ES.182A Topic 44 Notes Jeremy Orloff

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1 E.182A Topic 44 Notes Jeremy Orloff 44 urface integrals and flux Note: Much of these notes are taken directly from the upplementary Notes V8, V9 by Arthur Mattuck. urface integrals are another natural integral. They are like double integrals, except the area elements are little (infinitesimal) areas on the surface. After that, we can do our usual summing and limiting to get an integral. One of our main uses of surface integrals will be to compute the flux of a field through a surface. You might recall that when we introduced flux we started with a 3D example because it was more intuitive. The example was about the rate that falling water flows through a window. The units were volume/time. In this topic we will start with a quick discussion of vector fields in space. There is no conceptual difference between them and vector fields in the plane. After that, we will look at normals to surfaces. Next, we will learn about surface integrals and the surface area element d. Finally we learn about flux and the combined normal and area element n d Vector fields in space A vector field in space is exactly analogous to one in the plane. We will usually use the following notation: F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P (x, y, z)k = M, N, P. uch a function assigns the vector F(x, y, z ) to a point (x, y, z ) where M, N, and P are all defined. We place the vector so its tail is at (x, y, z ), and in this way get the vector field. uch a field in space looks a little like the interior of a haystack. As before, we say F is continuous in some domain D of 3-space (we will often use domain rather than region, when referring to a portion of 3-space) if M, N, and P are continuous in that domain. We say F is continuously differentiable in the domain D if all nine first partial derivatives M x, M y, M z ; N x, N y, N z ; P x, P y, P z exist and are continuous in D. As before, we give two physical interpretations for such a vector field. The three-dimensional force fields of different sorts gravitational, electrostatic, electromagnetic all give rise to such a vector field: at the point (x, y, z ) we place the vector having the direction and magnitude of the force which the field would exert on a unit test particle placed at the point. The three-dimensional flow fields and velocity fields arising from the motion of a fluid in space are the other standard example. We assume the motion is steady-state, i.e. at each point in space the the fluid has a velocity which doesn t change over time. 1

2 44 URFACE INTEGRAL AND FLUX 2 As in two dimensions, we allow singularities. For fluid flow these will often behave like sources and sinks places where fluid is being added to or removed from the flow. Obviously, we can no longer appeal to people standing overhead pouring fluid in at various points (they would have to be aliens in four-space), but we could think of thin pipes inserted into the domain at various points adding or removing fluid. Example Find the three-dimensional electrostatic force field F arising from a unit positive charge placed at the origin, given that in suitable units F is directed radially outward from the origin and has magnitude 1/ρ 2, where ρ is the distance from the origin. answer: The vector xi + yj + zk with tail at (x, y, z) is directed radially outward and has magnitude ρ. Therefore F = xi + yj + zk ρ 3, ρ = x 2 + y 2 + z 2 Example (a) Find the velocity field of a fluid rotating with constant angular velocity ω around the z-axis, in the direction given by the right-hand rule (right-hand fingers curl in direction of flow when thumb points in the k-direction). (b) Find the analogous field if the flow is rotating about the y-axis. answer: (a) The flow doesn t depend on z, i.e it is always in a horizontal plane. o, this is really just a two-dimensional problem. For a point on a horizontal circle of radius r centered on the z-axis, the direction should be tangential and the magnitude should be rω (velocity = radius times angular velocity). o, F(x, y, z) = ω y x. z z x y x y Left: constant angular velocity around the z-axis. Right: around the y-axis. (b) This is basically the same problem as part (a). We have to exchange y and z, j and k, and then get the sign of the tangential field correct. By the right-hand rule, the flow on the positive x-axis points down. Thus F(x, y, z) = ω(zi xk).

3 44 URFACE INTEGRAL AND FLUX urfaces At a point on a surface there are two choices of normal. Choosing one of them is called a choice of orientation. Depending on the type of surface we have different names for the choice of normal. Upward normals Downward normals Outward normals Inward normals Note that to have inward or outward normals the surface needs to be closed, so that it has an inside and an outside Parametrization of surfaces Recall that for curves we have two types of parametrization. 1. x = x(t), y = y(t) e.g., x = cos(t), y = sin(t), t from to π/2. 2. x = x, y = f(x) e.g., x = x, y = x 2, x from to 1. urfaces require 2 parameters. Example Cylinder of radius a. x = a cos(θ), y = a sin(θ), z = z. (Parameters are θ and z.) 2. phere of radius a. x = a sin(φ) cos(θ), y = a sin(φ) sin(θ), z = a cos(φ). (Parameters are φ and θ.) 3. urface = graph of z = f(x, y). Here the parameters are just x and y: x = x, y = y, z = f(x, y).

4 44 URFACE INTEGRAL AND FLUX 4 Warning. Pay attention to when the radius is a constant, as on the surfaces in the previous examples 1 and 2, and when it varies, as in a triple integral over volume urface integrals As always, we construct a surface integral from a sum. We divide the surface into little curvy patches and sum over all the patches. z Patch i at (x i, y i, z i ) x f(x, y, z) d f(x i, y i, z i ) i, where i = area of the ith patch. As usual, the integral is the limit as the patches all shrink to points. We will wait till we ve defined flux to work examples of surface integrals. y 44.5 Flux Definition. The flux of a vector field F through a surface is defined as Flux of F through = F n d. Here n is a choice of unit normals on, i.e. an orientation. If F is a velocity field then the units of flux are volume/time. The geometric argument that this measures the rate that fluid passes through the surface is the same as we saw in two dimensions with flux across curves. We will derive the basic formula for 3D flux in an appendix at the end of these notes.

