Vector Calculus handout The Fundamental Theorem of Line Integrals Theorem 1 (The Fundamental Theorem of Line Integrals). Let C be a smooth curve given by a vector function r(t), where a t b, and let f be a differentiable function of two or three variables whose gradient vector f is continuous on C. Then f d r = f ( r(b)) f ( r(a)). C Note that the answer doesn t depend on the path, only the endpoints! o, for any two paths C 1 and C 2 having the same initial point and same terminal point, f d r = f d r. C 1 C 1 This leads to... efinition 1. Let F be a continuous vector field with domain. The line integral C F d r is said to be independent of path (or path independent if C 1 F d r = C 1 F d r for any two paths C1 and C 2 in that have the same initial point and same terminal point. efinition 2. A curve C is said to be closed if its initial point and terminal point are the same. Theorem 2. C F d r is path independent in if and only if C F d r = 0 for every closed path C in. Recall that a vector field is conservative if there exists a scalar function f such that f = F. Theorem 3. Let F be a continuous vector field on an open connected region. F d r is path independent if and only if F is conservative. C Then the line integral Why do we care? If F is conservative, then that can make evaluating C F d r much easier (if C is closed we get 0 immediately, if not then we could still choose a nicer path). But it would be nice to have some way to check to see if F is conservative without going back to the definition: Theorem 4. Let F = P î + Q ĵ be a vector field on an open, simply-connected region such that P and Q have continuous first-order partial derivatives. Then F is conservative if and only if on. P y = Q x 1
Curl and ivergence For three-dimensional vector fields, we need another way to check for a conservative vector field. Recall the del operator that we defined when we introduced the gradient: = x î + y ĵ + z ˆk = x, y, z We use this to define the curl and divergence of a three-dimensional vector field F = P î + Q ĵ + R ˆk: curl F = F î ĵ ˆk = x y z P Q R div F = F Theorem 5. Let F = P î + Q ĵ + R ˆk be a vector field on an open, simply-connected region such that P, Q, and R have continuous first-order partial derivatives. Then F is conservative if and only if curl F = 0 on. Example 1. etermine whether or not the following vector fields are conservative. If so, find f(x, y) such that f = F. (a) F (x, y) = y, x (b) F (x, y) = y, 1 (c) F (x, y) = y 2, 2xy 2
(d) F (x, y) = y 3 + 1, 3xy 2 + 1 (e) F (x, y) = 2xye x2y, x 2 e x2 y (f) F (x, y, z) = sin(y), x cos(y), 1 (g) F (x, y, z) = 2xy, x 2 + 2yz, y 2 3
We can combine these definitions and theorems as follows: Theorem 6. Let F be a continuous vector field with continuous first partial derivatives in an open connected region, and let C be a [piecewise] smooth curve in given by r(t). The following are equivalent: 1. F is conservative. 2. P y = Q x (for F = P î + Q ĵ) or curl F = 0 (for F = P î + Q ĵ + R ˆk) on. 3. There exists f such that f = F (which then allows us to use the Fundamental Theorem). 4. C F d r is independent of path. 5. C F d r = 0 for every closed curve C in. Example 2. Evaluate C F d r, where F = y 3 + 1, 3xy 2 + 1 and C is the semicircular path from (0, 0) to (2, 0) given by r(t) = 1 cos(t), sin(t), 0 t π. Example 3. Evaluate C F d r, where F = y 2, 2xy and C is the parabolic path from (4, 0) to (1, 3) given by r(t) = 4 t, 4t t 2, 0 t 3. 4
Example 4. Evaluate C F d r, where F = 2xy, x 2 + 2yz, y 2 and C is the path given by r(t) = t 2, t cos(t), e t sin(t), 0 t 3π. Green s Theorem Green s Theorem relates a line integral around a simple closed curve C to a double integral over the region in the plane bounded by C: Theorem 7 (Green s Theorem). Let C be a positively oriented, piecewise-smooth, simple, closed curve in R 2 and let be the region bounded by C, and let F = P î + Q ĵ be a continuous vector field such that P and Q have continuous partial derivatives on an open region containing. Then ( ) F d r = P dx + Q dy C C ( Q = x P ) da y = ( curl F ) ˆk da 5
