The Central Idea
A line integral accumulates along a curve.
A surface integral accumulates across a surface.
The central strategy is
\[ \boxed{ \text{surface} \longrightarrow \text{two parameters} \longrightarrow \text{double integral} }. \]The main new challenge is measuring a tiny patch of curved surface correctly.
Parameterized Surfaces
A surface \(S\) can often be described by a vector-valued function of two variables:
\[ \boxed{ \mathbf{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle }. \]The pair \((u,v)\) ranges over a region \(D\) in the parameter plane.
As \((u,v)\) moves through \(D\), the point \(\mathbf{r}(u,v)\) sweeps out the surface.
Why Two Parameters Are Needed
A curve is one-dimensional, so one parameter is enough to move along it.
A surface is two-dimensional, so two independent parameters are generally required.
One parameter moves in one tangent direction and the other moves in a second tangent direction.
Example: A Plane
Consider the plane
\[ z=2x+3y. \]Let
\[ x=u, \qquad y=v. \]Then
\[ z=2u+3v. \]A parameterization is
\[ \boxed{ \mathbf{r}(u,v) = \langle u, v, 2u+3v \rangle }. \]Example: A Cylinder
The cylinder
\[ x^2+y^2=R^2 \]can be parameterized by angle \(\theta\) and height \(z\):
\[ \boxed{ \mathbf{r}(\theta,z) = \langle R\cos\theta, R\sin\theta, z \rangle }. \]Here one parameter moves around the cylinder and the other moves vertically.
Example: A Sphere
A sphere of radius \(R\) centered at the origin can be parameterized by spherical angles:
\[ \boxed{ \mathbf{r}(\theta,\phi) = \langle R\sin\phi\cos\theta, R\sin\phi\sin\theta, R\cos\phi \rangle }. \]Common bounds are
\[ 0\le\theta\le2\pi, \qquad 0\le\phi\le\pi. \]Tangent Vectors to a Surface
Differentiate the parameterization with respect to each parameter:
\[ \boxed{ \mathbf{r}_u = \frac{\partial\mathbf{r}}{\partial u} } \]and
\[ \boxed{ \mathbf{r}_v = \frac{\partial\mathbf{r}}{\partial v} }. \]These vectors are tangent to the surface.
Why Two Tangent Vectors Matter
Near a regular point, the surface behaves approximately like a plane.
The vectors
\[ \mathbf{r}_u \]and
\[ \mathbf{r}_v \]span the local tangent plane when they are not parallel.
The Cross Product Gives a Normal Vector
The cross product
\[ \boxed{ \mathbf{r}_u\times\mathbf{r}_v } \]is perpendicular to both tangent vectors.
Therefore, it is normal to the surface.
Reversing the order gives the opposite orientation:
\[ \mathbf{r}_v\times\mathbf{r}_u = - \mathbf{r}_u\times\mathbf{r}_v. \]The Surface-Area Element
A small rectangle of parameter area
\[ du\,dv \]maps approximately to a small parallelogram on the surface.
Its area is approximately
\[ \|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv. \]Therefore,
\[ \boxed{ dS = \|\mathbf{r}_u\times\mathbf{r}_v\|\,du\,dv }. \]Why the Cross-Product Magnitude Appears
The magnitude of a cross product equals the area of the parallelogram spanned by the two vectors:
\[ \|\mathbf{a}\times\mathbf{b}\| = \|\mathbf{a}\| \|\mathbf{b}\| \sin\theta. \]Thus,
\[ \|\mathbf{r}_u\times\mathbf{r}_v\| \]measures how a tiny parameter rectangle stretches into a tiny surface patch.
Surface Area as a Surface Integral
The area of a surface is
\[ \boxed{ A = \iint_S 1\,dS }. \]With a parameterization,
\[ \boxed{ A = \iint_D \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv }. \]Scalar Surface Integrals
If \(f(x,y,z)\) is a scalar field, then its surface integral over \(S\) is
\[ \boxed{ \iint_S f\,dS }. \]Using a parameterization,
\[ \boxed{ \iint_S f\,dS = \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv }. \]The Structure Mirrors Scalar Line Integrals
Recall the scalar line integral:
\[ \int_C f\,ds = \int f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \,dt. \]The scalar surface integral has the parallel form
\[ \iint_S f\,dS = \iint f(\mathbf{r}(u,v)) \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv. \]In each case, a geometric scaling factor converts parameter measure into actual geometric measure.
