Lecture 23. Surface integrals, Stokes theorem, and the divergence theorem. Dan Nichols

Transcription

1 Lecture 23 urface integrals, tokes theorem, and the divergence theorem an Nichols MATH 233, pring 218 University of Massachusetts April 26, 218 (2) Last time: Green s theorem Theorem (Green s theorem) Let be a simply connected region in the xy-plane and let be its boundary, a positively oriented piecewise-smooth simple closed curve. If P (x, y) and Q(x, y) have continuous partial derivatives in and around, then ˆ P dx + Q dy ( Q x P ) y da. Idea: take a line integral over an annoying simple closed curve and turn it into an easier double integral over the region enclosed by that curve. (Or sometimes the other way around.)

2 (3) Green s theorem: example Example 1: Let C be the triangle shown in the diagram below. Compute the line integral ˆ (sec x + y) dx + xe y dy. C Q x P y ey 1 y 6 ˆ C (sec x + y) dx + xe y dy (e y 1) da ˆ 6 ˆ 5 x/3 x/2 (e y 1) dy dx 5e 3 + 3e x C (4) Curl and divergence Let F(x, y, z) P (x, y, z), Q(x, y, z), R(x, y, z) be a vector field in R 3. Two operators that help us understand local behavior: (curl F)(x, y, z) F x, y, P (x, y, z), Q(x, y, z), R(x, y, z) z R y Q z, P z R x, Q x P y (div F)(x, y, z) F x, y, P (x, y, z), Q(x, y, z), R(x, y, z) z P x + Q y + R z (curl F)(x, y, z) tells us how/if the field F induces rotation at point (x, y, z). (div F)(x, y, z) tells us how/if F induces expansion or contraction at (x, y, z). Both are types of derivatives; they are linear and obey a product rule.

3 (5) Parametric surfaces efinition A parametric surface is the set of points in R 3 with position vectors given by a vector function of two variables defined on some region in the uv-plane. r(u, v) x(u, v), y(u, v), z(u, v) Jeff Knisley, ETU (6) Parametric surfaces: example Example 2: What does this parametric surface look like? ee what happens to vertical and horizontal lines in the uv-plane o the three component functions x, y, and z satisfy some implicit equation? r(u, v) cos u, 12 sin u, v Image of line v c in uv-plane: r(u) cos u, 1 2 sin u, c. This is an ellipse in the plane z c. Image of line u c in uv-plane: r(v) cos c, sin c, v. This is a vertical line through point (cos c, sin c, ). o this surface is made of horizontal ellipses and vertical lines. All horizontal traces are identical ellipses. Component functions x, y, z satisfy implicit equation x 2 + 4y 2 1. The vector function r(u, v) is a parametrization of this elliptic cylinder.

4 (7) Parametric surfaces There are many parametric surfaces which (unlike the previous examples) can t easily be described explicitly or implicitly, like a torus of major radius R and minor radius r. r(u, v) R + r cos(v) cos(u), R + r cos(v) sin(u), r sin(v) Image of v c is a big horizontal circle Image of u c is a smaller vertical circle u 2π, v 2π gives you the whole torus. (8) Parametric surfaces Another weird example: Klein bottle. This surface is non-orientable, meaning it has only one side instead of two. x(u, v) cos(u) (3 + cos(u/2) sin(v) sin(u/2) sin(2v)) y(u, v) sin(u) (3 + cos(u/2) sin(v) sin(u/2) sin(2v)) z(u, v) sin(u/2) sin(v) sin(u/2) sin(2v) credit wikipedia

5 (9) Closed surfaces and surface boundaries Let be a surface defined by r(u, v) on a region in the uv-plane. The boundary of is a space curve that makes up the edge of the surface. Just as with planar regions, we use the notation to mean the boundary of surface Usually (but not always) is the image of. Closed surfaces (e.g. unit sphere, torus) have no boundary. urface shown here is r(u, v) v cos u, v sin u, v 2, [, 2π] [, 1] is unit circle in the plane z 1 (1) Closed surfaces and surface boundaries: example Example 3: Let be the surface parametrized by the vector function r(u, v) v cos(2πu), v sin(2πu), v u on the region [, 1] [, 1]. escribe the boundary. Hint: is (probably) the image of under r(u, v). Look at each of the four pieces of and see what r(u, v) does to it... 1 v r(u, v) u z y 1 x

