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An infinitesimal parameter rectangle becomes a parallelogram with edges ${\mathbf{r}}_{s}ds$ and ${\mathbf{r}}_{t}dt$. Its area is the Euclidean norm of their cross product, giving the surface area elementThus, for a regular parametrization covering the surface once except possibly on sets of area zero,For several patches, add their surface integrals while avoiding overlap; repeated coverage would otherwise count area with multiplicity.

Let $\Omega \subset {\mathbb{R}}^3$ be a bounded region with a piecewise smooth closed boundary, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. With $\mathbf{n}$ the outward unit normal vector, the divergence theorem statesThe boundary surface integral is the total outward flux. All boundary components are included; on the boundary of a cavity, the outward unit normal vector points into the cavity.

Let $S$ be a piecewise smooth oriented surface with piecewise smooth boundary $\partial S$, and let $\mathbf{F}$ be a continuously differentiable vector field on a neighbourhood of $S$. The Stokes theorem statesThe boundary orientation is induced by the chosen unit normal vector $\mathbf{n}$ through the right-hand rule: with the right thumb along $\mathbf{n}$, the curled fingers indicate positive boundary traversal. Equivalently, when walking along the boundary with one's head pointing along $\mathbf{n}$, the surface lies to the left. An upward oriented disk therefore has counterclockwise boundary traversal when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

Choose the outward orientation on the upper half of the ellipsoid, giving counterclockwise boundary orientation on $C$ as viewed from $z>0$. Parametrize the boundary circle byThen $\mathbf{u}(\mathbf{c}(t))=(-\sin t,\cos t,0)={\mathbf{c}}^{\prime }(t)$, so the line integral isFor a direct surface integral, use the parametrized surfaceIts outward vector surface element is supplied by the cross productUsing the curl from a,The two integrals agree for every $\alpha >0$, verifying the Stokes theorem. Only the vertical component of the vector surface element enters, which explains the absence of $\alpha$ in the answer.

For each of the three constant Cartesian unit vectors ${\mathbf{e}}_j$, apply the divergence theorem to the vector field $\varphi {\mathbf{e}}_j$. Its divergence is ${\partial }_j\varphi$, soThese are precisely the three components ofThus the volume integral of the gradient is the boundary surface integral weighted by the vector surface element $d\mathbf{S}=\mathbf{n}dS$. The regularity needed here is that $\varphi$ be continuously differentiable on a neighbourhood of the closure of $R$.

One sufficient version of the divergence theorem is the following. Let $R$ be a bounded region in ${\mathbb{R}}^3$ with a closed, piecewise smooth surface as its boundary, and let $\mathbf{F}$ be a continuously differentiable function with values in ${\mathbb{R}}^3$ on an open neighbourhood of the closure of $R$. With $\mathbf{n}$ the outward unit normal vector,The boundary surface integral is the total outward flux integral. If the boundary has several components, each has the outward orientation relative to the material of $R$; inner components therefore point into the holes. Edges and isolated vertices may be omitted when assigning the unit normal vector, since they have zero surface area.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.

Let $\Sigma$ be a compact, piecewise smooth oriented surface with piecewise smooth boundary, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of $\Sigma$. The Stokes theorem statesThe boundary orientation must agree with the chosen unit normal vector by the right-hand rule. For an upward orientation on a horizontal disk, its boundary is traversed counterclockwise when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

Differentiate the ellipsoidal parametrization. Its outward vector surface element isFor $a,b,c>0$, its dot product with $\mathbf{r}$ simplifies to $abc\sin \theta$. Thus the requested direct surface integral isAs a check, the divergence of $\mathbf{r}$ is three, and the ellipsoid has volume $4\pi abc/3$, so the divergence theorem gives the same value.

Compute the curl component by component from the given vector field:For example, $B_x={\partial }_y{A}_z-{\partial }_z{A}_y=0-(-2xz)$ and $B_z={\partial }_x{A}_y-{\partial }_y{A}_x=(1-z^2)-(z^2-1)$. Its divergence isas expected because the divergence of a curl is zero.

