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 [latex]S[/latex] be a piecewise smooth surface with parameterization [latex]{\bf{r}}(u,v)=\langle{x}(u,v),y(u,v)z(u,v)\rangle[/latex] with parameter domain [latex]D[/latex] and let [latex]f(x, y, z)[/latex] be a function with a domain that contains [latex]S[/latex]. For now, assume the parameter domain [latex]D[/latex] 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 [latex]D[/latex] into subrectangles [latex]D_{ij}[/latex] with horizontal width [latex]\Delta{u}[/latex] and vertical length [latex]\Delta{v}[/latex]. Suppose that [latex]i[/latex] ranges from [latex]1[/latex] to [latex]m[/latex] and [latex]j[/latex] ranges from [latex]1[/latex] to [latex]n[/latex] so that [latex]D[/latex] is subdivided into [latex]mn[/latex] rectangles. This division of [latex]D[/latex] into subrectangles gives a corresponding division of [latex]S[/latex] into pieces [latex]S_{ij}[/latex]. Choose point [latex]P_{ij}[/latex] in each piece [latex]S_{ij}[/latex], evaluate [latex]P_{ij}[/latex] at [latex]f[/latex], and multiply by area [latex]S_{ij}[/latex] to form the Riemann sum
[latex]\large{\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^nf(P_{ij})\Delta{S_{ij}}}[/latex].
To define a surface integral of a scalar-valued function, we let the areas of the pieces of [latex]S[/latex] shrink to zero by taking a limit.
definition
The surface integral of a scalar-valued function of [latex]f[/latex] over a piecewise smooth surface [latex]S[/latex] is
[latex]\large{\displaystyle\iint_Sf(x,y,z)dS=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^nf(P_{ij})\Delta{S_{ij}}}[/latex].
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 [latex]\Delta{S_{ij}}[/latex] with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors [latex]{\bf{t}}_u[/latex] and [latex]{\bf{t}}_v[/latex]:
[latex]\large{{\bf{t}}_u=\left\langle\frac{\partial{x}}{\partial{u}},\frac{\partial{y}}{\partial{u}},\frac{\partial{z}}{\partial{u}}\right\rangle\text{ and }{\bf{t}}_v=\left\langle\frac{\partial{x}}{\partial{v}},\frac{\partial{y}}{\partial{v}},\frac{\partial{z}}{\partial{v}}\right\rangle}[/latex].
From the material we have already studied, we know that
[latex]\large{\Delta{S}_{ij}\approx||{\bf{t}}_u(P_{ij})\times{\bf{t}}_v(P_{ij})||\Delta{u}\Delta{v}}[/latex].
Therefore,
[latex]\large{\displaystyle\iint_Sf(x,y,z)dS\approx\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^nf(P_{ij})||{\bf{t}}_u(P_{ij})\times{\bf{t}}_v(P_{ij})||\Delta{u}\Delta{v}}[/latex].
This approximation becomes arbitrarily close to [latex]\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^nf(P_{ij})\Delta{S}_{ij}[/latex] as we increase the number of pieces [latex]S_{ij}[/latex] by letting [latex]m[/latex] and [latex]n[/latex] go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:
[latex]\large{\displaystyle\iint_Sf(x,y,z)dS=\displaystyle\iint_Df({\bf{r}}(u,v)||{\bf{t}}_u\times{\bf{t}}_v||}dA[/latex].
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:
[latex]\large{\displaystyle\iint_Cf(x,y,z)ds=\displaystyle\int_a^bf({\bf{r}}(t))||{\bf{r}}^\prime(t)||dt}[/latex].
In this case, vector [latex]{\bf{t}}_u\times{\bf{t}}_v[/latex] is perpendicular to the surface, whereas vector [latex]{\bf{r}}^\prime[/latex] is tangent to the curve.
Example: Calculating a surface integral
Calculate surface integral [latex]\displaystyle\iint_S5dS[/latex], where [latex]S[/latex] is the surface with parameterization [latex]{\bf{r}}(u,v)=\langle{v},u^2,v\rangle[/latex] for [latex]0\leq{u}\leq2[/latex] and [latex]0\leq{v}\leq{u}[/latex].
Example: calculating the surface integral of a cylinder
Calculate surface integral [latex]\displaystyle\iint_S(x+y^2)dS[/latex], where [latex]S[/latex] is cylinder [latex]x^2+y^2=4[/latex], [latex]0\leq{z}\leq3[/latex] (Figure 1).
try it
Calculate [latex]\displaystyle\iint_S(x^2-z)dS[/latex], where [latex]S[/latex] is the surface with parameterization [latex]{\bf{r}}(u,v)=\langle{u},u^2+v^2,1\rangle[/latex], [latex]0\leq{u}\leq2, \ 0\leq{v}\leq3[/latex].
Example: calculating the surface integral of a piece of a sphere
Calculate surface integral [latex]\displaystyle\iint_Sf(x,y,z)dS[/latex], where [latex]f(x, y, z)=z^{2}[/latex] and [latex]S[/latex] is the surface that consists of the piece of sphere [latex]x^{2}+y^{2}+z^{2}=4[/latex] that lies on or above plane [latex]z=1[/latex] and the disk that is enclosed by intersection plane [latex]z=1[/latex] and the given sphere (Figure 2).
try it
Calculate line integral [latex]\displaystyle\iint_S(x-y)dS[/latex], where [latex]S[/latex] is cylinder [latex]x^2+y^2=1[/latex],[latex]0\leq{z}\leq2[/latex], including the circular top and bottom.
Watch the following video to see the worked solution to the above Try It
Example: calculating the mass of a sheet
A flat sheet of metal has the shape of surface [latex]z=1+x+2y[/latex] that lies above rectangle [latex]0\leq{x}\leq4[/latex] and [latex]0\leq{y}\leq2[/latex]. If the density of the sheet is given by [latex]\rho(x,y,z)=x^2yz[/latex], what is the mass of the sheet?
try it
A piece of metal has a shape that is modeled by paraboloid [latex]z=x^2+y^2[/latex], [latex]0\leq{z}\leq4[/latex], and the density of the metal is given by [latex]\rho(x,y,z)=z+1[/latex]. Find the mass of the piece of metal.