Worked Scalar Integrals on Curved Patches

Integrating a Scalar Function over a Curved Patch

A scalar surface integral combines two kinds of information: the values of a function across a surface, and the area of the small pieces of that surface. On a curved patch, those small pieces do not generally have the same area as rectangles in the parameter domain. A parametrization lets us account for the difference systematically.

The method is to substitute the parametrized coordinates into the scalar function, calculate the surface area factor, and integrate over bounds that describe the patch. The details of these steps matter: the correct function values alone do not give the correct surface integral unless they are weighted by surface area.

Definition: Let a surface patch $S$ be parametrized by $\mathbf r(u,v)$, for $(u,v)$ in a region $D$ of the parameter plane. If $\mathbf r$ is smooth and regular on the interior of $D$, and the parametrization covers the patch without overlapping its interior, then the scalar surface integral of $f$ is $$ \iint_S f\,dS = \iint_D f(\mathbf r(u,v)) \left\lVert \mathbf r_u(u,v)\times \mathbf r_v(u,v)\right\rVert\,du\,dv. $$ Here $f$ is evaluated at points of the surface, and $dS$ denotes surface area.

The regularity condition means that $\mathbf r_u\times\mathbf r_v\ne\mathbf 0$ in the interior: the two parameter directions span a genuine tangent plane there. A parametrization may degenerate at a boundary point, such as the apex of a cone, if that point does not prevent the integral from existing. If a patch has multiple parameter pieces, integrate over the pieces and add the results.

Why the Cross Product Gives the Area Factor

For small changes $\Delta u$ and $\Delta v$, the surface near $\mathbf r(u,v)$ is approximated by a tangent parallelogram. Its two side vectors are approximately $\mathbf r_u\Delta u$ and $\mathbf r_v\Delta v$. The area of a parallelogram with side vectors $\mathbf a$ and $\mathbf b$ is $\lVert\mathbf a\times\mathbf b\rVert$. Therefore the approximate surface area of this small piece is $\lVert\mathbf r_u\times\mathbf r_v\rVert\Delta u\Delta v$.

If $f$ is continuous, its value on a sufficiently small surface piece is approximately $f(\mathbf r(u,v))$. The contribution of that piece to the integral is consequently approximated by $f(\mathbf r(u,v))\lVert\mathbf r_u\times\mathbf r_v\rVert\Delta u\Delta v$. Adding these contributions and taking the limit of finer subdivisions gives the formula in the definition. Thus the cross-product norm is not an extra factor chosen by convention: it converts parameter-plane area into actual surface area.

Key takeaway: To evaluate a scalar surface integral, use both the substituted scalar function $f(\mathbf r(u,v))$ and the area factor $\lVert\mathbf r_u\times\mathbf r_v\rVert$. The bounds must describe the parameter region for the specific patch.

Worked Example: A Spherical Cap

Worked Example: A Scalar Integral on a Spherical Cap

Find $\iint_S (x^2+y^2)\,dS$, where $S$ is the cap of the sphere of radius $2$ above the plane $z=1$.

Use spherical parameters $\mathbf r(\phi,\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)$. The cap condition $z\ge 1$ becomes $2\cos\phi\ge1$, so $0\le\phi\le\pi/3$. One full turn around the axis gives $0\le\theta\le2\pi$.

Substituting the coordinates into the scalar function gives $x^2+y^2=4\sin^2\phi(\cos^2\theta+\sin^2\theta)=4\sin^2\phi$.

The parameter derivatives are $\mathbf r_\phi=(2\cos\phi\cos\theta,2\cos\phi\sin\theta,-2\sin\phi)$ and $\mathbf r_\theta=(-2\sin\phi\sin\theta,2\sin\phi\cos\theta,0)$. Their cross product is $(4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi)$, whose norm is $4\sin\phi$ on these bounds. Therefore

\begin{aligned} \iint_S (x^2+y^2)\,dS &=\int_0^{2\pi}\int_0^{\pi/3} (4\sin^2\phi)(4\sin\phi)\,d\phi\,d\theta\\ &=2\pi\cdot16\int_0^{\pi/3}\sin^3\phi\,d\phi\\ &=32\pi\left[-\cos\phi+\frac{\cos^3\phi}{3}\right]_0^{\pi/3}\\ &=32\pi\left[\left(-\frac12+\frac1{24}\right) -\left(-1+\frac13\right)\right]\\ &=32\pi\left(\frac5{24}\right) =\frac{20\pi}{3}. \end{aligned}

Check the antiderivative: differentiating $-\cos\phi+\cos^3\phi/3$ gives $\sin\phi-\cos^2\phi\sin\phi=\sin^3\phi$, as required. The bounds also describe the intended cap: $\phi=0$ is the top of the sphere and $\phi=\pi/3$ is its circular boundary.

Worked Example: A Quarter-Cylinder

Worked Example: Integrating on a Cylindrical Patch

Find $\iint_S (x+z^2)\,dS$ on the quarter of the cylinder $x^2+y^2=4$ specified by $x\ge0$, $y\ge0$, and $0\le z\le1$.

Parametrize the patch by $\mathbf r(\theta,z)=(2\cos\theta,2\sin\theta,z)$. The first-quadrant condition gives $0\le\theta\le\pi/2$; the height bounds give $0\le z\le1$. The function becomes $x+z^2=2\cos\theta+z^2$.