5 44 URFACE INTEGRAL AND FLUX Finding n d Recall that for line integrals we had n ds = dy, dx. A key thing to notice is that this doesn t give n or ds separately (although we could find them). eparately they can be a little complicated, but the combination n ds is simple. Another method we sometimes used was to use geometry, i.e., look at a picture to determine the normal. Then we found ds separately if necessary. omething similar happens with n d on surfaces. postpone an explanation to the appendices. Here we will give the formula, but Theorem. (Formula for finding n d) If the surface is the graph of z = f(x, y) then Upward normal: n d = f x, f y, 1 dx dy (points generally up). (1) Downward normal: n d = f x, f y, 1 dx dy (points generally down). (2) (You should recall that in the first half of 18.2 we found that a normal to is f x, f y, 1.) Examples of flux Example uppose z = f(x, y) = x 2 + y and is the graph of f above the unit square in the xy-plane. Let F = zi + xk. Find the upward flux of F through. answer: Finding the upward flux means we need to use the upward pointing normals. Here are the steps: 1. Find n d: We use the formula in Equation 1. n d = f x, f y, 1 dx dy = 2x, 1, 1 dx dy. 2. Compute F n d = z,, x 2x, 1, 1 dx dy = ( 2xz + x) dx dy. 3. Find the limits of integration: ince n d uses x and y the region of integration is R, which is the projection (shadow) of on the xy-plane. In this example, we are given explicitly that R is the unit square. o, Inner variable: y from to 1. Outer variable: x from to 1.

6 44 URFACE INTEGRAL AND FLUX 6 4. Compute the flux integral Flux = Inner integral: 1 F n d = Outer integral: Upward flux = R 2xz + x dx dy = 1 1 2x(x 2 + y) + x dy = 2x 3 x + x = 2x x 3 dx = 1 2. Note, this implies the downward flux = 1/2. 2x(x 2 + y) + x dy dx. Remember: upward and downward flux are about the choice of n not F. Example Without computation find the rightward flux of F = j through the surface shown. answer: The surface is the unit square in the xz-plane, so the unit rightward normal is n = j. Thus, F n = j j = 1. Now we can compute flux: F n d = d = area = 1. Example Let F = x, y, z and =sphere of radius 2 centered on. Find the outward flux. x, y, z answer: On the sphere we have the outward normal is n = 2 (unit vector). x, y, z o, on the sphere, F n = x, y, z = x2 + y 2 + z 2 = (More simply, F is radial and so is n, so F n = F = 2 on the sphere.) Thus, flux = F n d = 2 area = 32π. Example (Cylinder) Find the flux of F = z, x, y outward through the piece of the cylinder x 2 + y 2 = a 2 in the first octant below z = h.

7 44 URFACE INTEGRAL AND FLUX 7 answer: The outward normal is n = picture.) xi + yj a and the area element d = a dz dθ. (ee zx + xy Thus, F n d = a dz dθ = (zx + xy) dz dθ. a Parametrization of : parameters z, θ: x = a cos(θ), y = a sin(θ), z = z. Limits: θ π/2, z h. o, flux = π/2 h Inner integral: h (zx + xy) dz dθ = Outer integral :... flux = a 2 h. π/2 h az cos(θ) + a 2 cos(θ) sin(θ) dz dθ. az cos(θ) + a 2 cos(θ) sin(θ) dz = ah2 2 cos(θ) + ah2 cos(θ) sin(θ). Example (phere) Find the flux of F = xz, yz, z 2 outward through the part of the sphere x 2 + y 2 + z 2 = a 2 in the first octant. answer: The area element is d = a 2 sin(φ) dφ dθ. This can be seen by computing the edge lengths of the curvy rectangle in the figure below. Another way to think about this is that dv = ρ 2 sin(φ) dρ dφ dθ = ρ 2 sin(φ) dφ dθ dρ. Which can be viewed as area thickness. x, y, z As always on the sphere, the unit outward normal is radial: n =. a Thus, F n = x2 z + y 2 z + z 2 z = az. a π/2 π/2 π/2 π/2 o, flux = F n d = az a 2 sin(φ) dφ dθ = a 4 cos(φ) sin(φ) dφ dθ. This is easy to compute: flux = a4 π 4.

8 44 URFACE INTEGRAL AND FLUX A general formula for n d uppose is a surface parametrized by x and y. necessarily of unit length). Then Here n is the upward unit normal. n d = N dx dy. N k The following example explains where this formula comes from. Also suppose N is any normal (not Example On the sphere x 2 + y 2 + z 2 = a 2 we know N = x, y, z is a normal. o, x n d = z, y z, 1 dx dy. Dividing by N k converts the last coordinate in the vector to 1. This is exactly what we saw in Equation 1. aid another way, the formula above is exactly what we get if let z = a 2 x 2 y 2, so, n d = z x, z y, 1 dx dy Using d by itself Once we have a formula for n d it is easy to get a formula for d. For example, if our surface is the graph of z = f(x, y) then n d = f x, f y, 1 dx dy. ince n is a unit vector we have d = n d = 1 + fx 2 + fy 2 dx dy. TO DO: Add examples, e.g. average value of f = 44.8 Appendix: the basic formula for flux To be added. f d area of Appendix: the formula for n d To be fleshed out. Just as we did in the first half of 18.2A, we use the tangent plane approximation: dz = f x dx + f y dy. In the figure at below the patch of surface is the graph of the small dx dy rectangle in the plane. A is the vector along the graph of the dx edge and B is along the graph of the dy edge. We have n d = A B since both vectors have the same magnitude and direction.

9 44 URFACE INTEGRAL AND FLUX 9 A has no j component: A = dx,, f x dx. B has no i component: B =, dy, f y dy. A B = f x dx dy, f y dx dy, dx dy = f x, f y, 1 dx dy. QED