6.6 Surface integrals  (Page 7/27)

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.

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 $\text{Δ}{S}_{ij}$ with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors ${\text{t}}_u$ and ${\text{t}}_v\text{:}$

From the material we have already studied, we know that

Therefore,

This approximation becomes arbitrarily close to $\underset{m,n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^m{\displaystyle \sum _{j=1}^{n}f\left(P_{ij}\right)\text{Δ}{S}_{ij}}}$ as we increase the number of pieces $S_{ij}$ by letting m and n go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

[link] allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to [link] for line integrals:

In this case, vector ${\text{t}}_u\times {\text{t}}_v$ is perpendicular to the surface, whereas vector $r^{\prime }\left(t\right)$ is tangent to the curve.

Calculating the surface integral of a cylinder

Calculate surface integral ${\displaystyle {\iint }_S\left(x+y^2\right)dS},$ where S is cylinder $x^2+y^2=4,0\le z\le 3$ ( [link] ).

To calculate the surface integral, we first need a parameterization of the cylinder. Following [link] , a parameterization is

$\text{r}\left(u,v\right)=⟨\text{cos}u,\text{sin}u,v⟩,0\le u\le 2\pi ,0\le v\le 3.$

The tangent vectors are ${\text{t}}_u=⟨\text{sin}u,\text{cos}u,0⟩$ and ${\text{t}}_v=⟨0,0,1⟩.$ Then,

${\text{t}}_u\times {\text{t}}_v=\left|\begin{array}{ccc}\hfill \text{i}\hfill & \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill \text{−}\text{sin}u\hfill & \hfill \text{cos}u\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=⟨\text{cos}u,\text{sin}u,0⟩$

and $\Vert {\text{t}}_u\times {\text{t}}_v\Vert =\sqrt{{\text{cos}}^{2}u+{\text{sin}}^{2}u}=1.$ By [link] ,

$\begin{array}{}\\ \\ \\ {\displaystyle {\iint }_{S}f(x,y,z)dS={\displaystyle {\iint }_{D}f\left(\text{r}(u,v)\right)\Vert {\text{t}}_u\times {\text{t}}_v\Vert }}dA\hfill \\ ={\displaystyle {\int }_0^3{\displaystyle {\int }_0^{2\pi }\left(\text{cos}u+{\text{sin}}^{2}u\right)dudv}}\hfill \\ ={{\displaystyle {\int }_0^3\left[\text{sin}u+\frac{u}{2}-\frac{\text{sin}(2u)}{4}\right]}}_0^{2\pi }dv={\displaystyle {\int }_0^3\pi dv}=3\pi .\hfill \end{array}$

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