Surface integrals and flux

Parametrization and area

A smooth parametrized surface is described by $\mathbf r(u,v)$ on a parameter region $S$. At a regular point $\mathbf r_u\times\mathbf r_v\ne0$. The cross product is normal to the tangent plane, and its magnitude gives local area scaling.

Definition

Scalar surface integral and flux

For a suitable piecewise smooth regular surface,

$$\iint_{\Sigma} f\,dS=\iint_S f(\mathbf r(u,v))\|\mathbf r_u\times\mathbf r_v\|\,du\,dv.$$

Choosing the orientation represented by $\mathbf r_u\times\mathbf r_v$, the flux of $\mathbf F$ is

$$\iint_{\Sigma}\mathbf F\cdot\mathbf n\,dS =\iint_S\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv.$$

Reversing the normal changes flux sign but not scalar surface area. If the parametrization covers a region multiple times, the displayed integral counts each traversal unless corrected.

Why a regular change of parameters preserves the integral

For this verification take compact Jordan parameter regions inside open coordinate domains, regular $C^1$ patches that cover their interiors once, and integrands continuous on the patch. Let the two compact parameter regions be $S$ and $\widetilde S$, with $\psi(S)=\widetilde S$. Suppose $r=\widetilde r\circ\psi$, where $\psi$ is a $C^1$ diffeomorphism on open neighborhoods of these regions. Write $\psi(u,v)=(s(u,v),t(u,v))$. The chain rule and bilinearity of the cross product give

$$r_u\times r_v =(s_u\widetilde r_s+t_u\widetilde r_t)\times(s_v\widetilde r_s+t_v\widetilde r_t) =\det D\psi\,(\widetilde r_s\times\widetilde r_t).$$

Taking lengths produces the absolute Jacobian of the two-dimensional change of variables. Applying the proved compact Jordan substitution theorem to the continuous parameter integrand therefore makes the two scalar surface integrals equal. For flux, retain the vector identity rather than its length: an orientation-preserving parameter change has $\det D\psi>0$ on each component and preserves the chosen normal. A negative determinant on a component reverses that component's normal and flux sign; when signs differ between components there is no single global minus sign. This proves invariance under the stated admissible parameter changes; it does not identify an arbitrary many-to-one parametrization with once-only geometric area.

Worked example

A tilted graph

For $\Sigma:z=x+y$ above $0\le x,y\le1$, take $\mathbf r(x,y)=(x,y,x+y)$. Then $\mathbf r_x\times\mathbf r_y=(-1,-1,1)$ is upward, and $dS=\sqrt3\,dx\,dy$. Its area is $\sqrt3$. For $\mathbf F=(0,0,z)$, upward flux is

$$\int_0^1\int_0^1(x+y)\,dy\,dx=1.$$

Downward flux is $-1$; scalar area stays $\sqrt3$.

For a graph $z=g(x,y)$, the upward oriented area vector is $(-g_x,-g_y,1)\,dx\,dy$, while scalar area uses $\sqrt{1+g_x^2+g_y^2}\,dx\,dy$. A unit normal and $dS$ may be used separately, but their product must recover the correct oriented area vector.

Worked example

Flux through a sphere by parametrization

On the sphere of radius $R>0$, let $\mathbf F=(x,y,z)$. The outward unit normal is $\mathbf n=\mathbf r/R$, so $\mathbf F\cdot\mathbf n=R$. The surface area is $4\pi R^2$, yielding outward flux $4\pi R^3$. Alternatively, the spherical parametrization has $\|\mathbf r_\phi\times\mathbf r_\theta\|=R^2\sin\phi$ and the outward orientation, with $0\le\phi\le\pi$, $0\le\theta\le2\pi$.

Proof of the sphere area used above

For the sphere of radius $R>0$, put

$$s(\phi,\theta)=(R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi).$$

Direct differentiation gives $\|s_\phi\times s_\theta\|=R^2\sin\phi$, positive when $0<\phi<\pi$. On a compact truncated rectangle $\delta\le\phi\le\pi-\delta$, $-\pi+\eta\le\theta\le\pi-\eta$, where $0<\delta<\pi/3$ and $0<\eta<\pi/2$, this is a regular once-covered patch. Its area is

$$A_{\delta,\eta} =R^2(2\pi-2\eta)\int_\delta^{\pi-\delta}\sin\phi\,d\phi =2R^2(2\pi-2\eta)\cos\delta.$$

It remains to control the omitted surface area, rather than merely volume. Near either pole the sphere is the regular graph $z=\pm\sqrt{R^2-x^2-y^2}$. On the disk of radius $R\sin\delta$, its graph area factor is $R/\sqrt{R^2-x^2-y^2}\le1/\cos\delta\le2$. Each omitted cap therefore has area at most $2\pi R^2\sin^2\delta$, tending to zero. This bound uses the graph formula and planar disk area, not the sphere area being proved.

For the remaining angular strip, use the alternative nonsingular azimuth interval $[\pi-\eta,\pi+\eta]$, still with $\delta\le\phi\le\pi-\delta$. It is a regular patch whose area is at most $2\eta R^2\int_0^\pi\sin\phi\,d\phi=4\eta R^2$. Thus the total omitted area is bounded by $4\pi R^2\sin^2\delta+4\eta R^2\to0$. The finitely many shared patch edges have zero parameter area and do not alter the sum. Passing to the limit proves $\operatorname{Area}(S_R)=4\pi R^2$. Multiplying by the constant outward normal component $R$ proves the earlier position-field flux $4\pi R^3$ as well.

Method comparison

For a graph, project onto the coordinate plane with the simplest region. For a sphere or cylinder, use a parametrization adapted to the symmetry. A scalar integral measures accumulated surface density; flux measures the signed normal component of a vector field. For a closed surface with outward orientation, the divergence theorem may simplify flux, but the hypotheses and closure must be checked.

Practice

1. Find the area of the plane $z=2x$ above $[0,1]^2$.
2. Find upward flux of $(0,0,1)$ through the graph $z=x^2+y^2$ over the unit disk.
3. If a surface normal is reversed, what happens to its scalar surface integral and to its flux?
4. Explain why a parametrization that covers a surface twice gives twice the scalar integral when used without correction.

Full solutions

1. $\sqrt{1+2^2+0^2}=\sqrt5$ is the area factor, so the area is $\sqrt5$.
2. The upward area vector is $(-2x,-2y,1)\,dx\,dy$. Its dot product with $(0,0,1)$ is $1$, so the flux is the disk area $\pi$.
3. The scalar integral is unchanged; the flux changes sign.
4. The integral sums over the parameter domain. If almost every surface point has two regular preimages, both copies contribute the same scalar density and area, hence double counting.

Synthesis

The tangent vectors determine both surface area and oriented normal. Flux is sensitive to orientation; scalar surface integrals are not. Always check coverage and geometry before calculating.