6 (11) Closed surfaces and surface boundaries: example Left: u, v 1. r(, v) hv,, vi. Line segment from (,, ) to (1,, 1). Top: v 1, u 1. r(u, 1) hcos(2πu), sin(2πu), 1 ui. Linear helix from (1,, 1) to (1,, ) going once clockwise around the cylinder x2 + y 2 1. Right: u 1, v 1. r(1, v) hv,, v 1i. Line segment from (,, 1) to (1,, ). Bottom: v, u 1. r(u, ) h,, ui. Line segment from (,, ) to (,, 1). (12) urface integral of a scalar function efinition Let be a surface parametrized by r(u, v) on a region in the uv-plane. The surface integral of a scalar function f (x, y, z) on with respect to surface area is written f (x, y, z) d. When is a rectangle, the surface integral can be defined as a limit of double Riemann sums over. The blue part is the (approximate) area of one patch of the surface. (Remember cross product magnitude area of parallelogram) lim lim m n m X n X f (r(ui, vj ))kru (xi, yj ) rv (ui, vj )k u v i1 j1 This limit of double Riemann sums can be written as a double integral: f (x, y, z)d f (r(u, v))kru (u, v) rv (u, v)k da. You can then use Fubini s theorem to turn it into an iterated integral and evaluate.

7 (13) urface integral of a scalar function Note: r u (u, v) is the partial derivative with respect to u of the vector-valued function r(u, v) x(u, v), y(u, v), z(u, v). That is, r u (u, v) x u (u, v), y u (u, v), z u (u, v). imilarly r v (u, v) is the partial derivative of r with respect to v. 1 d is the surface area of, in the same way that 1 da is the area of a planar region. d da is the surface area derivative, meaning what you integrate to get surface area: d da r u(u, v) r v (u, v) At each point (u, v) in, this tells you how much the transformation r is stretching or shrinking space. (14) urface integral of a scalar function: example Example 4: Let be the surface parametrized by r(u, v) u, v, 1 u + v on the rectangle {(u, v) : u 2, 1 v 4}. Compute xz d. Use formula from previous slide to turn this surface integral into a double integral: f(x, y, z)d f(r(u, v)) r u (u, v) r v (u, v) da f(r(u, v)) u(1 u v) u u 2 uv r u (u, v) 1,, 1 r v (u, v), 1, 1 r u (u, v) r v (u, v) 1, 1, 1 3 f(r(u, v)) r u (u, v) r v (u, v) da ˆ 2 ˆ 4 1 (u u 2 uv) 3 dv du

8 (15) urface integral of a scalar function: example Example 4: (cont.) f(r(u, v)) r u (u, v) r v (u, v) da ˆ 2 ˆ ˆ 2 ˆ 2 (u u 2 uv) 3 dv du (uv u 2 v 12 ) 4 uv2 du ( 3u 2 92 u ) du 3 ( u 3 94 ) 2 u2 3 ( 8 9) (16) urface integral of a vector field efinition Let be an orientable surface in R 3 parametrized by r(u, v) on and let F(x, y, z) be a continuous vector field on. The surface integral of F over is F d F n d where n is the unit normal vector field on. This is also called the flux of F across. Imagine the field F describes the velocity of a fluid, and the surface is a permeable sheet that the fluid can pass through... Then F d tells you the total amount of fluid passing through. Many applications of this throughout physics & engineering: electric fields, etc...

9 (17) urface integral of a vector field At every point r(u, v), the tangent plane of contains the vectors r u (u, v) and r v (u, v). We can cross these to get a normal vector. o the unit normal field on is Therefore F d F n d n(u, v) r u(u, v) r v (u, v) r u (u, v) r v (u, v). F(r(u, v)) r u (u, v) r v (u, v) r u (u, v) r v (u, v) r u(u, v) r v (u, v) da F(r(u, v)) (r u (u, v) r v (u, v)) da. With this formula, we can turn a vector field surface integral into a double integral and then evaluate it using Fubini s theorem. (18) urface integral of a vector field: example Example 5: Let be the surface parametrized by r(u, v) v, u v, u on [, 2] [, 1] in the uv-plane. Compute F d, where F(x, y, z) 2, x, 1. F d F(r(u, v)) (r u (u, v) r v (u, v)) da. F(r(u, v)) 2, v, 1 r u (u, v), 1, 1 r v (u, v) 1, 1, r u (u, v) r v (u, v) 1, 1, 1 F d ˆ 2 ˆ 1 ˆ 2 2, v, 1 1, 1, 1 da (1 + v) dv du 3 du 3. 2

