Background for urface Integration 1 urface Integrals We have seen in previous work how to define and compute line integrals in R 2. You should remember the basic surface integrals that we will need to compute and how they are in fact computed. You can remember most of what you need to remember if you really understand what we did with line integrals. To help, we want to look at some facts you have seen before and relate them to the problems we now need to solve. First, we mention just what forms we have to compute when we are using The Gauss Theorem or tokes Theorem. We will need to consider two cases: (1) The surface integral f d where is a surface given parametrically by := {(x(u, v), y(u, v), z(u, v)) (u, v) B} B being a domain in R 2, and where f : R is a given scalar field defined on the surface. (2) The surface integral f df where f is a vector field defined on and is a surface which is parameterized as in (1) above. We will see how to compute these integrals in a later section. First, let us concentrate on R 2 and go back to look at the second form of Green s theorem in the plane. In the original statement of the theorem, we were to integrate a vector field of the form f = (a(x, y), b(x, y)) and found that a(x, y) dx + b(x, y) dy = [b x (x, y) a y (x, y)] da. C A 1
If we then apply this theorem to the vector field g = ( b(x, y), a(x, y)) we find that Green s theorem in the plane has the second form: [ b(x, y) dx + a(x, y) dy = (a, b) (dy, dx) = [a x (x, y) + b y (x, y)] da. C C B Notice that in the middle integral above, the second factor is (dy, dx). If we think of the curve as being parameterized by x = x(t), y = y(t), α t β, then the vector t := (dx/dt, dy/dt) is tangent to the curve. Then taking the dot product of this tangent vector with the vector n := ( dy/dt, dx/dt) yields zero so that the two vectors are perpendicular. Notice that, because of our notation, we need to remember what is not written down explicitly. Both the tangent vector t and the normal vector n depend on the point at which they are calculated, and hence on the parameter value t corresponding to that point on the curve! The vector n is normal to the curve. If C is a closed curve with the usual counterclockwise orientation so that if we walk around the curve, the interior region is on our left, the vector n we want is the exterior normal. In fact, we want the exterior unit normal vector i.e., the exterior normal vector having unit length. 2 ome Elementary Facts about Curves In order to understand the statements of the theorems that we will use in practice, we need to understand something about normal vectors to curves and surfaces. It will help to start with a short review of elementary facts regarding the so- called differential geometry of curves in the plane and in R 3. Most of these facts were discussed in detail in your third- semester calculus class if not before. Again, if r(t) := (x(t), y(t)), α t β is the position vector that describes a plane curve parametrically, then the unit tangent vector, which points in the direction of increasing t-values i.e., in the same direction as the orientation of the curve, is given by T(t) = ṙ(t), ṙ(t) 0. ṙ(t) ( ) If ϕ = arccos ṙ(t) i, then we may write the unit tangent vector T as T = (cos ϕ, sin ϕ) ṙ(t) and we can define the interior unit normal by rotation clockwise through π, so that 2 ˆν = (cos ϕ + π 2, sin ϕ + π 2 ) = ( sin ϕ, cos ϕ). 2
This vector points toward the instantaneous center of curvature of the curve and hence, for a closed curve C, is the interior unit normal. The exterior unit normal is then ˆn = (sin ϕ, cos ϕ). Let us check that this gives the expected result in the case of a circle of radius a. Example uppose that we have a parameterization of the circle given by x(t) = a cos 2t y(t) = a sin 2t Then r(t) = (a cos 2t, a sin 2t), ṙ(t) = ( 2a sin 2t, 2a cos 2t), so that ṙ(t) = 4a 2 sin 2 2t + 4a 2 cos 2 2t = 2a, and the unit tangent vector T(t) is given by T(t) = ( sin 2t, cos 2t) the interior normal ν by ν = ( cos 2t, sin 2t), and so the exterior normal is ˆn = ( 1)ν = (cos 2t, sin 2t) which agrees with the fact that the tangent line to a circle is perpendicular to the radius of the circle i.e., the line through the point of tangency and the center of the circle. Clearly the exterior normal should point away from the center of the circle (which in this case is the center of curvature, and has the same direction as the radius vector. A complete discussion of curvature and the direction of the so-called principle normal vector which we have called ν, depends on the idea of parameterizing a given curve in terms of the arc-length parameter, s which is defined, given any choice of starting point on the curve, in terms of the given parameterization r(t) = (x(t), y(t)) by t (dx ) 2 ( ) 2 dy s(t) := + du. du du t 0 You may see such a discussion in book of Thomas & Finney (ection 11.4). 3 urfaces In three-dimensional Euclidean space. R 3, surfaces can be described most usefully in three different ways: 3
