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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 two independently computed answers agree:This verifies the divergence theorem. Its hypotheses hold because the field is smooth and the prism has a closed piecewise smooth boundary with consistently outward normals.

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.

Write $q(x)=1/[x(1+(\ln x{)}^2)]$. The vector field is undefined at $x=0$ because $q$ diverges there, but this component is tangential to that face. The face fluxes have finite improper integral limits. In particular, the substitution $s=\ln x$ givesUsing the outward unit normal on each face, the flux contributions areHenceFor a precise justification at the singular face, first use the truncated cube $\epsilon \le x\le 1$. Its opposite $y$-face integrals cancel, the $x$-faces contribute $e^{-1}-e^{-\epsilon }$, and the upper $z$-face contributes $1-\epsilon$. The limit as $\epsilon \downarrow 0$ is $e^{-1}$. On each truncated cube the divergence theorem gives the same answer, since $\nabla \cdot \mathbf{F}=1-e^{-x}$. Thus the result is a flux of a vector field with an integrable tangential singularity, rather than an application of the classical smooth-field theorem directly at $x=0$.

Let $V$ be the enclosed region with piecewise smooth closed boundary $S$, oriented by its outward unit normal $\mathbf{n}$. For a continuously differentiable vector field on a neighbourhood of $V$, the divergence theorem statesThe outward flux integral equals the volume integral of the divergence.

Add the four outward fluxes:For a check, $\nabla \cdot \mathbf{F}=2(x+y+z)$. The divergence theorem gives the same result from

The divergence of the position vector field is $\nabla \cdot \mathbf{r}=3$. At height $z$, the horizontal cross-section is a disk of area $\pi (1-z^2)$. By the divergence theorem, the outward flux integral is three times the enclosed volume:ThereforeA direct check separates the two pieces of the surface integral. On the unit sphere, the outward unit normal is $\mathbf{r}$, so $\mathbf{r}\cdot \mathbf{n}=1$ and the curved contribution is $2\pi (z_0+1)$. On the top disk, $\mathbf{n}={\mathbf{e}}_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same flux.

Green identity. With $n$ the outward unit normal, the product rule gives $\nabla \cdot (u\nabla v)=u{\nabla }^{2}v+\nabla u\cdot \nabla v$. Integrating over $V$ and applying the divergence theorem yields Green's first identity:

Uniqueness for nonzero mass. Suppose two real solutions have the same prescribed boundary data, and let $w$ be their difference. The spatial equation called the Klein-Gordon equation in the question is the modified Helmholtz equation, so ${\nabla }^{2}w=m^{2}w$. With $u=v=w$, Green's first identity becomesFor identical Dirichlet boundary conditions, $w=0$ on $S$; for identical Neumann boundary conditions, ${\partial }_{n}w=0$ there. Either makes the right-hand side zero. When $m$ is real and nonzero, both integrands on the left are nonnegative, forcing $w=0$. ThusThis is the required uniqueness assertion; existence is understood for admissible smooth data on a suitable bounded region. Complex-valued solutions obey the same result by treating their real and imaginary parts separately.

Zero mass. For $m=0$, the argument gives $\nabla w=0$, hence $w=C$ on the connected region. Dirichlet boundary conditions still force $C=0$, but Neumann boundary conditions leave an arbitrary additive constant:The prescribed normal data in the zero-mass case must also satisfy ${\int }_S{\partial }_n\Phi dS=0$, by the divergence theorem. For nonzero $m$, that flux can instead equal $m^2{\int }_V\Phi dV$.

Energy minimization. Let $\eta =w-v$ for the two functions in the last part. They share their boundary values, so $\eta =0$ on $S$. Since $v$ is a harmonic function, Green's first identity givesExpand the square in the Dirichlet energy of $w=v+\eta$:The cross term vanishes, soEquality requires $w-v$ to be constant, and the shared boundary values then make it zero. This is the Dirichlet principle: among functions with fixed boundary values, the harmonic function minimizes Dirichlet energy.

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

For a bounded region $R$ with piecewise smooth boundary and a continuously differentiable vector field, the divergence theorem gives ${\int }_R\nabla \cdot \mathbf{F}dV={\int }_{\partial R}\mathbf{F}\cdot \mathbf{n}dS$, with outward unit normal $\mathbf{n}$. Apply it to $\mathbf{F}=u\nabla v-v\nabla u$. The cross terms cancel, giving the Green's second identityThe $m^{2}uv$ terms cancel identically. Sending $R\to \infty$ proves the whole-space identity provided the boundary integral tends to zero. The boundary term at infinity in Green's second identity requires adequate derivative decay as well as field decay; mere pointwise convergence $u,v\to 0$ does not suffice. The screened kernels used below satisfy the required decay.

