10 Parametric Surfaces and Surface Integrals

Parametrizing a Surface

A surface in three-dimensional space is two-dimensional, so its points generally require two parameters. A parametrized surface is written as

where $(u,v)$ ranges over a region $D$ in the parameter plane. The region $D$ is the parameter domain, and each point of $D$ maps to a point on the surface.

Holding one parameter constant produces coordinate curves. Setting $u$ equal to a constant lets $v$ vary, while setting $v$ equal to a constant lets $u$ vary. These curves describe how the parameter domain traces the surface.

Common choices include:

• For a graph $z=f(x,y)$, use $\mathbf r(x,y)=\langle x,y,f(x,y)\rangle$.

• For the lateral surface of a cylinder of radius $a$, use $\mathbf r(\theta,z)=\langle a\cos\theta,a\sin\theta,z\rangle$.

• For a sphere of radius $a$, use $\mathbf r(\phi,\theta)=\langle a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi\rangle$, with $0\le\phi\le\pi$ and $0\le\theta\le 2\pi$.

A regular parametrization is one for which the two parameter directions are not parallel, so they determine a genuine tangent plane.

Takeaway: Begin every surface calculation by identifying a parametrization and its parameter domain.

Tangent Planes and Normals

The tangent directions of $\mathbf r(u,v)$ are the tangent vector fields

Their cross product gives a normal vector:

A parametrization is regular at a point when

At $P=\mathbf r(u_0,v_0)=(x_0,y_0,z_0)$, the tangent plane is

The two unit normal choices are

For a graph $z=f(x,y)$,

This vector points upward because its third component is positive. The downward normal is the negative of the upward normal.

Takeaway: Use the cross product to obtain both the tangent-plane normal and the oriented area direction.

Surface Area

A small parameter rectangle with side lengths $\Delta u$ and $\Delta v$ is approximated on the surface by a parallelogram spanned by $\mathbf r_u\Delta u$ and $\mathbf r_v\Delta v$. Its approximate area is

Therefore, the surface-area element is

and the total area is

For a graph $z=f(x,y)$, this becomes

For example, consider $z=x+2y$ above $0\le x\le1$ and $0\le y\le2$. Since $f_x=1$ and $f_y=2$,

Takeaway: Surface area uses the magnitude $\|\mathbf r_u\times\mathbf r_v\|$, not the oriented cross product itself.

Scalar Surface Integrals

A scalar surface integral has the form $\iint_S g\,dS$, where $g(x,y,z)$ is a scalar function. Under the parametrization $\mathbf r(u,v)$, compute it by

For a graph $z=f(x,y)$, the formula is

If $g=1$, the result is surface area. If $g$ is a surface density, the result is total mass.

For the plane $z=x+y$ over the unit square $0\le x,y\le1$, with density $\rho(x,y,z)=z$, one has $f_x=f_y=1$ and $dS=\sqrt{3}\,dx\,dy$. Substituting $z=x+y$ gives

Because only a magnitude appears in the area factor, reversing the orientation does not change a scalar surface integral.

Takeaway: Substitute the surface into the scalar function, then multiply by the surface-area element.

Flux Through an Oriented Surface

Let $\mathbf F(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$ be a vector field on an oriented surface. Its flux is

Using a parametrization, the computational formula is

The cross product must have the required orientation. If the opposite orientation is needed, use $\mathbf r_v\times\mathbf r_u$, which is the negative of $\mathbf r_u\times\mathbf r_v$.

For a graph $z=f(x,y)$ with upward orientation,

Thus,

For example, let $\mathbf F=\langle x,y,z\rangle$ and $z=1-x-y$ over $D=\{(x,y):x\ge0,\ y\ge0,\ x+y\le1\}$. Since $f_x=f_y=-1$, the upward vector area element is $\langle1,1,1\rangle\,dx\,dy$. On the surface,

Therefore, the upward flux equals the area of the triangular domain:

Takeaway: Flux uses the oriented cross product directly; replacing it with its magnitude loses the direction information.

A Reliable Solution Strategy

A reliable workflow keeps the geometry and the integration factor separate:

1. Describe the surface and specify its parameter domain.

2. Choose $\mathbf r(u,v)$, or use a graph representation when the surface is given by $z=f(x,y)$.

3. Compute $\mathbf r_u$ and $\mathbf r_v$.

4. Compute $\mathbf r_u\times\mathbf r_v$.

5. Check whether its direction matches the required orientation; reverse it if necessary.

6. Substitute the parametrization into the integrand.

7. Use $\|\mathbf r_u\times\mathbf r_v\|$ for a scalar surface integral and $\mathbf r_u\times\mathbf r_v$ itself for flux.

8. Integrate over the parameter domain.

The central distinction is:

• Scalar surface integrals measure quantities distributed over area and are independent of orientation.

• Flux integrals measure passage through a surface and change sign when orientation is reversed.

Final takeaway: Identify the surface, compute its tangent directions, choose the correct area factor, and verify orientation before integrating.