(1) In parametric form using two parameters u, v D R 2, in the form F(u, v) = (x(u, v), y(u, v), z(u, v), (u, v)) D; (2) In explicit form as the graph of a function, z = f(x, y), (x, y) D; (3) In implicit form as level surfaces of a single function f(x, y, z) = 0. These various forms can be illustrated easily by looking at the familiar situation of a plane through a point with a given normal direction n. Example: Let P be a plane in R 3 which passes through the point (1, 2, 1) and having the normal vector N = (2, 1, 3). Then any vector v, parallel to a line in the plane, must have components which satisfy the equation n v = n (1, 2, 1). A quick calculation shows that this requirement reduces to the equation 2x + y + 3z = 1. This last equation clearly gives the form of the planar surface as the level surface of a given function, namely P = {(x, y, z) 2x + y + 3z 1 = 0}. If we set u = x and v = z for paramters, then we can rewrite the equation of the plane in vector form: P : F(u, v) = (u, 1 2u 3v, v), (u, v) R 2 = (0, 1, 0) + u(1, 2, 0) + v(0, 3, 1). This is the parametric form for the plane P. Finally, solving for the unknown z in the original equation for the plane, we arrive at the form which expresses P as the graph of a function z = f(x, y) = 2 3 x 1 3 y + 1 3. A more complicated example is the following three representations of a torus (a donut 4
shaped surface generated by rotating a circle of radius a with center (R, 0) R 2, with R > a around the z-axis. Example: Let P be a point on the circle that generates the torus. If θ is the angle between the center (R, 0) of the circle and the point P, then the (x, y, z)-coordinates of P are (R + a cos θ, 0, a sin θ). (Draw a sketch!) Introducing the angle ψ as the angle of rotation from the x-axis around the z-axis, every point on the torus can be represented (as in cylindrical coordinates) as P (ψ, θ) = x((ψ, θ), y(ψ, θ), z(ψ, θ)) = ((r + a cos θ) cos(ψ), (r + a sin θ) sin ψ, a sin θ) with (ψ, θ) [0, 2π] [0, 2π] and R > a > 0 fixed. This gives a parametric representation of the surface of the torus. In order to get an explicit form for the equations, we must treat the surface in two separate pieces. Then we get the two halves of the torus: the upper half z 1 = a 2 ( x 2 + y 2 R) 2, and the lower half z 2 = a 2 ( x 2 + y 2 R) 2, where x and y satisfy the inequalities (R a) 2 x 2 + y 2 (R + a) 2. And implicit representation is simply ( x 2 + y 2 R) 2 + z 2 a 2 = 0. 4 Normals to a urface In order to discuss the normal vector to a surface,, we need to define it in terms of the tangent plane to the surface. Let us suppose that the surface is given implicitly as f(x, y, z) = c. If r = (x(t), y(t), z(t)) is a smooth curve that lies in the surface then f(x(t), y(t), z(t)) = c. The key observation is that the gradient vector of f is orthogonal to the tangent vector of the curve, namely ṙ. Now, differentiating and using the chain rule, we have or which reduces simply to d dc f(x(t), y(t), z(t)) = dt dt = 0 f dx x dt + f dy y dt + f dz z dt = 0 f dr dt = 0. 5
This means that at every point along the curve, f is orthogonal to the velocity vector along the curve. If we fix a point P 0 on the surface, then we can consider the set of all smooth curves passing through the point P 0. Each will have a tangent vector at the point P 0 and all of these tangent vectors are orthogonal to f. Thus all these tangent vectors lie in a plane containing the point P 0 and with normal vector f. The tangent plane containing this point P 0 = (x 0, y 0, z 0 ) therefore has the equation ( f x (P 0), f y (P 0), f ) z (P 0) ((x x 0 ), (y y 0 ), (z z 0 )) = 0. In the case where the surface is given in explicit form z = f(x, y) we simply notice that this equation is equivalent to the preceeding by setting F (x, y, z) = f(x, y) z and considering the zero-level set of F. ince F = ( f(x, y) z, x f(x, yu) z, y ) f(x, y) z = z ( f x, f ) y, 1, the equation of the plane tangent to the surface at P 0 is f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) (z z 0 ) = 0. We remark that in this last case, the normal vector to the surface at the point P 0 is n = ( f x, f ) y, 1. 5 Orientation and the Exterior Normal In the previous section, we have seen how we can identify a vector normal to a given surface. Actually we have defined two normals, pointing in opposite directions. It is almost always the convention, that in the formulas we take the exterior unit normal to the surface and we wish to understand just which of the two vectors, say f or f will be what we call the exterior normal. In the case of a simply connected bounded surface, we have the geometric notion that the exterior normal points into the unbounded region since the closed surface divides R 3 into two simply connected components, one bounded (the interior) and one unbounded (the exterior). The choice of exterior normal in more general cases depends on the notion of the orientation of the surface. Let us see how this notion arises in the simplest cases. uppose, 6