Take $m>0$. Apply the identity with $u=\varphi$ and $v=G(\cdot ,\mathbf{r})$, treating the source through its Dirac delta function, or excising a small source sphere and taking its radius to zero. It gives $\varphi (\mathbf{r})-\int G({\mathbf{r}}^{\prime },\mathbf{r})\rho ({\mathbf{r}}^{\prime })d{V}^{\prime }=0$. Symmetry of the radial Green function then yields the requested representation.

For separation $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|>0$, the modified Helmholtz equation reduces to $(sG{)}^{\prime \prime }-m^2(sG)=0$. Decay selects $G=C{e}^{-ms}/s$. Integrating the equation over a small sphere gives $4\pi s^2{G}^{\prime }\to 1$, so $C=-1/(4\pi )$. ThusFor $\rho =e^{-\beta r}/r$ with $\beta >0$, put $h=r\varphi$. Then $h^{\prime \prime }-m^{2}h=e^{-\beta r}$. When $\beta \ne m$, decay removes $e^{mr}$, while $h(0)=0$ removes an additional point source at the origin. HenceSince $G$ has a minus sign, the positive Yukawa kernel convolution isThe coincident-rate limit $\beta =m$ is $e^{-mr}/(2m)$ for the convolution, and the origin limit for general rates is $1/(m+\beta )$. Both are finite; the apparent denominators in the distinct-rate formula are removable limits.

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.

Away from its source, a radial two-dimensional Green's function satisfies $G^{\prime \prime }+G^{\prime }/r=0$, so $G=C\log r+C_0$. The divergence theorem on a disk of radius $\rho$ around the source requires $\int {\nabla }^{2}GdA=1$, hence $2\pi \rho G^{\prime }(\rho )=1$ and $C=1/(2\pi )$. Choosing the additive constant givesThe equality with the Dirac delta is distributional; away from the source the Laplacian vanishes.

For the wedge, use the finite image construction for a planar wedge. In complex coordinates $z_0=r_0{e}^{i{\theta }_0}$, place three positive sources at $z_0{e}^{2k\pi i/3}$ and three negative sources at $\overline{z_0}{e}^{2k\pi i/3}$, $k=0,1,2$. The positive source at $k=0$ is the actual charge. The five images are thereforeOn either grounded ray, reflection pairs each positive source with an equally distant negative source, so their logarithmic potentials cancel. All images lie outside the open physical wedge. Summing their contributions with coefficient $-e/(2\pi {\epsilon }_0)$ gives the required potential. The figure shows the bisecting-source configuration.

For $r_0=1$, ${\theta }_0=\pi /6$, the positive points are the three roots of $z^3=i$ and the negative points are the roots of $z^3=-i$. Thus the logarithmic sum isBoth boundaries have $\sin (3\theta )=0$, verifying the zero potential directly. An arbitrary reference length inside the individual logarithms cancels because the six signed charges sum to zero.

The divergence is $y^2+z^2+x^2$. Over the cube $[-1/2,1/2{]}^3$, each squared-coordinate term has integral ${\int }_{-1/2}^{1/2}{s}^{2}ds=1/12$, while the other two coordinate integrals have length one. ThusCompute the surface flux separately. On $x=1/2$, the outward normal is $e_x$ and $F\cdot n=y^2/2$, giving flux $1/24$. On $x=-1/2$, the normal is $-e_x$, so $F\cdot n=-x{y}^2=y^2/2$, again giving $1/24$. The two $y$ faces similarly give $z^2/2$ each, and the two $z$ faces give $x^2/2$ each. All six face integrals therefore giveThe independently evaluated values agree, verifying the divergence theorem for this field and cube.

Write $R=|x-x_0|$. A radial function harmonic away from the source satisfies $(R^2{G}^{\prime }{)}^{\prime }=0$, hence $G=A+B/R$. Decay at infinity forces $A=0$. Integrating $\Delta G=\delta (x-x_0)$ over a small sphere, the divergence theorem requires flux one:Therefore the fundamental solution in this sign convention isAdding point sources outside the domain changes only a harmonic function inside it. The method of images selects their positions and strengths to cancel boundary values. The sign reverses for the alternative convention $-\Delta G=\delta$.

Use separation of variables, $\Phi =F(r)H(\theta )$, in the Laplace equation in polar coordinates. Dividing by $FH/r^2$ givesSingle-valued $2\pi$-periodic angular factors have integer $n\ge 0$. For $n\ge 1$, they are $\cos n\theta ,\sin n\theta$, and the radial equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.

Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine Fourier coefficients vanish. With the convention $f(\theta )=a_0+{\sum }_{n\ge 1}{a}_n\cos n\theta$, its Fourier coefficients areThus the bounded interior harmonic extension isFor $r<R$ the series and its derivatives converge locally uniformly, so it is harmonic. It is a rotated and scaled instance of the harmonic measure of a semicircle in a disk. For example, summing the power series for the complex arctangent, or differentiating the following expression, gives the equivalent real formThis also proves boundedness directly.

At the two jumps, the boundary interpretation matters. At every continuity point the radial limit equals the prescribed value. At $\theta =\pm \pi /2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne 0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded Dirichlet problem imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.