The derivatives are $\mathbf r_\theta=(-2\sin\theta,2\cos\theta,0)$ and $\mathbf r_z=(0,0,1)$. Thus $\mathbf r_\theta\times\mathbf r_z=(2\cos\theta,2\sin\theta,0)$, with norm $2$. The integral is

\begin{aligned} \iint_S (x+z^2)\,dS &=\int_0^{\pi/2}\int_0^1(2\cos\theta+z^2)\,2\,dz\,d\theta\\ &=\int_0^{\pi/2}\left[4z\cos\theta+\frac{2z^3}{3}\right]_0^1\,d\theta\\ &=\int_0^{\pi/2}\left(4\cos\theta+\frac23\right)\,d\theta\\ &=\left[4\sin\theta+\frac{2\theta}{3}\right]_0^{\pi/2}\\ &=4+\frac{\pi}{3}. \end{aligned}

For a check, differentiating $4\sin\theta+2\theta/3$ gives $4\cos\theta+2/3$, the integrand in the final one-variable integral. The angular interval covers exactly one quarter of the cylinder, while the $z$-interval covers its full specified height.

Worked Example: A Cone Including Its Apex

Worked Example: A Scalar Integral on a Conical Patch

Find $\iint_S (1+z)\,dS$ on the part of the cone $z=\sqrt{x^2+y^2}$ between $z=0$ and $z=2$.

Use $\mathbf r(u,\theta)=(u\cos\theta,u\sin\theta,u)$, with $0\le u\le2$ and $0\le\theta\le2\pi$. Since $z=u$, the scalar function becomes $1+u$. The parameter derivatives are $\mathbf r_u=(\cos\theta,\sin\theta,1)$ and $\mathbf r_\theta=(-u\sin\theta,u\cos\theta,0)$. Their cross product is $(-u\cos\theta,-u\sin\theta,u)$, whose norm is $\sqrt{2}\,u$.

The parametrization is regular for $u>0$, but degenerates at $u=0$, the apex. The area factor and integrand are bounded there, and the integral over $0\le u\le2$ is finite. The single degenerate point does not change the integral. Hence

\begin{aligned} \iint_S (1+z)\,dS &=\int_0^{2\pi}\int_0^2(1+u)\sqrt{2}\,u\,du\,d\theta\\ &=2\pi\sqrt{2}\int_0^2(u+u^2)\,du\\ &=2\pi\sqrt{2}\left[\frac{u^2}{2}+\frac{u^3}{3}\right]_0^2\\ &=2\pi\sqrt{2}\left(2+\frac83\right) =\frac{28\pi\sqrt{2}}{3}. \end{aligned}

Differentiating $u^2/2+u^3/3$ gives $u+u^2$, confirming the antiderivative. The parameter bounds cover the cone from its apex to height $2$ and make a full revolution about the vertical axis.

A Common Error: Integrating Only in the Parameter Plane

For the quarter-cylinder example, a tempting but incorrect calculation is to substitute $x=2\cos\theta$ and then integrate the function over the parameter rectangle without the area factor: $\int_0^{\pi/2}\int_0^1(2\cos\theta+z^2)\,dz\,d\theta$. This gives $\int_0^{\pi/2}(2\cos\theta+1/3)\,d\theta=2+\pi/6$, not the surface integral.

The calculation omits $\lVert\mathbf r_\theta\times\mathbf r_z\rVert=2$. The actual surface has twice the area represented by the corresponding parameter rectangle, so the correct integral is twice the omitted-factor result: $2(2+\pi/6)=4+\pi/3$. Always compute the area factor before integrating; it can vary over the parameter domain, so it cannot in general be replaced by a single constant.

A Reliable Setup

For a scalar integral on a curved patch, write down the geometry and the parametrization before evaluating anything. Then check that the bounds cover the intended patch and that the scalar function is evaluated at $\mathbf r(u,v)$, not left in unrelated Cartesian coordinates. The following sequence keeps these tasks distinct:

Describe the patch.
Identify its boundary and determine the parameter ranges that cover it.

Substitute into the scalar.
Compute $f(\mathbf r(u,v))$ using the coordinate formulas of the parametrization.

Find the area factor.
Calculate $\mathbf r_u$, $\mathbf r_v$, and $\lVert\mathbf r_u\times\mathbf r_v\rVert$.

Integrate and check.
Multiply the substituted scalar by the area factor, integrate over the bounds, and verify the antiderivative and domain.

These integrals are useful whenever a quantity is distributed over a curved surface. The scalar may represent a density, a weighting function, or simply a function whose total over the patch is being measured. The same setup handles each case; what changes is the meaning of $f$, not the area correction.

Check Your Understanding

For each question, identify the parametrization, the parameter bounds, and the area factor before deciding how to evaluate the integral.

On a sphere of radius $3$, what parameter bounds describe the cap above $z=3/2$ when using $\phi$ measured from the positive $z$-axis?
For $\mathbf r(\theta,z)=(a\cos\theta,a\sin\theta,z)$, calculate $\lVert\mathbf r_\theta\times\mathbf r_z\rVert$, assuming $a>0$.
On the cone $\mathbf r(u,\theta)=(u\cos\theta,u\sin\theta,u)$, what scalar function results from substituting the coordinates into $x^2+y^2$?
Why does a scalar surface integral require an area factor in addition to the substituted function?
What condition on the parameter bounds describes a quarter-cylinder in the first quadrant, and what angular interval expresses it?