two planar half-disks on $x=0$ and $y=0$, forming the quarter unit ball $V$. Its total closed flux iswhere $r$ in this integral is cylindrical radius. The $x=0$ face has outward normal $-{\mathbf{e}}_x$ and zero flux because $P_x=0$ there. The $y=0$ face has outward normal $-{\mathbf{e}}_y$, so its flux isSubtracting these planar contributions from the closed flux gives the requested open-patch answerAs a direct check, on the unit sphere the outward normal is $(x,y,z)$ and $\mathbf{P}\cdot \mathbf{n}=xy(x^2-2z^2)$. With spherical angles $0\le \vartheta \le \pi$, $0\le \phi \le \pi /2$, its surface integral is $\frac{1}{4}{\int }_0^{\pi }{\sin }^5\vartheta d\vartheta -{\int }_0^{\pi }{\sin }^3\vartheta {\cos }^2\vartheta d\vartheta =\frac{1}{4}(16/15)-4/15=0$. The curved patch's vanishing flux is a cancellation, not an inference that its flux density vanishes everywhere.

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.

For the translating sphere, write its center as $\mathbf{X}(t)=a_1\cos (\omega t)\mathbf{e}$, with $\mathbf{e}$ a unit vector, and its velocity as $\mathbf{V}=\dot{\mathbf{X}}$. Let $\mathcal{V}=4\pi a^3/3$, $\eta =a\omega /c_0\ll 1$ and $d=a_1/a\ll 1$. Impermeability gives $u_n=\mathbf{V}\cdot {\mathbf{n}}_s$. Since its volume is constant,in the leading compact formula. The individual local thickness sources have scale ${\rho }_0(a_1/r){\eta }^2$, but their net monopole cancels.

At first order in amplitude and leading order in $\eta$, the exterior incompressible velocity potential is $\varphi =-a^3\mathbf{V}\cdot {\mathbf{r}}_s/(2r_s^3)$, where ${\mathbf{r}}_s$ is measured from the mean center. It has ${\partial }_{r_s}\varphi =\mathbf{V}\cdot {\mathbf{n}}_s$ at $r_s=a$. Linear Bernoulli equation gives $p^{\prime }=-{\rho }_0{\varphi }_t=({\rho }_{0}a/2)\dot{\mathbf{V}}\cdot {\mathbf{n}}_s$ there. Since $\int n_{s,i}{n}_{s,j}dS=(4\pi a^2/3){\delta }_{ij}$,This is the added mass of a sphere. The explicit leading loading contribution isIt scales as ${\rho }_0(a_1/r){\eta }^3$. In homentropic small-amplitude flow the volume stress is quadratic to leading order, $T_{ij}=O({\rho }_0{V}^2)$ over volume $O(a^3)$, so the volume acoustic quadrupole has scaleIt is smaller than the linear dipole by $O(d\eta )=O(V/c_0)$ and can contain a second harmonic.

For completeness, the exact thickness-source integral has a nonzero first delay correction even though $m=0$. Its first spatial moment isExpand its integrand at ${\tau }_y=\tau +\mathbf{n}\cdot \mathbf{y}/c_0$. The leading nonzero translating-sphere acoustic thickness dipole is thereforeIt too scales as ${\rho }_0(a_1/r){\eta }^3$. Thus zero net-volume monopole does not mean zero thickness radiation. The full leading linear sound, summing thickness and loading, isThe leading compact equation gives the requested zero $\dot{m}$ and added-mass $\dot{f}$ terms; this extra delay calculation is necessary to obtain the complete linear field when its nominal monopole has cancelled.

The divergence theorem states that a continuously differentiable vector field on a bounded region with piecewise smooth boundary satisfieswith the unit normal pointing out of $V$. Apply it separately to $\mathbf{u}=f{\mathbf{e}}_i$, where ${\mathbf{e}}_i$ is each constant Cartesian basis vector. Since $\nabla \cdot (f{\mathbf{e}}_i)={\partial }_{i}f$ and $(f{\mathbf{e}}_i)\cdot \mathbf{n}=f{n}_i$, the three component identities combine to give

For the specified scalar field, $\nabla f=(z,0,x)$. The $x$ term integrates to zero by reflection symmetry. At height $z$, the region has a disk cross-section of area $\pi (a^2-cz)$, and $0\le z\le a^2/c$. Therefore the volume integral isTo check the surface integral independently, split the boundary into the base disk and the curved graph $z=g(x,y)=(a^2-x^2-y^2)/c$. On the base $f=0$, so its contribution vanishes. The region lies below the curved graph, whose outward vector surface element of a graph isOver the disk $x^2+y^2\le a^2$, the $y$ and $z$ components of $fd\mathbf{S}$ are odd in $x$ and integrate to zero. Its $x$ component gives, in polar coordinates,Thus the two explicitly evaluated integrals agree, with the outward orientation included.