Example 5. Evaluate C x2 y 2 dx + xy dy where C is the positively oriented curve consisting of the arc of the parabola y = x 2 from (0, 0) to (1, 1) and the line segments from (1, 1) to (0, 1) and from (0, 1) to (0, 0). Parametric urfaces Given a vector valued or parametric function in one variable, say t, we can trace out a curve in R 2 or R 3 : r(t) = x(t) î + y(t) ĵ[+z(t) ˆk]. But with two parameters we can trace out a surface (in R 3 ): r(u, v) = x(u, v) î + y(u, v) ĵ + z(u, v) ˆk. Given a function z = f(x, y), one way to get a parameterization of the surface is to simply let u = x, v = y, and then let z = f(u, v) (similar construction for surfaces defined by y = f(x, z) or x = f(y, z)). Tangent Planes to Parametric urfaces Recall that to find the equation of a plane, we need to know a normal vector to the plane n = a, b, c and a point in the plane (x 0, y 0, z 0 ). Then the equation of the plane is a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. Now, given a parameterization r(u, v) = x(u, v) î + y(u, v) ĵ + z(u, v) ˆk of a surface, to find an equation of a tangent plane we still need a normal vector and a point. The point will either be given as (x 0, y 0, z 0 ), or we will be given (u 0, v 0 ) which we ll then plug into r to find the coordinates of the point. Our normal vector is n = r u (u 0, v 0 ) r v (u 0, v 0 ) (or, for this type of problem, we could also use r u (u 0, v 0 ) r v (u 0, v 0 )) where r u = x u î + y u ĵ + z u ˆk and r v = x v î + y v ĵ + z v ˆk Example 6. Find the equation of the plane tangent to the surface at the point (1, 1). r(u, v) = u î + v ĵ + uv ˆk 6
Example 7. Find the equation of the plane tangent to the surface at the point (0, 6, 4). r(u, v) = 3u cos(v) î + 3u sin(v) ĵ + u 2 ˆk urface Area of Parametric urfaces Recall that the area of a parallelogram with sides given by a and b is a b. This leads to the following: Theorem 8. Let is a smooth parametric surface given by r(u, v) = x(u, v) î + y(u, v) ĵ + z(u, v) ˆk, (u, v), such that is covered just once as (u, v) ranges throughout. Then the surface area of is A() = r u r v da If we have a surface given by z = f(x, y) and view x and y as the parameters, then we have r(x, y) = x, y, f(x, y) which means r x = 1, 0, f x and r y = 0, 1, f y, so î ĵ ˆk r u r v = 1 0 f x 0 1 f y = f x, f y, 1 and r u r v = f x 2 + f y 2 + 1 Thus the surface area is A() = f x 2 + f y 2 + 1 da (imilar formulae if the surface is given by y = f(x, z) or x = f(y, z).) 7
Example 8. Find the surface area of for 0 u 1, 0 v 2π. r(u, v) = 2u cos(v) î + 2u sin(v) ĵ + u 2 ˆk Example 9. Find the surface area of the portion of the hemisphere z = 25 x 2 y 2 that lies inside the cylinder x 2 + y 2 = 9. 8
urface Integrals Previously, we had that the line integral of f(x, y, z) along a curve C given by r(t) = x(t), y(t), z(t), a t b, is b (dx ) 2 ( ) 2 ( ) 2 dy dz b f (x(t), y(t), z(t)) + + dt = or f ( r(t)) r (t) dt dt dt dt a Now we extend this definition to define the surface integral of f(x, y, z) over a surface given by r(u, v) = x(u, v) î + y(u, v) ĵ + z(u, v) ˆk where (u, v), as f(x, y, z) d = f ( r(u, v)) r u r v da Example 10. Evaluate the surface integral xy d where is given by r(u, v) = 2 cos(u) î+2 sin(u) ĵ+v ˆk with 0 u π 2 and 0 v 2. a If is a surface given by a function z = g(x, y) then we can let x and y be the parameters and find r x r y as we did previously to get f(x, y, z) d = f (x, y, g(x, y)) (g x ) 2 + (g y ) 2 + 1 2 da (imilar formulae if the surface is given by y = g(x, z) or x = g(y, z).) Example 11. Evaluate ( x 2 + y 2 + z 2) d where is the portion of the plane z = x+2 that is bounded by x = 1, x = 1, y = 0, and y = 2. 9
Orientable urfaces Orientable surfaces have two sides, and we can distinguish between the two based on which way a normal vector is pointing. For a closed surface (such as a sphere, cube, or any region E over which we evaluated a triple integral) we can have a normal vector pointing outward from the surface (this is called positive orientation) or it can point inward (this is called negative orientation). ome surfaces are non-orientable, such as a Möbius strip. Now that we have the notion of orientability, we can talk about... urface Integrals Over Oriented urfaces Let F = P î + Q ĵ + R ˆk be a continuous vector field, defined on an oriented surface with unit normal vector n. The surface integral of F over is F d = F n d = F ( r u r v ) da If is given by the graph of a function z = g(x, y), then F g x, g y, 1 da = ( P g x Q g y + R) da (oriented upward) F d = F g x, g y, 1 da = (P g x + Q g y R) da (oriented downward) where is the projection of z = g(x, y) onto the xy-plane. Example 12. Evaluate the surface integral F d, where F = x, y, z and is the part of the paraboloid z = 4 x 2 y 2 that lies above the xy-plane, and has upward orientation. Example 13. Evaluate the surface integral F d, where F = 3z, 4, y and is the part of the plane x + y + z = 1 in the first octant, with downward orientation. 10