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Magnetic flux measures the signed amount of magnetic field crossing an oriented surface $S$. Choose a unit normal $\hat{\mathbf{n}}$ and use the oriented surface element $d\mathbf{S}=\hat{\mathbf{n}}dA$. The defining surface integral isFor a uniform magnetic field across a flat surface of area $A$, the dot product reduces this towhere $\theta$ is the angle between the magnetic field and the chosen unit normal, rather than the plane of the surface. The SI unit is the weber, $1\mathrm{Wb}=1\mathrm{T}{\mathrm{m}}^2$. ^[ref].

For a stipulated magnetic field $B=2\mathrm{T}$ and area $A=3{\mathrm{m}}^2$, the magnetic flux is $6\mathrm{Wb}$ at $\theta =0$, $3\mathrm{Wb}$ at $\theta =\pi /3$, and zero at $\theta =\pi /2$. In the last case the magnetic field lies in the plane of the surface, so it does not cross it. Reversing the chosen unit normal reverses the sign of magnetic flux without changing the physical magnetic field; at $\theta =\pi$ the same example gives $-6\mathrm{Wb}$.

The geometric idea is projected area: tilting a flat surface reduces the area seen perpendicular to the magnetic field by the factor $\cos \theta$. This explains how rotation alone can change magnetic flux even when the magnetic field is constant; Faraday's law of induction turns such changes into an electromotive force. ^[ref].

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

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.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field on a neighborhood of it, Stokes theorem statesThe oriented surface element and boundary direction must agree by the right-hand rule. Every boundary component is included; for an upward-oriented graph over an annulus the outer boundary is anticlockwise and the inner boundary clockwise when viewed from above.

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.

Let $a$ be any constant vector and take $u=\varphi a$. The product rule for divergence gives $\nabla \cdot u=a\cdot \nabla \varphi$. The divergence theorem on a bounded region with piecewise smooth boundary, oriented outward, therefore yieldsSince $a$ is arbitrary, equality of all components provesHere $\varphi$ is continuously differentiable on a neighbourhood of the closed region.

For the side of this right circular cone, the parameter tangents are $x_r=(\cos \theta ,\sin \theta ,\sqrt{3})$ and $x_{\theta }=(-r\sin \theta ,r\cos \theta ,0)$. Their cross product in the outward order isThe sign is outward because the solid right circular cone lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the oriented surface element.

To check the integral identity, the closed boundary must include the top Euclidean disk $z=1$, radius $1/\sqrt{3}$, as well as the curved side. For $\varphi =z^2$, horizontal components cancel on integrating $\theta$. The side contribution isOn the top Euclidean disk $\varphi =1$ and $d\mathbf{S}=e_{z}dA$, so its contribution is $(\pi /3)e_z$. The total is $(\pi /6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla \varphi =2z{e}_z$, givingThus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\epsilon$ and let $\epsilon \downarrow 0$, with the extra boundary contribution vanishing.

Work with real-valued smooth functions on the bounded region enclosed by $S$, with the oriented surface element pointing outward. Set $\psi =\varphi -{\varphi }_1$. It is a harmonic function in $V$ and has zero Dirichlet boundary conditions on $S$. Applying Green's first identity to $\psi$ with itself givesBoth terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla \psi =0$ throughout $V$. Consequently $\psi$ is constant on each connected component; every bounded component meets the prescribed boundary, where that constant is zero. ThereforeThis is the energy proof of Uniqueness of the Dirichlet problem. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.

Using the Levi-Civita symbol and the epsilon-delta identity,ThereforeOne particularly simple vector potential for $u$ isDirect differentiation gives $\nabla \cdot A=0$ and $\nabla \times A=(y^2,z^2,x^2)=u$.

Parametrize the curved cone byThe outward oriented surface element isSubstitution and integration over $0\le z\le h$, $0\le \phi <2\pi$ give

Because $\nabla \cdot u=0$, the divergence theorem predicts that the two outward fluxes sum to zero. On the top disk $S_2$, $d\mathbf{S}=(0,0,1)dxdy$ and $u_z=x^2$, so, with $R=h\tan \alpha$,as predicted.

The outward orientation on $S_2$ induces the stated anticlockwise orientation on $C$, whereas the outward orientation on $S_1$ induces the reverse orientation. Since $u=\nabla \times A$, Stokes theorem predictsOn $C$, put $(x,y,z)=(R\cos \phi ,R\sin \phi ,h)$. Thenverifying both identities.

For an arbitrary constant vector $\mathbf{c}$, the product rule for divergence givesThe divergence theorem therefore impliesSince this holds for every $\mathbf{c}$,

For $\varphi =x+y$, $\nabla \varphi =(1,1,0)$. The ball of radius $a$ has volume $4\pi a^3/3$, soOn the sphere, use $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$ and the stated outward oriented surface element. The first component of the surface integral isand symmetry gives the same second component and a zero third component. The two sides agree.

Both patches project onto the unit disk $D$ in the $xy$-plane. Their outward oriented surface elements areThe cone's apex has zero area and does not affect the surface integral. For $k=0$, the vector field is $(y,x,1{)}^T$, soThe terms involving $xy$ vanish by reflection symmetry of the disk. Thus the total outward flux isThis calculation evaluates the two surface integrals directly.

On $OAC$, $y=0$ and the outward unit normal is $-{\mathbf{e}}_y$. Its flux isOn $OBC$, $x=0$ and the outward unit normal is $-{\mathbf{e}}_x$. HenceOn $ABC$, write $z=1-x-y$ over $x\ge 0$, $y\ge 0$, $x+y\le 1$. The oriented surface element is $(1,1,1{)}^{T}dxdy$, pointing away from the origin. Here $x+y+z=1$, so $\mathbf{F}=(1,1,0{)}^T$. ThusThe unnormalized normal vector in the oriented surface element already incorporates the factor $\sqrt{3}$ from the surface area; no additional factor is needed.