Differentiate the position vector with respect to each of the spherical polar coordinates. The derivatives areTheir lengths give the scale factors of orthogonal coordinates:Dividing the derivatives by these lengths yieldsEach has unit length. For example, ${\mathbf{e}}_r\cdot {\mathbf{e}}_{\theta }=\sin \theta \cos \theta ({\cos }^2\varphi +{\sin }^2\varphi )-\cos \theta \sin \theta =0$; the other two dot products vanish because their first two terms cancel. Thus these are mutually orthogonal vectors, and the total differential has the required form. The coordinates degenerate at $r=0$ and on the polar axis, but this does not affect an area integral.

On the parametrized surface $\theta =\alpha$, use $r,\varphi$ as parameters. The surface integral area element isFor $0\le \alpha \le \pi$, its lateral area, with no closing disk, is thereforeAt $\alpha =0,\pi$ the surface collapses to a line and its area is zero.

For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

Take the normal vector outward from the cylinder-and-cone solid. The surface consists of the cylindrical side and conical roof, with no bottom disk; this convention fixes the otherwise unspecified orientation of a surface. The boundary is the radius-two circle at $z=-2$. Reversing every normal vector reverses the final sign.

For $\mathbf{F}=(y{z}^2,0,0)$, its curl is $(0,2yz,-z^2)$. For the cylindrical parametrized surface, use ${\mathbf{r}}_c=(2\cos \theta ,2\sin \theta ,z)$, with $-2\le z\le 2$, and use ${\mathbf{r}}_{c,\theta }\times {\mathbf{r}}_{c,z}=(2\cos \theta ,2\sin \theta ,0)$ as the outward area vector. Its flux isFor the cone, let $q=4-z$, $2\le z\le 4$ and ${\mathbf{r}}_k=(q\cos \theta ,q\sin \theta ,z)$. Its outward area vector is ${\mathbf{r}}_{k,\theta }\times {\mathbf{r}}_{k,z}=(q\cos \theta ,q\sin \theta ,q)$. ThusThe total surface integral is therefore $-16\pi$ with the outward convention, or $+16\pi$ for the opposite convention.

For Stokes theorem, the induced bottom-circle orientation is counterclockwise as viewed from above, opposite to the orientation of an outward bottom cap. Use $\mathbf{r}=(2\cos \theta ,2\sin \theta ,-2)$ with increasing $\theta$. Thenagreeing with the two parametrized fluxes. The seam at $z=2$ is internal and cancels between the two pieces; the cone tip does not supply another boundary curve.

Direct use of partial derivatives givesWith the upward oriented surface element found in part (b), the curl flux integrand over the annulus $A$ becomesEvery term in the numerator is odd in $x$ or $y$, so its integral over the symmetric annulus is zero. Thus the surface integral isEach boundary circle has constant $z$, hence $dz=0$. Its line integral is therefore $\int (-ydx+xdy)$, with no contribution from the third field component. An anticlockwise radius-$r$ circle gives $2\pi r^2$. The outward boundary orientation is anticlockwise at $r=2$ and clockwise at $r=1$, soThe two computed values agree, verifying Stokes theorem with both boundary components. A downward choice would give $-6\pi$ on both sides.

At every regular point of the double circular cone, its normal is perpendicular to $\mathbf{F}$ by part (a). Therefore the integrand of the flux integral vanishes pointwise:The vertex, if included, is a single point of zero surface area and does not alter the surface integral. The answer is independent of both the subset and its orientation. Tangency, rather than a cancellation between different parts of the surface, explains the zero flux.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

In cylindrical coordinates, $\rho =(x^2+y^2{)}^{1/2}$ and $\mathbf{F}=(f(\rho )x/\rho ,f(\rho )y/\rho ,0)$. The chain rule gives ${\partial }_x\rho =x/\rho$ and ${\partial }_y\rho =y/\rho$. ThereforeThis is cylindrically radial divergence, valid for $\rho >0$.

The bounded domain is outside the unit cylinder, below the inverse-square surface and above the lower plane:The upper plane touches only the rim $\rho =1,z=1$; it is not an extra positive-area top face. Revolving the shaded meridional region below around the $z$ axis gives the three-dimensional domain.

With the Jacobian determinant $dV=\rho d\rho d\varphi dz$ of cylindrical coordinates,Zero divergence gives $(\rho f{)}^{\prime }=0$, hence the most general solution on the positive radial interval isAlthough this vector field is singular on the axis if $A\ne 0$, the entire closure of $V$ lies at $\rho \ge 1$, so it is smooth on a neighbourhood of $V$ and the divergence theorem applies.

Its volume integral of divergence is zero. Compute the outward surface integral directly. On the inner cylinder, $\mathbf{n}=-{\mathbf{e}}_{\rho }$, $dS=d\varphi dz$, and the flux is $-2\pi A(1-1/4)=-3\pi A/2$. On the curved roof $z=g(\rho )={\rho }^{-2}$, the outward vector surface element of a graph isTherefore its flux isThe horizontal bottom has zero flux and the top rim has zero surface area. Summing giveswhich verifies the divergence theorem with every boundary contribution accounted for.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.