For uniqueness let $w=\tilde{\Phi }-\Phi$. Formally the divergence theorem applied to $w\nabla w$ givesFor continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite Dirichlet energy there. Let $|w|\le M$ and choose a cutoff ${\eta }_{ϵ}$ which is zero within ${ϵ}^2$ of those points, one beyond $ϵ$, and linear in logarithmic distance between. Its gradient satisfies $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Applying the divergence theorem to ${\eta }_{ϵ}^{2}w\nabla w$ givesThe resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions.

For the exterior harmonic extension of disk boundary data, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients giveThis is ${\Phi }_{\mathrm{in}}(R^2/r,\theta )$, using inversion in a circle. Its limit at infinity is $C/2$. To see why that value is forced independently of this construction, average any bounded exterior harmonic function over a circle. The mean satisfies $(r{\overline{\Phi }}^{\prime }{)}^{\prime }=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.

Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar inversion in a circle, it is bounded and harmonic near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, ${\partial }_{r}w=O(r^{-2})$ in the original exterior. Apply the preceding divergence theorem energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is ${\int }_{r=L}w{\partial }_{r}wds=O(L^{-2})$. Letting $L\to \infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

The $k_0^{2}uv$ terms cancel in $u{L}_{H}v-v{L}_{H}u$. For a ball $B_R$, the product rule and divergence theorem give Green's second identity in the formThus the whole-space integral vanishes when this boundary term tends to zero. For example, uniform bounds $u,v=O(R^{-1-ϵ})$ and $\nabla u,\nabla v=O(R^{-1-ϵ})$, with $ϵ>0$, make it $O(R^{-2ϵ})$.

Pointwise vanishing alone, as stated in the source, is insufficient. Smooth fields agreeing with $u=\sin (k_{0}r)/r$ and $v=\cos (k_{0}r)/r$ outside a ball both tend to zero, but their distant boundary integral is $-4\pi k_0$. One may smooth them inside the ball without changing that flux. The requested zero identity therefore needs sufficient decay or boundary/radiation conditions that actually eliminate the boundary term.

For the generalized plane wave eigenfunctions, their orthogonality is distributional:Here $\int e^{iqx}dx=2\pi \delta (q)$ is the Fourier representation of the Dirac delta function. These eigenfunctions are not ordinary square-integrable functions or decaying test fields.

Expand $\Phi =\int A_k{\psi }_k{d}^{3}k$ and define $V_k=\int {\psi }_k^{\ast }(r^{\prime })V(r^{\prime })d^3{r}^{\prime }$. Since $L_H{\psi }_k=(k_0^2-k^2){\psi }_k$, projection of the source equation gives $(k_0^2-k^2)A_k=V_k$. Formally dividing and substituting back givesAt the shell $k=k_0$, a pole prescription is required; homogeneous solutions can otherwise be added. The formula is an inverse of the Helmholtz equation with that convention still to be selected.

Let $H$ denote the kernel in this source formula, so $L_{H}H=+{\delta }^{(3)}$. Put $l=|r-r^{\prime }|$ and align the Fourier polar axis with $r-r^{\prime }$. Angular integration givesConsequently the radial Fourier representation of the Helmholtz Green function givesThe last Green function written in the PDF has the opposite denominator. It is obtained by setting $G=-H$, the common convention $L_{H}G=-{\delta }^{(3)}$:Thus the two displayed source and Green formulas are consistent only when this minus sign is retained; $G$ cannot also be the positive-delta inverse kernel.

Finally the integral is not an ordinary improper integral across $k=k_0$. A Cauchy principal value yields $G_{\mathrm{PV}}(l)=\cos (k_{0}l)/(4\pi l)$. The outgoing prescription $k^2-k_0^2-i0$ yields the outgoing Green function for the three-dimensional Helmholtz equation, $G_{+}(l)=e^{i{k}_{0}l}/(4\pi l)$; the opposite sign gives the incoming wave. The Sokhotski–Plemelj formula accounts for their difference. The source does not specify which prescription is intended, so the boxed radial expression must be read with one chosen consistently.

The divergence theorem states that a continuously differentiable vector field $\mathbf{F}$ on a volume $V$ with piecewise smooth closed boundary $S$ satisfieswhere $\hat{\mathbf{n}}$ is the outward unit normal. Take $\mathbf{F}=\nabla a\times \mathbf{B}$. Since the curl of a gradient is zero for a twice continuously differentiable scalar field, part (i) givesIntegrating this identity and applying the divergence theorem proves

Use the full field in the PDF, including the exponential term missing from the transcription. Its divergence isThe exponential in the third component has zero $z$ derivative. In cylindrical coordinates, the solid has $0\le r\le 2$ and $0\le z\le 4-r^2$. Therefore the divergence theorem gives the outward fluxThe integration includes the entire closed boundary, both cap and base.