Scalar Surface Integrals on Parametrized Patches

From a Scalar Field to an Integral on a Surface

A double integral accumulates values over a region in the plane. A scalar surface integral does the corresponding job over a surface in space: it accumulates the value of a scalar function at each point of the surface, with each small contribution weighted by the area of the surface there.

To calculate such an integral, we describe the surface by two parameters. The scalar function then becomes a function of those parameters, and the tangent vectors determine how parameter-domain area is stretched onto the surface. This gives a direct procedure: compose the scalar function with the parametrization, multiply by the surface-area factor, and integrate over the parameter domain.

Definition: Let \(D\) be a bounded region in the \(uv\)-plane and let \(\mathbf{r}:D\to\mathbb{R}^3\) be a continuously differentiable parametrization that is regular in the interior of \(D\), meaning \(\mathbf{r}_u\times\mathbf{r}_v\ne\mathbf{0}\). Suppose \(\mathbf{r}\) has no overlapping sheets in the interior and \(g\) is continuous on the resulting surface patch \(S=\mathbf{r}(D)\). The scalar surface integral is \[ \iint_S g\,dS = \iint_D g(\mathbf{r}(u,v))\,\|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv. \]

Here, \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are the tangent vectors obtained by differentiating the parametrization. The norm of their cross product is the surface-area factor. A patch may include boundary points where regularity fails without changing the integral, provided those points do not affect the surface area; in standard examples, regularity holds throughout the interior and the boundary has area zero.

Why the Formula Uses a Cross-Product Magnitude

Imagine a small rectangle in the parameter domain with side lengths \(\Delta u\) and \(\Delta v\). Near a point \((u,v)\), the parametrized surface is approximated by its tangent plane. The two sides of the rectangle are carried approximately to the vectors \(\mathbf{r}_u\Delta u\) and \(\mathbf{r}_v\Delta v\). These vectors span a parallelogram whose area is the magnitude of their cross product:

\|(\mathbf{r}_u\Delta u)\times(\mathbf{r}_v\Delta v)\| = \|\mathbf{r}_u\times\mathbf{r}_v\|\,\Delta u\,\Delta v.

Over a sufficiently small cell, the scalar function is approximately constant, with value \(g(\mathbf{r}(u,v))\). Thus the contribution from that cell is approximately its scalar value times the approximating surface area. Adding these contributions over a partition and taking the limit gives the parameter-domain double integral in the definition. Continuity ensures that these local approximations converge to the integral as the cells become smaller.

This explanation also clarifies why the norm is required. The cross product itself depends on the order of the tangent vectors: reversing that order changes its sign. But area is nonnegative and has no orientation. The magnitude gives the area of the tangent parallelogram regardless of that sign.

Key formula: For a regular parametrization, the surface-area element is \(dS=\|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv\). The scalar surface integral weights this area element by \(g(\mathbf{r}(u,v))\), not by \(g(u,v)\) unless \(g\) is already defined on parameter space.

A Reliable Setup

Keep the surface and parameter domain distinct. The function \(g\) takes points \((x,y,z)\) on the surface as inputs. To use a double integral in \(u\) and \(v\), first substitute the coordinates of \(\mathbf{r}(u,v)\) into \(g\). Then compute the tangent vectors and their cross product, take its magnitude, and use the given parameter domain for the bounds.

Identify the parameter domain.
Write the region \(D\) in the \(uv\)-plane, including its bounds or inequalities.

Compute the tangent vectors.
Differentiate \(\mathbf{r}(u,v)\) with respect to \(u\) and \(v\).

Find the area factor.
Calculate \(\|\mathbf{r}_u\times\mathbf{r}_v\|\).

Compose and integrate.
Form \(g(\mathbf{r}(u,v))\), multiply by the area factor, and integrate over \(D\).

Worked Examples

Worked Example: A Scalar Integral on a Plane Patch

Let \(\mathbf{r}(u,v)=(u+v,u-v,2u)\) for \(0\le u\le1\) and \(0\le v\le1\). Find \(\iint_S g\,dS\) when \(g(x,y,z)=x+z\).

The tangent vectors are

\mathbf{r}_u=(1,1,2), \qquad \mathbf{r}_v=(1,-1,0).

Their cross product and its magnitude are

\mathbf{r}_u\times\mathbf{r}_v = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&1&2\\ 1&-1&0 \end{vmatrix} =(2,2,-2), \qquad \|\mathbf{r}_u\times\mathbf{r}_v\| =\sqrt{4+4+4}=2\sqrt{3}.

On the parametrized surface, \(x=u+v\) and \(z=2u\), so

g(\mathbf{r}(u,v))=(u+v)+2u=3u+v.

Therefore,

\begin{aligned} \iint_S g\,dS &=\int_0^1\int_0^1 (3u+v)(2\sqrt{3})\,dv\,du\\ &=2\sqrt{3}\int_0^1\left[3uv+\frac{v^2}{2}\right]_{v=0}^{v=1}\,du\\ &=2\sqrt{3}\int_0^1\left(3u+\frac12\right)\,du\\ &=2\sqrt{3}\left[\frac{3u^2}{2}+\frac{u}{2}\right]_{u=0}^{u=1}\\ &=2\sqrt{3}\left(2\right)=4\sqrt{3}. \end{aligned}

Check: the composed scalar function is \(3u+v\), whose integral over the unit square is \(2\); the constant area factor is \(2\sqrt{3}\). Their product is \(4\sqrt{3}\), as computed.

