Chapter 17. Review. 1. Vector Fields (Section 17.1)

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1 hapter 17 Review 1. Vector Fields (Section 17.1) There isn t much I can say in this section. Most of the material has to do with sketching vector fields. Please provide some explanation to support your answers. I will talk more about this in the class. 2. Line Integrals (Section 17.2) (i) Line integral with respect to s: A line integral of a function is simply an integral evaluated along a given curve. A normal integral is typical evaluated over an interval. onsider a parameterized curve in R 2 given by r(t) = x(t), y(t), a t b and a function f defined on the curve. Then, the line integral of f along is given by f(x, y)ds = b Remember that s is the arc length parameter. a (dx ) 2 f(x(t), y(t)) + dt ( ) dy 2 dt. (2.1) dt (ii) Geometric meaning. Given a curve in the xy plane, construct a strip in R 3 over this curve such that the height above each point on the curve corresponds to the value of the function at that point. The line integral is then the area of one side of such a strip. (iii) Line integral with respect to x and y. Instead of integrating with respect to the arc length parameter, we can find line integrals with respect to x and y as follows: f(x, y)dx = f(x, y)dy = b a b a f(x(t), y(t))x (t)dt (2.2) f(x(t), y(t))y (t)dt. (2.3) Important: As you can see, if you take the line integral with respect to x, there s an x (t) on the right hand side. Similarly, for the line integral with respect to y, we get y (t) on the right hand (dx ) 2 ( ) 2 side. Notice that we need dt + dy dt if we want the line integral with respect to s. Notation: We shall use the following notation P(x, y)dx + Q(x, y)dy = P(x, y)dx + Q(x, y)dy. 1

2 2 Review (iv) Strategy. If you are asked to find the line integral of a given function along a curve, you first find a parameterization of the curve (that is, express it with respect to a parameter t and find its bounds a and b) and then use one of the formulas (2.1), (2.2) or (2.3), depending on the given problem. (v) Parameterizing curves. Given a curve, there are a number of ways to parameterize it. Let s look at a few cases. (a) Line segments. Example: Evaluate the line integrals c f(x, y)ds, f(x, y)dx and c f(x, y)dy where f(x, y) = xy and is the line segment joining (1, 1) and (3, 4). There are at least a couple of different ways to parameterize the given line segment: From alc III we know that a vector equation of the line segment joining x, y and x 1, y 1 is given by x, y = (1 t) x, y + t x 1, y 1, t 1. Equating the x and y components respectively, we get a parameterization for the line segment. That is, x = (1 t)x + tx 1 y = (1 t)y + ty 1, t 1. So for our example, a parameterization is given by x = (1 t)+3t = 1+2t and y = 5t 1, t 1. We can also simply find the equation of the line joining the two points, which is y = 5 2 x 7 2. Then let x = t, so y = 5 2 t 7 2. This is a parameterization where 1 t 3. Both the above parameterizations are correct and you can use either one to evaluate your line integral. If we use the first parameterization, we get Hence xyds = 1 (dx ) 2 + dt ( dy dt dx dt = 2 dy dt = 5 (1 + 2t)(5t 1) 29dt = 29 ) 2 = = (1t 2 + 3t 1)dt =

3 hapter 17 3 Similarly, xydx = xydy = 1 1 (1 + 2t)(5t 1)(2)dt = 2 (1 + 2t)(5t 1)(5)dt = (1t 2 + 3t 1)dt = (1t 2 + 3t 1)dt = You can use the second parameterization also and the answer will be the same. I find that in many cases (such as this) having the lower bound of t being makes the calculation a little simpler but the choice is yours. (b) General curves. If a curve is given by y = f(x), you can simply take x = t and y = f(t). Similarly, if the curve is given by x = f(y), you can let y = t and x = f(t). The bounds on t can be found by looking at the starting point and the end point on the curve. Example: Find y3 dx + x 2 dy where is the arc of the parabola x = 1 y 2 joining (, 1) to (, 1). A parameterization for the curve is y = t, x = 1 t 2, 1 t 1. We have dx dt = 2t dy dt = 1. Hence, y 3 dx + x 3 dy = 1 1 [ 2t 5 = 5 t 3 ( 2t)dt + ] 1 1 = [ t t3 3 + t (1 t 2 ) 2 (1)dt (t 4 2t 2 + 1)dt ] 1 1 = = (vi) Piecewise-smooth curves: If a curve is the union of a finite number of piecewise smooth curves, that is, if = n, then f(x, y)ds = f(x, y)ds + f(x, y)ds +... f(x, y)ds. 1 2 n So you simply find the line integral along each piece of the curve (and each piece will have a different parameterization) and add the result to get the line integral along the complete curve.

