Section 17.7: Surface Integrals. 1 Objectives. 2 Assignments. 3 Maple Commands. 4 Lecture. 4.1 Riemann definition

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1 ection 17.7: urface Integrals 1 Objectives 1. Compute surface integrals of function of three variables. Assignments 1. Read ection Problems: 5,7,11,1 3. Challenge: 17,3 4. Read ection Maple Commands 4 Lecture Line integrals are to arc length what surface integrals are to surface area. Thus, you should expect that when we find a surface integral we ll include some term that account for surface area, the same way we included an arclength when we calculated a line integral. 4.1 Riemann definition Let f be a function of three variables. We want to integrate f over general surfaces that are included in its domain (which is in IR 3 ). We begin by dividing the surface (which is in IR 3 ) into patches that we ll label ij. Each of these patches has an associated area that we ll call ij. Note that the area of these patches is a surface area, because these patches are in IR 3 as well. We evaluate f at a point Pij that is in ij (the same way that we have evaluated our function in subdivisions of our integration region previously). Then we form the Riemann sum to define the surface integral of f over as lim m i1 n j1 f(pij) ij f(x, y, z) d. We know how we can find ij, since it s a surface area. Let be defines by the functional relationship z g(x, y), where (x, y) are in some planar region. Think about the tangent plane approximation to our surface at the point P ij (x i, y j ). We can approximate the surface area of the patch ij by using the area of the 1

2 tangent plane approximation. Remember that the surface area of is approximated by the tangent plane approximation as ij T ij g x (x i, y j ) + g y (x i, y j ) + 1 A, where A is the rectangular region in the xy plane that includes (x i, y j ). (Review earlier material on finding surface areas to help you see this connection.) Therefore, if f is continuous on and g has continuous partial derivatives, we can define the surface integral as m n f(x, y, z) d lim f(pij ij lim i1 m i1 j1 n j1 f(x i, yj, g(x i, yj )) g x (x i, y j ) + g y (x i, y j ) + 1 A f(x, y, g(x, y)) g x (x, y) + g y (x, y) + 1 da f(x, y, g(x, y)) z x (x, y) + z y (x, y) + 1 da. Note the change in the notation in the last two equations. ince z g(x, y), you are just as likely to see g x here as you are to see z x ; but they are the same quantity. You differentiate the function describing your surface with respect to x. You can change this integral depending on how your surface is defined. If the surface is defines as y h(x, z), the integral changes to f(x, y, z) d f(x, h(x, z), z) y x (x, z) + y z (x, z) + 1 da Example 1: Problem We want to evaluate xy d, where is the triangular region with vertices (1,, ), (,, ), and (,, ). The equation for the plane associated with this triangular region is z x y. is the base of this region; I suggest you draw the triangular region to help you find the limits on. I find that x 1, and y x. Thus, the surface integral is evaluated as xy d xy ( ) + ( 1) + 1 da 1 x xy 6 dy dx 6 6. We didn t have z as part of our integrand (we had f(x, y, z) xy), so we didn t have to substitute our equation of the plane anywhere in our integrand; we only had to use it when finding the surface area and finding the limits of integration on. Class Questions:

3 1. Can you determine the equation of the plane I got above? Use vectors represented by two sides of the triangle and one of the vertices and use the equation for the plane covered in earlier material. (The vectors I used are < 1,, > and < 1,, >. I used the cross-product of these two vectors to find the unique normal vector that defines the plane.). oes the equation of the plane change if I choose a different set of sides of the triangles for the vectors or a different point on the triangle? 4. Applications One of the applications of surface integrals is in finding the mass of a thin sheet of material. We used it in our Maple Project to find the weight of our section of bridge. While bridge sections are not as thin as the aluminum foil example in the text, it is relatively thin if you consider the planar region that is spanned by the bridge section. (Think of the xy region beneath a bridge section, for example. The bridge is relatively thin compared to this.) If the density ρ is defined in terms of mass per unit area, then the mass of the thin sheet of material is m ρ(x, y, z) d. The coordinates of the center of mass are given as x 1 xρ(x, y, z) d m ȳ 1 yρ(x, y, z) d m z 1 zρ(x, y, z) d. m 4.3 Parametric surfaces We didn t cover any applications of surface area that dealt with parametric representations of surfaces, the subject of ection You should be aware of this material, in case it is needed in later classes. The whole idea is to represent your surface in terms of parametric equations. However, instead of having one variable, as is the case when representing curves by parametric equations, you have two. 4.4 Flux We can use surface integrals to define the flux of a vector field. If F is a continuous vector field defined on an oriented surface, with unit normal vector n, then the surface integral of F over is F d F n d, 3

4 which is also called the flux of F across. A surface is oriented with respect to a normal vector, much the same way that a spacecurve is oriented with respect to a normal and/or tangent vector. To think about the concept of oriented surfaces, think about the tangent plane at each point. The normal vector that determines the orientation of the surface is orthogonal to the tangent plane at the point (x, y, z). You actually have two normal vectors at this point, n 1 and n, where n 1 n. For a closed surface, which happens when the surface is the boundary of a solid, the positive orientation is the one for which the normal vectors point out from the solid. Normal vectors pointing inward give the negative orientation. How can we get these normal vectors? Think about the vectors associated with the tangent plane to the surface z g(x, y). We know that one vector is given by a i + g x k and a second is given by b j + g y k. (Think about earlier material on tangent planes; we recently found a formula for surface area using these vectors.) The normal vector to the tangent plane is a b g x i g y j + k. We can divide by the magnitude of this vector to make it a unit normal vector. Note that the coefficient of k is 1; this is the upward normal vector at the point (x, y, z); the downward normal vector is g x i + g y j k. Also, note that is z g(x, y), then we have f(x, y, z) z g(x, y) and f f < g x, g y, 1 > n. Thus, if F P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, we have that F d F n d ( F < g x, g y, 1 > g x + g y + 1 F < g x, g y, 1 > da ( P g x Qg y + R) da Example : Problem ) ( g x ) + ( g y ) + 1 da This problem asks you to find the flux of F given a surface with a downward orientation. The surface is described by the equation z 1 x y, so z x 1 and z 1. We can use the normal vector with downward orientation as n < y 1, 1, 1 >; or, we can set up the integral assuming upward orientation, that is, use n < 1, 1, 1 >, and take the negative of this integral. I ll do the latter. The vector field function is F(x, y, z) (xze y )i + ( xze y )j + zk. The integral to 4

5 evaluate is Class Exercises: F d 1 1 x 1 1 x 1 1 x [( xze y )( 1) ( xze y )( 1) + z] dy dx z dy dx (1 x y) dy dx (y xy y 1 x + x dx ) 1 x dx 1. How did I get the limits for the integral above?. Evaluate the integral using the negative normal vector. You should get the same answer. 4.5 Heat Flow Let u u(x, y, z) be a function describing the temperature at (x, y, z) in a material occupying a region containing the oriented surface. The quantity q k u is the heat flux and has units of energy per unit per time. The constant k is the thermal conductivity of the material. The amount of energy crossing is given by the flux of q over, or q n d k u n d The idea of flux can be expanded to handle lots of different physical situations. If the vector field F is a velocity vector field of a fluid with density 1, then the surface integral is actually the rate of flow of mass through the surface per unit time, or the mass flux. 5