15.6 Surface Integrals

The rest of this unit will focus on surface integrals and their applications. The type of problems will be concerned with fluid flow across surfaces in R³. 

Let z = g(x,y) be some surface S and R be its projection (like its 'shadow') in the xy plane. Suppose further that g, g_x and g_y are continuous at all points in R and that there is a function f that is defined at all points on the surface S. If we cut R up into small rectangles, and project those rectangles upward to S, it would form a partition of S. We could pick a point in each sub-region of S and form the sum  where DS_i is the surface area of the piece and
f(x_i,y_i,z_i) is the function evaluated at some point within S_i. Note that   where DA_i is the area of the rectangle in R underneath S_i . If we let the norm of the partition of R be such that  ||D|| approaches 0, then we define the 'surface integral of S' to be

Theorem 15.10 Evaluating a Surface Integral

Let S be a surface with equation z = g(x,y) and let R be it's projection (shadow) in the xy plane. If g, g_x and g_y are all continuous on R and f is continuous on S, then the surface integral of f over S is

Ex 1 Evaluate the surface integral given if S is the surface 2x - 2y + z = 10 where  and . 

If we write the planar equation as z = g(x,y) = 10 - 2x + 2y, we can see that g_x= -2 and g_y = 2, so we know that our integral can be written as below, given the above x and y limits:  or more simply , you should verify this yields 216 yourself, here is Mathcad's result: 

What the calculation found is the mass of the section of plane shown in the picture if it's density at (x,y,z) is given by x - 2y + z.

Ex 1-A Try to get the same result as in Ex 1 if we use a different axes, say let the region R be the shadow in the yz plane.

First, x = g(y,z) = 5 + y - z/2 so g_y = 1 and g_z = -1/2. We also need y and z limits of integration. The y limits remain the same. Note y=0 and x=0 forces z=10, x=2 and y=4 forces z=14 so, our z limits appear to be . The integral will be  simplifying to  or even more simply  which you should verify is also 216.

Ex 2 A cone shaped lamina S is given by  as shown in the picture below:  At each point  (x,y,z) on the cone, the density is proportional to the distance between the point and the z axis. Find the mass of the lamina.

To solve this, we need a density function, which based on the information given, would be  since the part under the radical is the distance to the z axis. Also, we need integration limits from the region in the xy plane that is the shadow of the surface S. R can be defined as  (if you don't see this, let z = 0 in the cone equation) The integral now would look like

whose integrand algebraically simplifies to   which would make for a tough integral. However, if we switch to polar coordinates, it becomes the following, (and note the simple limits of integration as well!) I used Mathcad to evaluate this, but you should verify we get the answer given here:  or  .

Parametric Surfaces and Surface Integrals

Suppose our surface is defined parametrically like those in lesson 5 of this unit. That is, suppose  defines a surface S defined over some region D (the shadow) in the uv plane. It can be shown that the surface integral of f(x,y,z) over S is given by the formula:

Note the similarity to the line integral over a path C described by a vector valued function r(t) in a vector field f(x,y,z) given here and discussed in Unit V lesson 2:

Ex 3  Let S be the first octant portion of the cylinder y²+z²=9 where . Evaluate the integral 

over the given surface in parametric form.

Now describes our surface in parametric form. Note and , and we have  and so  and our surface integral has become

and then evaluates as 12p + 36. You should verify this result. A sketch of the surface is given below.  

Orientation of a Surface

If we have what is called an oriented surface, it can be used to represent problems involving fluid flow. (Keep in mind, the fluid may be liquid or gas which may behave as a fluid under certain conditions.) A surface S is called orientable if a unit normal vector N can be defined at every non-boundary point of S in such a way that the set of normal vectors varies continuously over all S. If this is possible, then S is called an oriented surface. Note further that an orientable surface has two distinct sides. Ellipsoids, paraboloids, planes and most surfaces we have studied are orientable. A Mobius strip is not, since it has but one side. Here is a drawing of a Mobius strip. If you look at it carefully, you can see that you can draw a continuous line around the entire strip without removing your pencil, hence it has but one side.

Suppose we have a surface z = g(x,y) If we examine the function of three variables constructed from g, G(x,y,z) = z - g(x,y), we can orient the surface by use of the unit vector in the direction of the gradient vector at any point on the surface. Of course since we have a two sides surface, the downward orientation would be . Note a simpler form for each of these is  and .

If a smooth surface S is given in parametric form by the vector valued function , then unit normal vectors can be given by  and .

Flux Integrals

If a fluid has a continuous velocity vector field F and a an imaginary surface S is submerged in the fluid, then the volume of fluid crossing the surface S per unit of time is called the flux of F across S. If we cut the surface S into small units in the usual way, we can assume that F is nearly constant across DS. The amount of fluid crossing DS per unit of time is approximated by the volume of a column of height F^.N (the vector field's component normal to the surface) and base DS, i.e. the amount of volume of the fluid crossing this portion of the surface is . Adding all of these up for a surface S to get the amount of fluid for the entire surface gets us the flux integral.

Def Flux Integral

Let F=Mi+Nj+Pk where M, N, and P have continuous first partial derivatives on a surface S oriented by a unit normal vector N. The flux integral of F across S is given by

If the fluid is not uniformly dense, and r(x,y,z) is the density of the fluid at (x,y,z) then the integral 

represents the total mass of fluid crossing the surface S per time unit.

Theorem 15.11 Evaluating a Flux Integral

Let S be an oriented surface given by z = g(x,y) and R be the projection of S (shadow) onto the xy plane. Then the flux integrals (oriented upwards and downwards respectively) are

Ex 4  Set up the integral to find the flux of F through S Where N is the 'upper' unit normal vector to S where F = xi+yj and S is the plane 2x + 3y + z =  6 in the first octant.  

First z = 6 - 2x - 3y and g_x = -2, g_y = -3, so F^.N = (xi+yj)^.(2i+3j+k) = 2x+3y. Now we need to find limits of integration and an order of integration, so we need to find the region R in the xy plane. If z = 0, we have 2x+3y=6 or y=(-2/3)x+2. Look at the picture below to see the triangular region R under the surface. The required integral is  .  What would the result be if we used the lower unit normal vector to orient S? What does this say about the actual result regarding the meaning of flux?

Suppose the oriented surface is given by the vector valued function . Then we can define the flux integral as

Ex 5 Find the flux over the portion of the sphere S given by x²+y²+z²= 16 lying in the first octant where the force field F = xi+yj+zk. Assume the sphere is oriented with N outward.

Writing the surface parametrically, we have r(u,v)=4sin(u)cos(v)i+4sin(u)sin(v)j+4cos(u)k (See the conversion formula to spherical coordinates if necessary), which gives us r_u(u,v)=4cos(u)cos(v)i+4cos(u)sin(v)j-4sin(u)k and r_v(u,v)=-4sin(u)sin(v)i+4sin(u)cos(v)j.

So our integral is now

Ex 5A Do the above problem without the parametric representation of the sphere. (Note in Theorem 15.11 above that the integrand can be written as  or )

With an integrand such as this, a switch to polar coordinates would be wise.

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Assignment: Chapter 15 section 6 problems 1,11,23,29 (hint-need to find the flux across all six sides of the cube and add up.)