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Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

An elliptic paraboloid is a smooth bowl-shaped quadratic surface with elliptical horizontal sections. In the circular case $A=B$, it is a surface of revolution. For $z=x^2+y^2$, polar coordinates give the parametrization $(r\cos \theta ,r\sin \theta ,r^2)$ and upward vector area element $(-2r^2\cos \theta ,-2r^2\sin \theta ,r)drd\theta$. A band between two positive heights has two boundary circles and no cap.

For $h,R>0$ and $0<\alpha \le 2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\le r\le R$, $0\le \theta \le \alpha$ has volume $\alpha h{R}^2/4$ beneath it. Its surface area isTo derive this, parametrize the surface by $(r\cos \theta ,r\sin \theta ,h(1-r^2/R^2))$. The upward vector area element is $(2h{r}^2\cos \theta /R^2,2h{r}^2\sin \theta /R^2,r)drd\theta$; its magnitude integrates to the stated surface area. Its integral gives the vector areaIn the shallow limit $h/R\to 0$, the surface area is $\alpha R^2/2+\alpha h^2/2+O(h^4/R^2)$, approaching the area of its planar sector projection.

The divergence theorem states that a continuously differentiable vector field on a bounded region $V$ satisfieswith outward unit normal $\mathbf{n}$ and a sufficiently regular boundary; piecewise smooth boundaries are also allowed. For $\mathbf{u}=\mathbf{c}\Omega$ with constant $\mathbf{c}$, the left side is $\mathbf{c}\cdot {\int }_V\nabla \Omega dV$ and the right side is $\mathbf{c}\cdot {\int }_{\partial V}\Omega \mathbf{n}dS$. Equality for every constant vector gives the gradient volume-to-boundary identityFor the cone and its cap, the volume has $0\le z\le 1$ and $0\le r\le \sqrt{3}z$ in cylindrical coordinates. Its volume is ${\int }_0^1\pi (\sqrt{3}z{)}^{2}dz=\pi$. Since $\nabla \Omega =-{\mathbf{e}}_z$,The cap at $z=1$ has area $3\pi$, outward normal ${\mathbf{e}}_z$ and $\Omega =a-1$, so its contribution to the surface integral is $3\pi (a-1){\mathbf{e}}_z$.

Parameterize the side by $\mathbf{R}(r,\theta )=(r\cos \theta ,r\sin \theta ,r/\sqrt{3})$, $0\le r\le \sqrt{3}$, $0\le \theta <2\pi$. The outward vector area element ispointing radially outward and downward from the region above the cone. The radial part integrates to zero around the circle. The vertical part of the side integral isAdding cap and side givesThe apex and circular rim have zero surface area. Alternatively, removing a tiny apex neighborhood and taking its radius to zero justifies applying the smooth-boundary theorem there; the additional flux vanishes because the scalar field is bounded.

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 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 $-40\pi /27$.

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

Use cylindrical coordinates. The real square root restricts the projection to $0\le r\le 1$, and there $r-1\le 0\le \sqrt{1-r^2}$. Since $\nabla x={\mathbf{e}}_x$, the volume integral isFor an independent boundary calculation, split $\partial V$ into an upper hemisphere $S_{+}$ and a lower right circular conical surface $S_{-}$. They meet in the circle $r=1,z=0$ and there is no additional planar face.

On $S_{+}$, use the parametrizationThe outward unit normal vector is $\mathbf{R}$, and the surface area element is $\sin \alpha d\alpha d\theta$. The $y$ and $z$ components of ${\iint }_{S_{+}}x\mathbf{n}dS$ vanish by the angular integrals of $\sin \theta \cos \theta$ and $\cos \theta$. The $x$ component isOn $S_{-}$, the parametrization $(r\cos \theta ,r\sin \theta ,r-1)$ has outward vector area elementIts negative vertical component points out below the solid. Multiplying by $x=r\cos \theta$ and using the same angular symmetry in integration givesTherefore the independently evaluated surface integral isin agreement with the volume integral. The isolated apex of the cone does not contribute to the surface integral.

Take the dot product of the given field with the upward vector area element. The surface relation simplifies its integrand:The projected domain is exactly the rectangle $1/2\le x,y\le 1$, since $y>0$ determines a unique $z=(1-x^3)/y-x$. Therefore the surface integral is

Where $f_z\ne 0$, the implicit function theorem makes the surface a local graph $z=h(x,y)$ with $h_x=-f_x/f_z$ and $h_y=-f_y/f_z$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Therefore the vector area element of an implicit surface isDividing by $f_z$ fixes the upward orientation even if the initially chosen gradient points downward. The restriction excludes a locally vertical graph, where this particular projected-area representation cannot be used.

Choose $f=x^2(1+y)+y^{2}z-1$. Since $f_z=y^2>0$ on the stated open domain, its upward vector area element isThe graph becomes unbounded near parts of $y=0$; the flux integrals below are understood as limits from $y>0$.

Subtract the integrands before integrating: ${\mathbf{G}}_2-{\mathbf{G}}_1=(0,y^3,0)$. With the upward vector area element,This extends continuously to the lower edge, soThe individual flux integrals also converge: ${\mathbf{G}}_1\cdot d\mathbf{S}=x(3+2y)dxdy$ gives $I_1=5$, and hence $I_2=7$. Reversing the surface orientation changes both signs and leaves the requested magnitude unchanged.

The upward orientation of ${\mathbf{X}}_r\times {\mathbf{X}}_{\theta }$ corresponds to traversing the boundary from the apex to the starting rim, around the rim in increasing azimuth, and back to the apex. This is the indicated $ABCA$ orientation and is outward from the solid. Integrate that vector area element directly:The radial integrals are $2aR/3$ in the horizontal components and $b^2/2$ in the vertical component. Hence the vector area isThe absolute value is needed only because the original statement permits negative $b$. Reversing the boundary orientation reverses the entire vector. For a full rotation the horizontal components cancel, leaving $\pi b^2\mathbf{k}$, the signed projected area of the base.

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.

For the outward orientation of a spherical cap, its vector area element isIn the surface integral defining the vector area, the $x$ and $y$ components vanish by integration over a complete azimuth. The remaining component isConsequentlyThis also equals the area of the cap's projection onto the horizontal plane, directed upwards.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

The new cone can be parametrized byThe outward vector area element is $r(\frac{b}{a}\cos \theta ,1,\frac{b}{a}\sin \theta )drd\theta$. This is ${\mathbf{R}}_{\theta }^{\prime }\times {\mathbf{R}}_r^{\prime }drd\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its dot product with $\mathbf{F}({\mathbf{R}}^{\prime })$ is again $brdrd\theta$. HenceThis equality also follows because a rigid rotation takes the first cone into the second and leaves the radial vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ invariant.

For a regular implicit surface $f(x,y,z)=0$ with $f_z\ne 0$, the implicit function theorem gives a local graph $z=h(x,y)$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Differentiating $f(x,y,h(x,y))=0$ yields $h_x=-f_x/f_z$, $h_y=-f_y/f_z$, proving the formula. Division by $f_z$ makes the final component positive independently of the sign of the original gradient. At a point where $f_z=0$, another projection or parametrization is needed.