Summary of various integrals

Transcription

1 ummary of various integrals Here s an arbitrary compilation of information about integrals Moisés made on a cold ecember night. 1 General things o not mix scalars and vectors! In particular ome integrals are integrals of functions, and some are integrals of vector fields. on t interchange them. You wil never ever have to take an integral of a vector field, such as 1 0 (1, x2 ) dx or 1 i + x 2 j dx 0 (these don t make sense!!!). 2 Integrals on curves 2.1 Integral of a function on a curve Used for questions such as integral of a function on a curve, average of a function on a curve, center of mass on a curve. For functions only! Makes sense in 2 or 3 if you follow the blue instructions. The formula: If is a curve and f is a function, the integral of f on is f ds = t1 f( x(t)) x (t) dt x(t) is a parametrization, where the bounds are t t 1. f( x(t)) means: if f is a function of x and y (and z), such as x + sin y z 2, plug the parametrization x(t) into x and y (and z). x (t) is the velocity vector of the curve (i.e. the derivative). We take its length. Orientation doesn t matter. 2.2 Integral of a vector field along a curve Used for questions such as integral of a vector field along a curve, work, circulation. Makes sense in 2 or 3 if you follow the blue instructions. The formula: If is a curve and F = (P, Q, R) is a vector field, the integral of F along is F T ds = t1 F ( x(t)) x (t) dt 1

2 x(t) is a parametrization, where the bounds are t t 1. f( x(t)) means: if P, Q, R are functions of x and y (and z), such as x+sin y z 2, plug the parametrization x(t) into x and y (and z). x (t) is the velocity vector of the curve (i.e. the derivative). We dot it with F to obtain a scalar. Another formula computing the exact same thing: If is a curve and F = (P, Q, R) is a vector field, the integral of F along is F T t1 ds = P dx + Q dy+r dz x(t) is a parametrization, where the bounds are t t 1. P, Q, R are given in terms of x, y, z. Everywhere where we see thees variables x, y, z we need to plug in x(t). We interpret df(t) as f (t) dt. For example, d(t 2 ) = 2t dt and d sin t = cos t dt. Orientation changes the sign. If the curve is closed in 2: Green s theorem. If goes counterclockwise and is its interior, P dx + Q dy = Q x P y da If the curve is closed in 3: tokes theorem. If goes is the boundary of a surface, with a normal vector compatible with the orientation of, P dx + Q dy + R dz = R y Q z, P z R x, Q x P y N da If the field is conservative: Then it is the gradient of a function, F = f. In this case, F T ds = f(end Point) f(tart Point) 2.3 Flux of a vector field across a curve Used for questions such as flux through/into/out of a curve, how much fluid flows through curve. Makes sense in 2 only. The formula: If is a curve and F = (P, Q) is a vector field, the flux of F across is F N t1 ds = F ( x(t)) x (t) dt x(t) is a parametrization, where the bounds are t t 1. F ( x(t)) means: if P, Q are functions of x and y, such as x + sin y, plug the parametrization x(t) into x and y. x (t) is the velocity vector of the curve (i.e. the derivative). x (t) is my way of writing x (t) rotated 90. To rotate it 90, we interchange the coordinates and change one of the signs, so (a, b) becomes either (b, a) or ( b, a). We dot it with F to obtain a scalar. There s two choices of perpendicular vector and only one works. You need to figure out which one using which flux you want to compute (in? out? up? left?). You can draw a picture and plug in a particular value of t to figure this out. 2

3 Another formula computing the exact same thing: If is an oriented curve and F = (P, Q) is a vector field, the integral of F along going to the right of is the following F N ds = t1 P dy Q dx if has a direction, then it makes sense so talk about the right : for example if goes counterclockwise, then the right points out x(t) is a parametrization, where the bounds are t t 1. P, Q are given in terms of x, y. Everywhere where we see thees variables x, y we need to plug in x(t). We interpret df(t) as f (t) dt. For example, d(t 2 ) = 2t dt and d sin t = cos t dt. Orientation doesn t change the sign. However, change of the normal direction does! If you follow the second method, changing orientation changes the normal direction, so it does change the sign. If the curve is closed: Green s theorem for flux. If is closed and is its interior, F N Out! ds = P x + Q y da 3 Integrals on surfaces 3.1 Integral of a function on a surface Used for questions such as integral of a function on a surface, average of a function on a surface, center of mass on a surface. For functions only! Makes sense in 3, because we don t really have surfaces in 2. The formula: If is a surface and f is a function, the integral of f on is f ds = f( x(u, v)) x u x v du dv Bounds x(u, v) is a parametrization of, and it comes with some bounds that we make into the integral bounds. f( x(u, v)) means: if f is a function of x, y and z, such as x + sin y z 2, plug the parametrization x(u, v) into x, y and z. x u and x v are the derivatives of the parametrization with respect to u and v. We need to take the cross product of these two vectors. Then we take its length to get a number. 3.2 Flux of a vector field across a surface Used for questions such as flux through/into/out of a surface, how much fluid flows through surface. Makes sense in 3 only, because we don t really have surfaces in 2. 3