Mass of a Thin Surface
If a thin sheet lies along a surface \(S\) and has surface density
\[ \sigma(x,y,z), \]then its total mass is
\[ \boxed{ m = \iint_S \sigma\,dS }. \]Average Value on a Surface
If \(S\) has area \(A\), then the average value of a scalar field on the surface is
\[ \boxed{ f_{\text{avg}} = \frac{1}{A} \iint_S f\,dS }. \]This generalizes average value from intervals and curves to surfaces.
Example: Surface Area of a Plane Patch
Consider
\[ \mathbf{r}(u,v) = \langle u, v, 2u+3v \rangle \]over the rectangle
\[ 0\le u\le1, \qquad 0\le v\le2. \]Compute
\[ \mathbf{r}_u = \langle1,0,2\rangle \]and
\[ \mathbf{r}_v = \langle0,1,3\rangle. \]Then
\[ \mathbf{r}_u\times\mathbf{r}_v = \langle-2,-3,1\rangle. \]Its magnitude is
\[ \sqrt{4+9+1} = \sqrt{14}. \]Therefore,
\[ A = \int_0^1\int_0^2 \sqrt{14} \,dv\,du = \boxed{2\sqrt{14}}. \]Surfaces Given as Graphs
A common surface has the form
\[ z=g(x,y). \]A natural parameterization is
\[ \boxed{ \mathbf{r}(x,y) = \langle x, y, g(x,y) \rangle }. \]The parameter domain is the projection of the surface onto the \(xy\)-plane.
Surface-Area Formula for a Graph
For
\[ \mathbf{r}(x,y) = \langle x,y,g(x,y)\rangle, \]we have
\[ \mathbf{r}_x = \langle1,0,g_x\rangle \]and
\[ \mathbf{r}_y = \langle0,1,g_y\rangle. \]Their cross product is
\[ \mathbf{r}_x\times\mathbf{r}_y = \langle-g_x,-g_y,1\rangle. \]Therefore,
\[ \boxed{ dS = \sqrt{1+g_x^2+g_y^2}\,dx\,dy }. \]Scalar Surface Integral for a Graph
If
\[ z=g(x,y), \]then
\[ \boxed{ \iint_S f\,dS = \iint_D f(x,y,g(x,y)) \sqrt{1+g_x^2+g_y^2} \,dx\,dy }. \]Example: Surface Area of \(z=x+y\)
Let
\[ z=x+y \]over
\[ 0\le x\le1, \qquad 0\le y\le1. \]Here
\[ g_x=1, \qquad g_y=1. \]Therefore,
\[ dS = \sqrt{1+1+1}\,dx\,dy = \sqrt3\,dx\,dy. \]Hence,
\[ A = \int_0^1\int_0^1 \sqrt3 \,dy\,dx = \boxed{\sqrt3}. \]Example: Scalar Surface Integral on a Graph
Evaluate
\[ \iint_S z\,dS \]over the surface
\[ z=x+y, \qquad 0\le x\le1, \qquad 0\le y\le1. \]Since
\[ z=x+y \]and
\[ dS=\sqrt3\,dx\,dy, \]we obtain
\[ \iint_S z\,dS = \sqrt3 \int_0^1\int_0^1 (x+y) \,dy\,dx. \]The inner and outer integrations give
\[ \boxed{\sqrt3}. \]Surface Orientation
A surface is oriented when one of its two normal directions has been chosen consistently.
If
\[ \mathbf{n} \]is a chosen unit normal, then
\[ -\mathbf{n} \]is the opposite orientation.
Unit Normal Vector
If
\[ \mathbf{r}_u\times\mathbf{r}_v \ne \mathbf{0}, \]then a unit normal is
\[ \boxed{ \mathbf{n} = \frac{ \mathbf{r}_u\times\mathbf{r}_v }{ \|\mathbf{r}_u\times\mathbf{r}_v\| } }. \]Reversing the cross-product order gives the opposite unit normal.
Upward and Downward Orientation
For a surface written as
\[ z=g(x,y), \]an upward-pointing normal is
\[ \langle-g_x,-g_y,1\rangle, \]because its \(z\)-component is positive.
A downward-pointing normal is
\[ \langle g_x,g_y,-1\rangle. \]Outward Orientation
A closed surface encloses a volume.