4 4 Review Example: Evaluate y3 dx+x 2 dy where is the closed curve consisting of the line segment from (, 1) to (, 1) and the arc of the parabola x = 1 y 2 from (, 1) to (, 1). We have already calculated the line integral of the given function on the arc of the given parabola previously. Let 1 be the line segment joining (, 1) and (, 1) and 2 be the arc of the parabola given. A parameterization for 1 is given by x =, y = t, 1 t 1. onsequently, x (t) = and y (t) = 1. Hence y 3 dx + x 2 dy = 1 y 3 dx + 1 x 2 dy = (1)dt =. Hence y 3 dx + x 2 dy = + y 3 dx + x 2 dy = (vii) The same ideas are applicable in three dimensions. To evaluate the line integral of a function f of three variables along a curve in R 3 given by r(t) = x(t)i + y(t)j + z(t)k, a t b we have f(x, y, z)ds = b a (dx ) 2 f(x(t), y(t), z(t)) + dt ( ) dy 2 + dt ( ) dz 2 dt dt Similarly, f(x, y, z)dx = b a f(x(t), y(t), z(t))x (t)dt, etc. (viii) Line integrals of vector fields. The motivation here comes from Physics. A force F is really a vector field and the line integral of F along a curve is interpreted as the work done by the force in moving a particle along the curve. Let F be a vector field (on R 2 or R 3 ) along a curve given by r(t) (in R 2 or R 3 ), a t b. Then, the line integral of F along is given by F dr = b a F(r(t)) r (t)dt. We can relate this to a line integral of a function as follows. Let T = r (t) r (t) be the unit tangent vector to the curve at a point. Then F dr = F Tds.

5 hapter 17 5 Important: Given a vector field F = Pi + Qj along a curve in R 2, we have F dr = Pdx + Qdy. Similarly, for a vector field F = Pi + Qj + Rk and a curve in R 3, we have F r = Pdx + Qdy + Rdz. Example Evaluate F dr where F(x, y, z) = ez i + xzj + (x + y)k and is given by r(t) = ti + t 3 j + tk, t 1. Solution: We first evaluate Also, r (t) = i + 3t 2 j + k and Therefore, F(r(t)) = e t i + t 2 j + (t + t 3 )k. F(r(t)) r (t) = e t + 3t 4 + t + t 3. F r = 1 (e t + 3t 4 + t + t 3 )dt = e Fundamental Theorem for line integrals (Section 17.3) (i) Fundamental Theorem for line integrals. Let be a smooth curve given by the vector function r(t), a t b and let f be a differentiable function of two or three variables whose gradient vector f is continuous on. Then f dr = f(r(b)) f(r(a)). (ii) A line integral is independent of path if the value of the integral remains the same along all curves that have the same end points. (iii) A vector field F is conservative if we can find a function f such that F = f. The line integral of a conservative vector field along a curve is independent of path. This means that the line integral depends only on the end points of the curve. This follows from the Fundamental Theorem for line integrals. (iv) A curve r(t), a t b is called closed if its terminal point coincides with its initial point, that is, r(b) = r(a).

6 6 Review (v) Theorem: F dr is independent of path in (a region) if and only if F dr = for every closed path in. The fundamental theorem for line integrals tells us that the line integral of a conservative vector field along a closed curve is zero. (vi) efinitions: A region in the plane is open if it contains no boundary points. We say that is connected if any two points in can be joined by a path lying entirely in. (vii) Theorem: Suppose that F is a vector field that is continuous on an open connected region. If F dr is independent of path in, then F is conservative vector field on. (viii) Theorem: If F(x, y) = P(x, y)i + Q(x, y)j is a conservative vector field, where P and Q have continuous first-order partial derivatives on a domain, then on we have P y = Q x. (ix) efinition: A curve is called simple if it doesn t intersect itself between the end points. A simplyconnected region is a connected region such that every simple closed curve in encloses only points in. A simply-connected region has no holes. (x) Theorem: Let F = Pi + Qj be a vector field continuous on an open, simply-connected region. If P and Q have continuous first-order derivatives and on. Then, F is conservative. P y = Q x Example: etermine whether the vector field F = (2xe xy +x 2 ye xy )i+(x 3 e xy +2y)j is conservative or not. Solution: Let P(x, y) = 2xe xy + x 2 ye xy and Q(x, y) = x 3 e xy + 2y. We have Since P y = Q x P y = 2x2 e xy + x 2 e xy + x 3 ye xy = 3x 2 e xy + x 3 ye xy Q x = 3x2 e xy + x 3 ye xy., the vector field is conservative. (xi) Method for finding f such that F = f: Once we determine that F is conservative, it means that there exists a function f such that F = f. We can find this by partial integration. In our previous example, we let f be a function such that f x = 2xe xy + x 2 ye xy (3.1) f y = x 3 e xy + 2y. (3.2)