Worked Example: A Variable Area Factor That Cancels

Consider the patch \(\mathbf{r}(u,v)=(u,v,uv)\) on the rectangle \(0\le u\le1\), \(0\le v\le2\). Let \(g(x,y,z)=1/\sqrt{1+x^2+y^2}\). Evaluate \(\iint_S g\,dS\).

Differentiating gives

\mathbf{r}_u=(1,0,v), \qquad \mathbf{r}_v=(0,1,u).

Thus

\mathbf{r}_u\times\mathbf{r}_v=(-v,-u,1), \qquad \|\mathbf{r}_u\times\mathbf{r}_v\|=\sqrt{v^2+u^2+1}.

Because \(x=u\) and \(y=v\) on the patch, the scalar function becomes

g(\mathbf{r}(u,v))=\frac{1}{\sqrt{1+u^2+v^2}}.

The two factors cancel, so

\begin{aligned} \iint_S g\,dS &=\int_0^1\int_0^2 \frac{\sqrt{1+u^2+v^2}}{\sqrt{1+u^2+v^2}}\,dv\,du\\ &=\int_0^1\int_0^2 1\,dv\,du\\ &=\int_0^1 2\,du=2. \end{aligned}

Check: the integrand after including the area factor is exactly \(1\), and the rectangle \(D\) has area \((1-0)(2-0)=2\). Hence the result is \(2\).

Worked Example: A Triangular Parameter Domain

Let \(\mathbf{r}(u,v)=(u,v,u+v)\) on \(D=\{(u,v):u\ge0,\ v\ge0,\ u+v\le1\}\), and let \(g(x,y,z)=x+y\). Find the scalar surface integral.

The tangent vectors and their cross product are

\mathbf{r}_u=(1,0,1), \qquad \mathbf{r}_v=(0,1,1), \qquad \mathbf{r}_u\times\mathbf{r}_v=(-1,-1,1).

Thus \(\|\mathbf{r}_u\times\mathbf{r}_v\|=\sqrt{3}\), and \(g(\mathbf{r}(u,v))=u+v\). The triangular domain has bounds \(0\le u\le1\) and \(0\le v\le1-u\), so

\begin{aligned} \iint_S g\,dS &=\sqrt{3}\int_0^1\int_0^{1-u}(u+v)\,dv\,du\\ &=\sqrt{3}\int_0^1\left[u v+\frac{v^2}{2}\right]_{v=0}^{v=1-u}\,du\\ &=\sqrt{3}\int_0^1\left(u(1-u)+\frac{(1-u)^2}{2}\right)\,du\\ &=\sqrt{3}\int_0^1\left(\frac12-\frac{u^2}{2}\right)\,du\\ &=\sqrt{3}\left[\frac{u}{2}-\frac{u^3}{6}\right]_{u=0}^{u=1} =\frac{\sqrt{3}}{3}. \end{aligned}

Check: the integral of \(u+v\) over this triangle is \(1/3\), and the area factor is the constant \(\sqrt{3}\). Their product is \(\sqrt{3}/3\), agreeing with the evaluation.

Common Error: Forgetting the Area Factor

A frequent mistake is to integrate the composed scalar function over \(D\) and treat parameter-domain area as though it were surface area. The first example makes the error visible. Its composed function integrates to

\int_0^1\int_0^1(3u+v)\,dv\,du=2.

That is not the surface integral: it omits the factor \(2\sqrt{3}\). The correct calculation is

\iint_S g\,dS = (2\sqrt{3})\int_0^1\int_0^1(3u+v)\,dv\,du = (2\sqrt{3})(2) = 4\sqrt{3}.

The parameter domain tells you where to integrate, but its ordinary area does not generally equal the area of the corresponding surface patch. The cross-product magnitude accounts for that difference locally.

Interpreting the Result

If \(g=1\) everywhere on the patch, the scalar surface integral reduces to its surface area. If \(g\) is nonnegative, the integral accumulates nonnegative weighted area. A scalar function may also take negative values; in that case, the integral is a signed accumulation, with negative values subtracting from positive ones. In every case, it is the value \(g(\mathbf{r}(u,v))\) that supplies the scalar weight at a point represented by \((u,v)\).

The formula applies to a regular parametrized patch with a suitable parameter domain and an integrable scalar function; continuity on a bounded patch is a standard sufficient condition. If a parametrization covers an interior portion of the surface more than once, the parameter-domain integral counts each covering. For an ordinary single patch, use a parametrization without such overlapping sheets. The essential calculation remains the same: evaluate the scalar function on the surface, account for the local area scaling, and integrate over the parameters.

Check Your Understanding

Use the definition and the examples above to answer these questions.

For a parametrization \(\mathbf{r}(u,v)\), what geometric quantity does \(\|\mathbf{r}_u\times\mathbf{r}_v\|\) represent?
Why must the scalar function \(g(x,y,z)\) be composed with \(\mathbf{r}(u,v)\) before integrating over \(D\)?
What changes in the formula if the order of the tangent vectors in the cross product is reversed?
For \(\mathbf{r}(u,v)=(u,v,uv)\), what is the area factor?
Why is integrating \(g(\mathbf{r}(u,v))\) alone generally not enough to find a scalar surface integral?