Its standard positive orientation is usually the outward normal direction.
For a sphere centered at the origin, the outward normal points away from the center.
Open and Closed Surfaces
A surface with an exposed boundary is open.
Examples include:
- a disk;
- a hemisphere without its base;
- a rectangular patch of a plane;
- the side of a cylinder without end caps.
A closed surface has no boundary and completely encloses a volume.
Examples of Closed Surfaces
- a sphere;
- an ellipsoid;
- a cube;
- a cylinder including both end caps;
- any sufficiently regular sealed boundary of a solid region.
Closed surfaces become especially important for flux and the Divergence Theorem.
Vector Area Element
For an oriented surface, we define the vector area element
\[ \boxed{ d\mathbf{S} = \mathbf{n}\,dS }. \]Under a parameterization,
\[ \boxed{ d\mathbf{S} = (\mathbf{r}_u\times\mathbf{r}_v) \,du\,dv } \]when the cross product has the chosen orientation.
Scalar Area Element Versus Vector Area Element
The scalar surface element is
\[ dS. \]It measures area only.
The vector surface element is
\[ d\mathbf{S} = \mathbf{n}\,dS. \]It includes both area and orientation.
Surface Integrals and Flux
The scalar surface integral
\[ \iint_S f\,dS \]accumulates a scalar quantity over surface area.
A vector field can instead be integrated through an oriented surface:
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS. \]This is a flux integral and will be the subject of the next lesson.
Surface Parameterization Must Cover the Surface Correctly
A parameter domain should cover the intended surface exactly once whenever practical.
If the parameterization covers the surface twice, the integral may be doubled.
If part of the surface is omitted, the integral will be incomplete.
Regular Parameterizations
A surface parameterization is regular at a point when
\[ \boxed{ \mathbf{r}_u\times\mathbf{r}_v \ne \mathbf{0} }. \]This ensures the two tangent directions are locally independent.
When the cross product vanishes, the parameterization may become singular or degenerate there.
Parameterization Independence
A correctly defined scalar surface integral depends on the geometric surface, not on the particular regular parameterization chosen.
Different parameterizations may produce very different intermediate formulas but the same final value.
Choosing Coordinates That Match the Surface
Good parameter choices follow the geometry.
- Planes often use Cartesian parameters.
- Cylinders often use angle and height.
- Spheres often use two angles.
- Surfaces of revolution often use an axial parameter and an angle.
Matching coordinates to symmetry can simplify both the parameterization and the integral.
Surface Area of a Cylinder
Parameterize the lateral surface of a cylinder of radius \(R\) and height \(H\):
\[ \mathbf{r}(\theta,z) = \langle R\cos\theta, R\sin\theta, z \rangle, \]with
\[ 0\le\theta\le2\pi, \qquad 0\le z\le H. \]Then
\[ \mathbf{r}_\theta = \langle -R\sin\theta, R\cos\theta, 0 \rangle \]and
\[ \mathbf{r}_z = \langle0,0,1\rangle. \]Their cross-product magnitude is
\[ R. \]Therefore,
\[ A = \int_0^{2\pi}\int_0^H R \,dz\,d\theta = \boxed{2\pi RH}. \]Surface Area of a Sphere
For a sphere of radius \(R\),
\[ \mathbf{r}(\theta,\phi) = \langle R\sin\phi\cos\theta, R\sin\phi\sin\theta, R\cos\phi \rangle. \]The surface-area factor is
\[ \boxed{ \|\mathbf{r}_\theta\times\mathbf{r}_\phi\| = R^2\sin\phi }. \]Therefore,
\[ A = \int_0^{2\pi}\int_0^\pi R^2\sin\phi \,d\phi\,d\theta. \]Hence,
\[ \boxed{ A=4\pi R^2 }. \]Surface Density and Center of Mass
Surface integrals can also be used to compute moments and centers of mass for thin shells.
If the mass is
\[ m = \iint_S \sigma\,dS, \]then coordinate moments may involve integrals such as
\[ \iint_S x\sigma\,dS, \qquad \iint_S y\sigma\,dS, \qquad \iint_S z\sigma\,dS. \]These applications extend the same accumulation framework.
Units of a Surface Integral
In
\[ \iint_S f\,dS, \]the units are
\[ \boxed{ [f]\times[\text{area}] }. \]If \(f\) is surface density in kilograms per square meter, then the integral has units of kilograms.