4 The formula: If is a curve and F = (P, Q, R) is a vector field, the flux of F across is F N da = F ( x(u, v)) ( x u x v ) du dv Bounds x(u, v) is a parametrization of, and it comes with some bounds that we make into the integral bounds. F ( x(u, v)) means: if P, Q, R are functions of x, y and z, such as x+sin y z 2, plug the parametrization x(u, v) into x, y and z. x u and x v are the derivatives of the parametrization with respect to u and v. We need to take the cross product of these two vectors. Then we dot it with F to get a number. There s two choices of perpendicular vector and only one works. You need to figure out which one using which flux you want to compute (in? out? up? left?). You can draw a picture and plug in a particular value of u, v to figure this out. hange of the vector x u x v changes the sign!! If the surface is closed: divergence theorem. If is closed and is its interior, F N da = divf dv The divergence has the formula div P, Q, R = P, Q, R = P x + Q y + R z If the vector field is the curl of something: tokes Theorem. This means we can write F as F = R y Q z, P z R x, Q x P y = P, Q, R This is not very useful unless we can find P, Q, R. If goes is the boundary of a surface, with a normal vector compatible with the orientation of, P dx + Q dy + R dz = R y Q z, P z R x, Q x P y N da 4 Useful parametrizations 4.1 urves Line segments: ircles: The line segment from A to B (in 2 or 3) can always be parametrized as x(t) = (1 t)a + tb : 0 t 1 You can use any other parametrization as long as it s a linear function and x(0) = A and x(1) = B. This works in 2 and 3! The counterclockwise unit circle is given by: x(t) = cos t, sin t The counterclockwise circle of radius R is given by: x(t) = R cos t, R sin t 4

5 (Not as useful) The counterclockwise circle of radius R centered at (a, b) is given by: x(t) = a + R cos t, b + R sin t = a, b + R cos t, R sin t Graphs of functions: The graph of the function f is {y = f(x)}, and if x is between x 0 and x 1, it can be parametrized as x(t) = t, f(t) : x 0 t x 1 For example, the graph of f(x) = e x 2 with x between 1 and 3 is 4.2 urfaces pheres and pieces of spheres. coordinates: x(t) = t, e t 2 : 1 t 3 A whole sphere of radius R can be parametrized by using spherical x(θ, φ) = R cos θ sin φ R sin θ sin φ R cos φ : 0 θ 2π 0 φ π This is the whole sphere. A domain in the sphere can be parametrized changing the bounds. x θ x φ = R sin φ x Notice that left and right are both vectors. This normal vector points into the sphere. x θ x φ = R 2 sin φ ylinders and pieces of cylinders. A cylinder of radius R can be parametrized by using cylindrical coordinates: R cos θ 0 θ 2π x(θ, z) = R sin θ :? z? z This is the whole cylinder, and the bounds for z depend on the heights at which the cylinder starts and ends. x θ x z = R cos θ, R sin θ, 0 Notice that left and right are both vectors. This normal vector points out of the cylinder. Graphs of functions: parametrized as x θ x z = R The graph of the function f of two variables is {z = f(x, y)}, and it can be x(u, v) = u, v, f(u, v) The bounds for u and v will be the bounds we have for x and y. For example, the graph of f(x, y) = ye x 2 with x between 1 and 3 and y between 0 and 6 is On the graph of f(u, v), x(u, v) = u, v, ve u 2 : 1 u 3, 0 v 6 x u x v = f u, f v, 1 Notice that left and right are both vectors. This normal vector points up. x u x v = f 2 u + f 2 v + 1 5