Take $C$ counterclockwise when viewed from positive $z$, the disc $D$ upward oriented, and the hemisphere $H$ outward oriented. These choices give both surfaces the same positive boundary orientation. Parametrize $C$ by $r(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. On it, $A=(-\sin t,\cos t,0)$ and $dr=(-\sin t,\cos t,0)dt$, so the line integral isOn $D$, use polar coordinates $(x,y)=(r\cos \varphi ,r\sin \varphi )$ and $d\mathbf{S}=(0,0,1)rdrd\varphi$. Since $B_z=2$ there, the surface flux isOn the unit hemisphere, the outward unit normal vector is $n=(x,y,z)$. Since $x^2+y^2=1-z^2$, we have $B\cdot n=4z(1-z^2)$. In spherical polar coordinates, $z=\cos \theta$ and $dS=\sin \theta d\theta d\varphi$, giving the explicit surface integralIndeed the inner integral is $[{\sin }^4\theta {]}_0^{\pi /2}=1$.

These equalities illustrate Stokes theorem twice: $B=\nabla \times A$ and either upward $D$ or outward $H$ spans the oriented circle $C$, so each surface flux equals the same line integral. For the divergence theorem, close the upper half-ball by adjoining the downward oriented base, namely $-D$. Its outward surface flux isThe minus sign on the base is required because its outward normal vector points down. Reversing a specified orientation reverses that integral's sign.

Orient the three faces outward from the tetrahedron. The two coordinate-plane faces have normals parallel to $-{\mathbf{e}}_x$ or $-{\mathbf{e}}_y$, so their flux of the curl is zero. On the inclined face $z=1-x-y$, its upward vector area is $(1,1,1)dxdy$, with $x,y\ge 0$, $x+y\le 1$. Thus the direct surface integral isThis surface induces the counterclockwise boundary $O\to X\to Y\to O$ viewed from above, where $X=(1,0,0)$ and $Y=(0,1,0)$. At $z=0$, $\mathbf{A}=(2x-y,0,0)$. On $OX$, take $(x,y)=(t,0)$, giving integral ${\int }_0^{1}2tdt=1$. On $XY$, take $(x,y)=(1-t,t)$, giving ${\int }_0^1(-2+3t)dt=-1/2$. On $YO$, $dx=0$ and the integral vanishes. ThereforeBoth sides have been calculated directly, verifying the Stokes theorem with consistent orientations.

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

On the radius-$a$ sphere, the outward unit normal vector is $(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+2y^2+3z^2)/a$. For a direct surface integral, put $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\phi$. The integral isThe azimuthal integrals are $\pi ,\pi ,2\pi$, while ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. ThusThe flux of a linear field through a sphere also follows from the equality of the three spherical second moments, without using the divergence theorem.

The divergence is $\nabla \cdot \mathbf{F}=1+2+3=6$. Since the volume enclosed by a sphere is $4\pi a^3/3$,This agrees with the independently computed outward surface integral, as the divergence theorem requires.

Write $\mathbf{F}=\mathbf{r}{r}^{-\gamma -1}$. For $r>0$, direct Cartesian differentiation gives the divergence and curl of a radial vector field:The volume element in spherical coordinates therefore gives, for $\gamma \ne 2$,For the direct surface integral, the outer sphere has outward normal $+\mathbf{n}$ and contributes $4\pi R^{2-\gamma }$. The inner boundary's outward normal relative to the shell is $-\mathbf{n}$, so it contributes $-4\pi {ϵ}^{2-\gamma }$. HenceFor $\gamma =2$, the field has zero ordinary divergence everywhere in this shell. The separate surface contributions are $4\pi$ and $-4\pi$, so both full integrals are zero. This radial power-law flux through a spherical shell was obtained by direct differentiation and integration; the excluded origin is never part of the volume.

In cylindrical coordinates the spherical cap has $1<z<\sqrt{2}$, $0<r<\sqrt{2-z^2}$ and $0<\theta <2\pi$. Its mass density is $kz$ and the volume element is $rdrd\theta dz$. ThereforeThis is the mass of a spherical cap with height-proportional density with $R=\sqrt{2}$ and $c=1$.

For its curved surface area, write $z=\sqrt{2-r^2}$ with $0\le r\le 1$. The surface integral of one becomesIt agrees with the curved area of a spherical cap, $2\pi Rh$, for $h=\sqrt{2}-1$. The planar closing disk is not part of the requested curved surface.

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.

The scalar integrand still equals $10e$ throughout the unit sphere, and the upper hemisphere has the same surface area $2\pi$. Excluding its equator changes no area. ThereforeThere is no sign reversal between hemispheres because this scalar surface integral has no choice of normal orientation.