10 (19) tokes theorem at last Theorem (tokes theorem) Let be an oriented piecewise-smooth surface in R 3 bounded by a simple, closed, piecewise-smooth, positively oriented boundary curve. Let F(x, y, z) be a vector field with components differentiable on and around. Then F dr (curl F) d. That is, the line integral of F along the boundary of is equal to the surface integral of curl F over itself. Use this to trade a hard line integral for an easy surface integral (or occasionally the other way around) Also, there seems to be some kind of path-independence equivalent for curl fields... Textbook section 16.8 has a partial proof. Full proof is much too hard for MATH 233! (2) tokes theorem: example Example 6: Let C be the simple closed curve in R 3 consisting of four line segments connecting the four points (,, ), (2,, ), (2, 2, 1), and (, 2, 1) in that order. Let F(x, y, z) 1 z, x 2 + y, y. Compute C F dr. Bad method: split C into four line segments, parametrize and integrate over each one, then add the results together. Better method: use tokes theorem to turn this into a single surface integral, which becomes a double integral over a rectangle. (2,, ) x We can parametrize the plane segment bounded by C with the vector function r(u, v) u 2,, + v, 2, 1 2u, 2v, v z (, 2, 1) y (2, 2, 1) on [, 1] [, 1] in the uv-plane. Then r u (u, v) 2,, and r v (u, v), 2, 1.

11 (21) tokes theorem: example Example 6: (cont.) Integrating F(x, y, z) 1 z, x 2 + y, y over where is parametrized by r(u, v) 2u, 2v, v on [, 1] [, 1]. To use tokes theorem, we need the curl of F: i j k curl F F x y 1 z x 2 + y y z 1, 1, 2x. Have r u (u, v) 2,,, r v (u, v), 2, 1. Cross product: r u r v, 2, 4. tokes theorem: F dr C (curl F) d ˆ 1 ˆ 1 1, 1, 4u, 2, 4 da (2 + 16u) dv du 1, 1, 2x d ˆ 1 (2 + 16u) du 1. (22) tokes theorem and Green s theorem uppose is just a flat piece of the xy-plane (floor). Parametrize with the vector function r(u, v) u, v,... is the same as, just with different variable names is a simple closed curve on the floor Unit normal field is just n(u, v) k,, 1 And suppose F is flat on the xy-plane, i.e. F(x, y, ) P (x, y, ), Q(x, y, ),. Then curl F,, Q x P y. Then tokes theorem becomes F dr,, Q x P d y,, Q,, 1 da x P y ( Q x P y ) da. This is Green s theorem! It s just a special case of tokes theorem where everything is restricted to the xy-plane. Q x P y is the curl of a 2 field.

12 (23) Triple integrals Let E be a solid region of R 3. The triple integral of a function f(x, y, z) over E is f(x, y, z) dv. E When E is a box [a, b] [c, d] [r, s], we can use Fubini s theorem to turn a triple integral into an iterated integral in any order, for example E f(x, y, z) dv ˆ b ˆ d ˆ s a c r f(x, y, z) dz dy dx. The triple integral of f(x, y, z) 1 over E is the volume of E: V 1 dv. We can also integrate over non-box regions that are bounded in nice ways by nice surfaces, kind of like type I and II regions in R 2. E Theorem (ivergence theorem) (24) ivergence theorem Let E be a solid region of R 3 with closed positively-oriented boundary surface E. Let F(x, y, z) be a vector field with components differentiable on and around E. Then F d (div F) dv. E Left side: surface integral of vector field F flowing through closed surface E Right side: triple integral of scalar function div F over solid region E imilar to Green s theorem or tokes theorem trade an integral for another one of a different type. In this case, trade a vector field surface integral for a (hopefully easier) triple integral. Or sometimes the other way around. Physics connection: Gauss law (one of Maxwell s equations). Electric flux leaving a region is proportional to total charge inside. F d 1 ρ dv ε Equivalent to div F ρ/ε. Ω E Ω

13 Example 7: Let E be the cube-shaped region in R 3 defined by (25) ivergence theorem: example E [ 1, 1] [ 1, 1] [ 1, 1] {(x, y, z) : 1 x 1, 1 y 1, 1 z 1} and let F be the vector field F(x, y, z) 1 3 x3 + ln y, e z + yz, x 2 y 2. Calculate the flux of F over the closed boundary surface E. Bad method: split E into six squares (plane segments), do a surface integral over each one and add the results together. Good method: use divergence theorem and do an easy triple integral. F d (div F) dv E E (26) ivergence theorem: example Example 7: (cont.) Computing E F d where E [ 1, 1] [ 1, 1] [ 1, 1] and F(x, y, z) 1 3 x3 + ln y, e z + yz, x 2 y 2. div F F x, y, 1 z 3 x3 + ln y, e z + yz, x 2 y 2 ( ) 1 x 3 x3 + ln y + y (ez + yz) + z x2 y 2 ivergence theorem: F d E x 2 + z. E (div F) dv ˆ 1 ˆ ˆ 1 ˆ 1 1 ˆ ˆ 1 1 ( x 2 + z ) dz dy dx 2x 2 dy dx 4x 2 dx 8 3.