Common Error: Using One Parameter for a Surface
A general surface requires two parameters.
A one-parameter vector function traces a curve, not a two-dimensional surface.
Common Error: Forgetting the Cross-Product Magnitude
The scalar surface integral is not merely
\[ \iint_D f(\mathbf{r}(u,v)) \,du\,dv. \]It requires the geometric scale factor:
\[ \boxed{ \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv }. \]Common Error: Using the Cross Product Without Its Magnitude
For a scalar surface integral with \(dS\), use
\[ \boxed{ \|\mathbf{r}_u\times\mathbf{r}_v\| }. \]The cross product itself is vector-valued and belongs naturally to oriented surface or flux calculations.
Common Error: Ignoring Orientation in Vector Surface Problems
Scalar \(dS\) integrals do not depend on orientation.
Flux-type integrals do.
Reversing the normal direction reverses the sign of
\[ \mathbf{F}\cdot\mathbf{n}. \]Common Error: Projecting onto the Wrong Plane
When a surface is written as a graph, choose a projection that produces a single-valued description whenever possible.
A surface that is awkward as
\[ z=g(x,y) \]may be simpler as
\[ x=g(y,z) \]or
\[ y=g(x,z). \]Common Error: Incorrect Parameter Bounds
The parameter region \(D\) is part of the surface description.
Correct formulas with incorrect parameter bounds may integrate over the wrong portion of the surface.
A Reliable Procedure for Parameterized Surface Integrals
- Choose a parameterization \(\mathbf{r}(u,v)\).
- Determine the parameter domain \(D\).
- Compute \(\mathbf{r}_u\).
- Compute \(\mathbf{r}_v\).
- Compute \(\mathbf{r}_u\times\mathbf{r}_v\).
- Take its magnitude for a scalar \(dS\) integral.
- Substitute the parameterization into the scalar field.
- Integrate over \(D\).
- Interpret the result and units.
A Reliable Procedure for Graph Surfaces
- Write the surface as \(z=g(x,y)\) if convenient.
- Identify the projection region \(D\) in the \(xy\)-plane.
- Compute \(g_x\) and \(g_y\).
- Use \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\).
- Replace \(z\) in the integrand by \(g(x,y)\).
- Evaluate the resulting double integral.
Example: Full Parameterized Surface Workflow
Evaluate
\[ \iint_S z\,dS \]over the plane patch
\[ \mathbf{r}(u,v) = \langle u, v, u+2v \rangle, \]where
\[ 0\le u\le1, \qquad 0\le v\le1. \]Compute
\[ \mathbf{r}_u = \langle1,0,1\rangle \]and
\[ \mathbf{r}_v = \langle0,1,2\rangle. \]Then
\[ \mathbf{r}_u\times\mathbf{r}_v = \langle-1,-2,1\rangle. \]Its magnitude is
\[ \sqrt6. \]Along the surface,
\[ z=u+2v. \]Therefore,
\[ \iint_S z\,dS = \int_0^1\int_0^1 (u+2v)\sqrt6 \,dv\,du. \]Hence,
\[ \boxed{ \frac{3\sqrt6}{2} }. \]Example: Full Graph-Surface Workflow
Find the area of
\[ z=x^2+y^2 \]above the disk
\[ x^2+y^2\le1. \]We have
\[ g_x=2x, \qquad g_y=2y. \]Therefore,
\[ dS = \sqrt{1+4x^2+4y^2} \,dx\,dy. \]Because the region is circular, polar coordinates are natural:
\[ x=r\cos\theta, \qquad y=r\sin\theta. \]Then
\[ dA=r\,dr\,d\theta, \]so
\[ A = \int_0^{2\pi}\int_0^1 \sqrt{1+4r^2}\,r \,dr\,d\theta. \]The inner integral gives
\[ \frac{1}{12} \left[ (1+4r^2)^{3/2} \right]_0^1. \]Therefore,
\[ \boxed{ A = \frac{\pi}{6} \left( 5\sqrt5-1 \right) }. \]Practice 1: Parameterize a Plane
Parameterize the plane
\[ z=x+2y. \]Practice 2: Tangent Vectors
For
\[ \mathbf{r}(u,v) = \langle u, v, uv \rangle, \]find \(\mathbf{r}_u\) and \(\mathbf{r}_v\).