then, that we start with a function f defined on an interval [a, b] R. We know that we are led in a natural way to the statement that b f(x) dx = a a b f(x) dx. That is, the interval of integration [a, b] is given a natural direction or orientation by the usual order relation. If we reverse the orientation, that is, if we describe the interval in the opposite direction, the value of the integral is multiplied by 1. We have seen this same consideration arise in the case of line integrals both in R 2 and in R 3. We assign a definite direction to the curve, by convention we take the counterclockwise direction as the positive orientation and if we compute a line integral along the curve in the clockwise direction, that is if we reverse the orientation, the value of the line integral is changed by a factor of 1. In the same way, we introduce the definitions for any double integral. We say that a region D with boundary D is positively oriented if the bounding curve D is oriented counter-clockwise, and negatively oriented otherwise. We then state the definitions f(x, y) dxdy = f(x, y) dxdy D +D f(x, y) dxdy = f(x, y) dxdy D D On the other hand, we can assign an orientation to a region without reference to the bounding curve. Indeed, if we take any point in the region D we can assign a sense of rotation about that point by e.g., describing a small circle with that point as center and lying entirely in the region D and then choosing a direction to that circle. We then say that the region D is oriented is such a sense of rotation is assigned at each point of the region and, if as we traverse any smooth curve in the region, the sense of rotation is the same at any point along that curve. We can now see how to assign an orientation to a surface in R 3. On a surface, we first assign an orientation to a point on that surface by surrounding that point with a smooth curve lying in the surface. We then assign an orientation to that curve. Then, if we move that point along any smooth curve lying in the surface and move the surrounding curve with it, we assign a sense of rotation to every point on the surface. The question now is, what is a positively oriented surface? We start by choosing either of the normal vectors and declaring it to be positive (which one we actually pick is irrelevant. We say that a surface is positively oriented provided that the orientation given by the above procedure and the positive normal form a right handed screw. Otherwise 7
we say that the surface is negatively oriented. In other words, if a surface together with its orientation can be continusouly deformed in such a way that it becomes the positively oriented (x, y)-plane and at the same time the direction of the positive normal becomes the direction of the positive z-axis, then we say that the surface is positively oriented. 6 Unit Exterior Normals and Parameterizations As we will see in the first of the examples below, we will often compute integrals, as for example F ˆn d by parameterizing the surface, say in the (x, y)-plane, or by some other convenient parameterization. Let us look at a specific case: Example: We consider the example of the torus that we looked at earlier. As in that earlier example, we can write the parameterization as a transformation from (ψ, θ)-coordinates to (x, y, z)- coordinates. 7 Two Concrete Examples Example: Let be the closed surface of the cap of a paraboloid of revolution described by the set {(x, y, z) x 2 + y 2 1, 1 z (x 2 + y 2 )}. The cap of the parabloid is the surface 1 = {(x, y, z) x 2 + y 2 1, z = (x 2 + y 2 )} and the bottom of the surface is 2 = {(x, y, z) x 2 + y 2 1, z = 1}. The problem is to integrate the vector field f = (xy 2 z, x 2 yz, x) over the surface, that is to compute the integral f ˆn d which, according to the Divergence Theorem is also f ˆn d = div f dv V We consider the integrals around the cap and on the base separately. tarting with the cap, 1, the vector field has the form f = (xyz( x 2 y 2 ), x 2 y(x 2 + y 2 ), (x 2 + y 2 )) 8
Using the parameterized form of the surface: x = x, y = y, z = x 2 y 2, for x 2 +y 2 1 we can compute the normal to the cap by computing two tangent vectors in the directions of the coordinate axes and then taking the cross product. Note that as the rule for taking the cross product gives a right-handed coordinate system, the normal we get in this case is the exterior normal. pecifically, r x = r y = ( x x, y x, z ) = (1, 0, 2x), x ( x y, y y, z ) y = (0, 1, 2y) Then an exterior normal is ν = r x r y = ( 2x, 2y, 1) which has length ν = 4x2 + 4y 2 + 1 = 5 and a simple computation yields (x 2 +y 2 )=1 f ˆn = 1 5 o we have f ˆn d = 1 1 5 [ 2x 2 y 2 (x 2 + y 2 ) + x 2 y 2 (x 2 + y 2 ) x 2 y 2] ) = 1 5 ( x 2 y 2 ). D 2π ( 1 ) (x 2 + y 2 ) da = r 2 rdr dθ = π 0 0 2 5. For the other piece of the surface, the bottom of the cap, we can introduce the parameterization x = x, y = y, z = 1 and compute two tangent vectors as above. Then r x = (1, 0, 0), r y = (0, 1, 0) and so r x r y = (0, 0, 1). But this is the inner unit normal and we must take ˆn = (0, 0, 1). Then f ˆn d = ( xy 2, x 2 y, 1) (0, 0, 1) da = π. 2 Hence the total value of the surface integral is D 9