14 (27) Course summary We ve learned the basics of geometry and calculus in three dimensions ometimes this is actual three dimensional space Other times, it s some abstract space representing three variables Much of what we learned also applies in higher dimensions In real life, quantities usually depend on multiple variables. Physics, chemistry, statistics, economics, social sciences, etc. (28) Curves and surfaces A curve is the path of something moving through space. (1-dimensional) We describe a curve in R 3 using a vector function, i.e. r(t) x(t), y(t), z(t). A surface is like a folded sheet in space. (2-dimensional) escribe a surface in R 3 using an implicit equation F (x, y, z)... or as the graph of an explicit function z f(x, y)... or as a parametric surface r(u, v). A plane is a completely flat surface no curvature at all. These are all examples of manifolds: pieces of space that are curved or folded, but not bent or ripped. Curves are 1 manifolds, surfaces are 2 manifolds, etc.

15 (29) ifferential and integral calculus In different contexts, there are different generalizations of the calculus I and II concepts of differentiation and integration. ifferentiating something means zooming in and measuring how it is changing at a specific point. d Examples: f(x, y, z), dtr(t), curl F(x, y, z), div F(x, y, z) Integrating something means adding up the total change that happens over a piece of space or time. Examples: double/triple integrals, line integrals, surface integrals Ultimate goal of calculus: be able to compute various kinds of derivatives to get information about various types of functions and objects. Be able to integrate various types of functions over manifolds of various dimensions in various ways. (3) True tokes theorem You can really boil all of calculus down to just one theorem: Theorem (tokes theorem) ˆ M ˆ dω The integral of the derivative of ω over a manifold M is equal to the integral of ω over the boundary of M. M is a manifold, like a curve, surface, or region of space M is the boundary of M ω is (sort of) a function M dω is the derivative of ω (whatever that means) This is sometimes called the true tokes theorem. ω

16 (31) True tokes theorem Most of the essential theorems we ve learned in calculus so far are just special cases of the true tokes theorem: FToC: integration over an interval [a, b] on the real number line, so the boundary is two endpoints {a, b}. ˆ f (t) dt f(b) f(a). [a,b] The right-hand side is actually the integral of f (without dt) over this finite set of two points: [a,b] f. The minus sign is there to preserve orientation. FT of line integrals: integration over a curve C, so the boundary is the endpoints of the curve. ˆ f dr f(r(b)) f(r(a)). Again, the right-hand side is C C f. (32) True tokes theorem Green s theorem: integration over a region in R 2, so the boundary is a simple closed curve. ( Q x P ) da F dr y (Classical) tokes theorem: integration over a surface in R 3, so the boundary is a closed curve. (curl F) d F dr ivergence theorem: Volume integral over a solid region E in R 3. The boundary of E is a closed surface E. (div F) dv F d E E

17 (33) The end To learn about vector calculus in more detail (e.g. divergence, curl, surface integrals, triple integrals), take MATH 425 (Advanced Multivariate Calculus) To learn more about curvature and the geometry of curves and surfaces, take MATH 563 (ifferential Geometry) To learn more about the theory of limits, derivatives, and integrals, take MATH 523 (Introduction to Modern Analysis) Also, you ve reached the end of the math you can do without learning Linear Algebra (MATH 235, MATH 545) (34) Final exam topics Will be on the final? YE: ouble integrals with polar coordinates Vector fields Line integrals Green s theorem Everything from the first two exams (potentially) NO: Triple integrals Curl, divergence Parametric surfaces urface integrals tokes theorem ivergence theorem

18 (35) Homework Paper Homework #23 due Tuesday Homework 16.4 due Wednesday night Final exam: Monday 5/7, 8:-1: AM, Boyden gym. We ll use Tuesday s class to review for the final I ll also schedule some extra office hours next week I posted an answer key for the practice problems on my website Tutoring center (LGRT 14) is open through Tuesday 5/1