Practice 3: Normal Vector
Using Practice 2, compute
\[ \mathbf{r}_u\times\mathbf{r}_v. \]Practice 4: Surface-Area Element
For
\[ z=3x+4y, \]find \(dS\) in terms of \(dx\,dy\).
Practice 5: Plane-Patch Area
Find the area of
\[ z=3x+4y \]above
\[ 0\le x\le2, \qquad 0\le y\le1. \]Practice 6: Scalar Surface Integral
Evaluate
\[ \iint_S 1\,dS \]for the surface in Practice 5.
Practice 7: Cylinder Area
Find the lateral surface area of a cylinder of radius \(3\) and height \(5\).
Practice 8: Orientation
For
\[ z=x^2+y^2, \]give an upward-pointing normal vector that does not need to be normalized.
Practice 9: Surface Mass
A flat sheet occupies
\[ 0\le x\le1, \qquad 0\le y\le2, \qquad z=0, \]and has density
\[ \sigma(x,y)=x+y. \]Find its mass.
Practice 10: Scalar or Vector Area Element
State whether each expression uses the scalar area element \(dS\) or an oriented vector area element:
- surface area;
- surface mass;
- average temperature on a surface;
- flux of a vector field through a surface.
Practice Answers
Practice 1
\[ \boxed{ \mathbf{r}(u,v) = \langle u, v, u+2v \rangle }. \]Practice 2
\[ \boxed{ \mathbf{r}_u = \langle1,0,v\rangle } \]and
\[ \boxed{ \mathbf{r}_v = \langle0,1,u\rangle }. \]Practice 3
\[ \mathbf{r}_u\times\mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 0 & v\\ 0 & 1 & u \end{vmatrix}. \]Therefore,
\[ \boxed{ \mathbf{r}_u\times\mathbf{r}_v = \langle -v,-u,1\rangle }. \]Practice 4
Here
\[ g_x=3, \qquad g_y=4. \]Therefore,
\[ dS = \sqrt{1+9+16}\,dx\,dy = \boxed{\sqrt{26}\,dx\,dy}. \]Practice 5
The projection rectangle has area
\[ 2. \]Since the surface-area factor is constant,
\[ A = 2\sqrt{26}. \]Thus,
\[ \boxed{ A=2\sqrt{26} }. \]Practice 6
Since
\[ \iint_S1\,dS \]is exactly surface area,
\[ \boxed{ 2\sqrt{26} }. \]Practice 7
\[ A=2\pi RH. \]Therefore,
\[ \boxed{ A = 2\pi(3)(5) = 30\pi }. \]Practice 8
For
\[ g(x,y)=x^2+y^2, \]we have
\[ g_x=2x, \qquad g_y=2y. \]An upward-pointing normal is
\[ \boxed{ \langle-2x,-2y,1\rangle }. \]Practice 9
Since the sheet is flat in the \(xy\)-plane,
\[ dS=dx\,dy. \]Thus,
\[ m = \int_0^1\int_0^2 (x+y) \,dy\,dx. \]Evaluating,
\[ \boxed{ m=3 }. \]Practice 10
- surface area: scalar \(dS\);
- surface mass: scalar \(dS\);
- average temperature: scalar \(dS\);
- flux: oriented vector area element or \(\mathbf{n}\,dS\).
What to Remember
A parameterized surface is
\[ \boxed{ \mathbf{r}(u,v) = \langle x(u,v),y(u,v),z(u,v)\rangle }. \]Its scalar area element is
\[ \boxed{ dS = \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv }. \]A scalar surface integral is
\[ \boxed{ \iint_S f\,dS = \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u\times\mathbf{r}_v\| \,du\,dv }. \]- Two parameters are generally required to describe a surface.
- \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are tangent vectors.
- Their cross product gives a normal direction.
- The magnitude of the cross product gives the surface-area scale factor.
- Scalar surface integrals do not depend on orientation.
- Vector surface integrals do depend on orientation.
- For \(z=g(x,y)\), \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\).
- Closed surfaces are especially important for flux and the Divergence Theorem.
Where We Go Next
Surface integrals provide the geometric machinery for accumulating across curved surfaces. The next step is to combine that area element with a vector field and ask how much of the field passes through the surface.
The next lesson, Flux, develops
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS, \]explains positive and negative flux, develops outward orientation for closed surfaces, and prepares directly for the Divergence Theorem.