Surface Integral of a Scalar-Valued Function

Learning Objectives

  • Use a surface integral to calculate the area of a given surface.

Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion.

Let S be a piecewise smooth surface with parameterization r(u,v)=x(u,v),y(u,v)z(u,v) with parameter domain D and let f(x,y,z) be a function with a domain that contains S. For now, assume the parameter domain D is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle D into subrectangles Dij with horizontal width Δu and vertical length Δv. Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of S into pieces Sij. Choose point Pij in each piece Sij, evaluate Pij at f, and multiply by area Sij to form the Riemann sum

i=1mj=1nf(Pij)ΔSij.

To define a surface integral of a scalar-valued function, we let the areas of the pieces of S shrink to zero by taking a limit.

definition


The surface integral of a scalar-valued function of f over a piecewise smooth surface S is

Sf(x,y,z)dS=limm,ni=1mj=1nf(Pij)ΔSij.

Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.

The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas ΔSij with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors tu and tv:

tu=xu,yu,zu and tv=xv,yv,zv.

From the material we have already studied, we know that

ΔSij||tu(Pij)×tv(Pij)||ΔuΔv.

Therefore,

Sf(x,y,z)dSlimm,ni=1mj=1nf(Pij)||tu(Pij)×tv(Pij)||ΔuΔv.

This approximation becomes arbitrarily close to limm,ni=1mj=1nf(Pij)ΔSij as we increase the number of pieces Sij by letting m and n go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

Sf(x,y,z)dS=Df(r(u,v)||tu×tv||dA.

The surface integral of a scalar-valued function allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to the surface integral equation for line integrals:

Cf(x,y,z)ds=abf(r(t))||r(t)||dt.

In this case, vector tu×tv is perpendicular to the surface, whereas vector r is tangent to the curve.

Example: Calculating a surface integral

Calculate surface integral S5dS, where S is the surface with parameterization r(u,v)=v,u2,v for 0u2 and 0vu.

Example: calculating the surface integral of a cylinder

Calculate surface integral S(x+y2)dS, where S is cylinder x2+y2=4, 0z3 (Figure 1).

<img src="/apps/archive/20220422.171947/resources/2de01bbf14da890ca1a4c3a6c3b1a93dca62f7da" data-media-type="image/jpeg" alt="A graph in three dimensions of a cylinder. The base of the cylinder is on the (x,z) plane, with center on the y axis. It stretches along the y axis." id="39">

Figure 1. Integrating function f(x,y,z)=x+y2 over a cylinder.

try it

Calculate S(x2z)dS, where S is the surface with parameterization r(u,v)=u,u2+v2,1, 0u2, 0v3.

Example: calculating the surface integral of a piece of a sphere

Calculate surface integral Sf(x,y,z)dS, where f(x,y,z)=z2 and S is the surface that consists of the piece of sphere x2+y2+z2=4 that lies on or above plane z=1 and the disk that is enclosed by intersection plane z=1 and the given sphere (Figure 2).

<img src="/apps/archive/20220422.171947/resources/8060a589ddea4ad92316c95589f1b486dffea464" data-media-type="image/jpeg" alt="A diagram in three dimensions of the upper half of a sphere. The center is at the origin, and the radius is 2. The top part above the plane z=1 is cut off and shaded; the rest is simply an outline of the hemisphere. The top section has center at (0,0,1) and radius of radical three." id="42">

Figure 2. Calculating a surface integral over surface S.

try it

Calculate line integral S(xy)dS, where S is cylinder x2+y2=1,0z2, including the circular top and bottom.

Watch the following video to see the worked solution to the above Try It

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface S and the density of the sheet at point (x,y,z) is ρ(x,y,z), then mass m of the sheet is m=Sρ(x,y,z)dS.

Example: calculating the mass of a sheet

A flat sheet of metal has the shape of surface z=1+x+2y that lies above rectangle 0x4 and 0y2. If the density of the sheet is given by ρ(x,y,z)=x2yz, what is the mass of the sheet?

try it

A piece of metal has a shape that is modeled by paraboloid z=x2+y2, 0z4, and the density of the metal is given by ρ(x,y,z)=z+1. Find the mass of the piece of metal.