Surface integral

A surface integral uses the area element induced by a parametrisation to integrate over a surface.

A scalar surface integral integrates a scalar field against the unsigned surface area element. For a regular parametrization $\mathbf{r}(u,v)$, it is $\int f(\mathbf{r}(u,v))\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv$. Reversing orientation does not change it. For example, the scalar surface integral of the constant one is the surface area; a signed flux instead integrates the normal component of a vector field.

Let $S$ be the full boundary of $x^2+y^2\le R^2$, $0\le z\le H-\alpha y$, with $H>|\alpha |R$. The bottom contribution is zero by symmetry. On the top, the integrand is the constant $H$ and the surface area element is $\sqrt{1+{\alpha }^2}dxdy$. On the side, set $y=R\sin \varphi$ and integrate in $z$ first: the result is $R[H^2-{\alpha }^2{R}^2{\sin }^2\varphi ]/2$. Integrating from zero to $2\pi$ gives the displayed scalar surface integral. No orientation signs enter; a scalar area integral is different from a flux.

Let $S$ be the whole boundary of $x^2+y^2\le R^2$, $a\le z-\alpha x\le b$, where $R>0$ and $a<b$. On the curved cylinder, the parametrization $(R\cos \theta ,R\sin \theta ,\alpha R\cos \theta +s)$ has surface area element $Rd\theta ds$ and $\alpha x-z=-s$. Each planar cap has area $\pi R^2\sqrt{1+{\alpha }^2}$. Summing the side and both caps gives the scalar surface integralThe surface area element is unsigned; cap orientation does not change the signs of the scalar integrand. This separates the geometric area factor from the field values and avoids mistaking a scalar surface integral for a flux.

For the outward unit normal on a three-dimensional sphere, the off-diagonal integrals vanish by reflection symmetry. The diagonal integrals are equal by rotational symmetry, and their sum is ${\int }_{S^2}|\mathbf{n}{|}^{2}d\Omega =4\pi$. This proves the displayed matrix identity, where ${\delta }_{ij}$ is the Kronecker delta. Contracting it with a constant vector gives ${\int }_{S^2}(\mathbf{U}\cdot \mathbf{n})\mathbf{n}d\Omega =(4\pi /3)\mathbf{U}$, useful for integrating far-field stress forces.

For a regular parametrization $\mathbf{X}(u,v)$ of a surface in three-dimensional Euclidean space, the scalar surface area element is the area of the infinitesimal parameter parallelogram after mapping. Its value is the displayed cross product magnitude. The vector surface area element additionally records the chosen orientation.

A vector area element combines the scalar surface area element with a chosen unit normal vector. Reversing the order of the two parameter directions reverses its sign. Flux through an oriented surface is the integral of the dot product of a vector field with this vector element. For a graph $z=f(x,y)$, the upward choice is $(-f_x,-f_y,1)dxdy$.

For a closed polyhedron, let $A_F$ be a face area and ${\mathbf{n}}_F$ its outward unit normal vector. The sum of the face area vectors is zero. To prove it, fix any constant vector $\mathbf{q}$ and apply the divergence theorem to the constant vector field $\mathbf{q}$: its divergence vanishes, while its outward flux is $\mathbf{q}\cdot {\sum }_F{A}_F{\mathbf{n}}_F$. This is zero for every $\mathbf{q}$, so the sum vanishes. For a tetrahedron, the same identity follows directly by expanding the four oriented face cross products; multiplying every face vector by two leaves the identity intact.

For a regular parametric surface $\mathbf{r}:U\to X$, the displayed surface integral counts each image point as many times as its preimages in $U$, apart from sets of zero surface area. It equals the ordinary integral over the image when the parametrization is injective except on such sets. A periodic parametrization integrated over more than one period can give a different answer from integration over its image set. For example, $(u,v)\mapsto (\cos u,\sin u,v)$ with $0\le u\le 4\pi$ and $0\le v\le 1$ gives parameter area $4\pi$, whereas the image circular cylinder has surface area $2\pi$.

Take the upward-oriented graph $z=c\sqrt{1-x^2/a^2-y^2/b^2}$ over $0\le x\le L_x$, $0\le y\le L_y$, where $a,b,c,L_x,L_y>0$, $A=L_x^2/a^2$, $B=L_y^2/b^2$ and $A+B<1$. The surface flux of the vector field $F=(-y,x,0)$ is the displayed formula. Indeed the vector surface element is $(-z_x,-z_y,1)dxdy$, soThe change of variables $u=x^2/a^2$, $v=y^2/b^2$ changes $xydxdy$ to $a^2{b}^{2}dudv/4$. Integrating $(1-u-v{)}^{-1/2}$ over $[0,A]\times [0,B]$ gives $\frac{4}{3}[1-(1-A{)}^{3/2}-(1-B{)}^{3/2}+(1-A-B{)}^{3/2}]$, proving the result. When $a=b$, the value of the vector field is a tangent vector to the ellipsoid at every point, and the surface flux vanishes pointwise.

On an origin-centered radius-$a$ sphere, $\mathbf{n}=\mathbf{x}/a$. Spherical symmetry gives ${\int }_{S_a}{x}_{i}dS=0$ and ${\int }_{S_a}{x}_i{x}_{j}dS=(4\pi a^4/3){\delta }_{ij}$: off-diagonal moments vanish by reflection, diagonal moments are equal, and their sum is $a^2$ times the surface area. Contracting these identities with $M_{ij}/a$ proves the displayed flux directly, including a zero contribution from a constant field $\mathbf{b}$. It agrees with the divergence theorem, since $\nabla \cdot (M\mathbf{x}+\mathbf{b})=\mathrm{tr}M$.

Parametrize the ellipsoid by $\mathbf{x}=(b\sin \theta \cos \varphi ,b\sin \theta \sin \varphi ,a\cos \theta )$, with positive semiaxes $a,b$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The cross product ${\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi }$ gives the displayed outward vector area element. For $0<\theta <\pi$, it is the positive multiple $a{b}^2\sin \theta$ of the outward normal direction $(x/b^2,y/b^2,z/a^2)$, fixing its orientation. A flux integral can therefore be evaluated directly as $\int \int \mathbf{F}(\mathbf{x})\cdot ({\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi })d\theta d\varphi$, without computing the length of the normal separately. The scalar surface area element is the Euclidean norm of this vector element.

Let a solid cone of volume $V$ have its planar base in a plane through the origin, and orient its curved surface $C$ outwards. The vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ has divergence three. Its flux through the base vanishes because the position vector lies in that plane and is orthogonal to its normal vector. The divergence theorem therefore gives the displayed curved-surface flux. For a circular base of radius $a$ and perpendicular height $b$, this is $\pi a^{2}b$.

The graph $z=g(x,y)$ has upward-oriented vector area element $d\mathbf{S}=(-g_x,-g_y,1)dxdy$ and scalar area element $dS=\sqrt{1+g_x^2+g_y^2}dxdy$. These follow from the cross product of the parametrization derivatives $(1,0,g_x)$ and $(0,1,g_y)$. The vector element is used in flux surface integrals and Stokes theorem; reversing orientation changes its sign.

For an oriented parametrized surface $r(u,v)$, the vector surface element is $d\mathbf{S}=(r_u\times r_v)dudv$, with the sign chosen to match the orientation.

A parametrised surface $r(u,v)$ has area element $|r_u\times r_v|dudv$.

The graph $z=f(x,y)$ has area element $\sqrt{1+f_x^2+f_y^2}dxdy$.

A flux integral ${\int }_{S}F\cdot ndS$ measures flow through an oriented surface.

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.