7 hapter 17 7 Now, we can solve this using two equivalent ways. We start by integrating each side of (3.1) with respect to x to get f(x, y) = (2xe xy + x 2 ye xy )dx + g(y) where g is a function of y. Now, you can see that we encounter integration by parts on the righthand side here. We could potentially carry on and find g in the end but it looks cumbersome. Start with (3.2) instead and integrate each side with respect to y to get f(x, y) = (x 3 e xy + 2y)dy = x 2 e xy + y 2 + h(x) (3.3) where h is a function of x. Next, taking the partial derivative of each side with respect to x, we get Now, equating this with the right-hand side of (3.1), we get f x = 2xe xy + x 2 ye xy + h (x). (3.4) h (x) = hence h(x) = c where c is a constant. Substituting back in (3.3), we get which is what we wanted to find. f(x, y) = x 2 e xy + y 2 + c (3.5) (xii) Using the fundamental theorem for line integrals on conservative vector fields. If we show that a vector field is conservative, the line integral becomes easy to evaluate. Example. Find F dr where F = (2xexy + x 2 ye xy )i + (x 3 e xy + 2y)j and is the curve given by r(t) = t 3 i t 4 j, 1 t 2. Here, we have r(1) = i j and r(2) = 8i 16j. Since we showed earlier that F is conservative, we have F dr = f(r(2)) f(r(1)) = (8) 2 e (8)(16) + ( 16) 2 (1) 2 e 1 1 = 64e 128 e where we have used the function f from (3.5). Example: Show that the line integral is independent of path and evaluate the integral. (1 ye x )dx + e x dy

8 8 Review where is any path from (, 1) to (1, 2). Solution: Let P(x, y) = 1 ye x and Q(x, y) = e x. Then, we have P y = e x = Q x and thus the vector field Pi + Qj is conservative and the given integral is path independent. This also means that there is a function f such that f x = (1 ye x ) and f y = e x. Integrating each side of the first equality with respect to x, we get f(x, y) = x + ye x + g(y) where g is a function of y. Taking partial derivative of each side with respect to y, we get f y = e x + g (y) and comparing with above. we get g (y) = and thus g(y) = c. This gives us f(x, y) = x + ye x + c. Also, by Fundamental Theorem for line integrals. we have (1 ye x )dx + e x dy = f(1, 2) f(, 1) = 1 + 2e 1 1 = 2e Green s Theorem (Section 17.4) (i) Orientation of a curve. A simple, closed curve in R 2 encloses a planer region. We say that has positive orientation if the enclosed region is always to the left as the curve is traversed. If is simply-connected, then positive orientation corresponds to a counterclockwise traversal of the curve. (ii) Green s Theorem. Let be a positively oriented, piecewise-smooth, simple closed curve in the plane and let be the region bounded by. If P and Q have continuous first-order partial derivatives, then ( Q Pdx + Qdy = x P ) da y (iii) Explanation. Green s Theorem gives a relationship between the line integral along a simple closed curve and the double integral on the enclosed region. This can often help simply line integral calculations. (iv) Notation. We often write Green s Theorem as Pdx + Qdy = ( Q x P ) da y to indicate that the line integral is being taken over a positively oriented closed curve. Since is the boundary of the region, this theorem is sometimes written as ( Q x P ) da = Pdx + Qdy y

9 hapter 17 9 where denotes the boundary curve of. This version is similar in flavor to the Fundamental Theorem of alculus. (v) Note that Green s Theorem is not very useful if F is conservative. This is because we already know that for a conservative vector field F, we have F dr = if is a closed curve. (vi) Example: Use Green s Theorem to evaluate 1 + x 3 dx + 2xydy, where is the triangle with vertices (, ), (1, ) and (1, 3). Solution: We let P(x, y) = 1 + x 3 and Q(x, y) = 2xy, thus Q x P y = 2y. Now, the enclosed region is given by = {(x, y) x 1, y 3x} which is a Type I region. Green s Theorem says that Pdx + Qdy = ( Q x P y ) da. Thus, 1 + x 3 dx + 2xydy = 2ydA = 1 3x 2ydydx = 1 9x 2 dx = 3. (vii) Green s Theorem can be extended to the case where the enclosed region is not simply-connected. Example: Evaluate y3 dx x 3 dy where are the two circles of radius 2 and radius 1 centered at the origin with positive orientation. Solution: Let P(x, y) = y 3 and Q(x, y) = x 3. Then, Q x = 3x2 and P y = 3y2. By Green s Theorem, we get y 3 dx x 3 ( dy = 3x 2 3y 2) da. The region enclosed by is the ring between the two disks. We can write it in polar coordinates as Thus, switching to polar coordinates, we get = {(r, θ) 1 r 2, θ 2π}. y 3 dx x 3 dy = 2π 2 1 ( 3r 2 )rdrdθ = 45π 2.

10 1 Review 5. url and ivergence (Section 17.5) (i) So far we only know how to check whether a vector field F = Pi + Qj on R 3 is conservative. We next define an operator that enables us to check for conservative vector fields on R 3. (ii) el/gradient operator. We define Given a function f of three variables, we have ( ) f f = i + x which is exactly the gradient of f. = i x + j y + k z. ( f y ) ( ) j + k. z (iii) url. Let F = Pi + Qj + Rk, where the partial derivatives of P, Q and R exist. The url of F is defined as url(f) = F. We can write this in terms of the following symbolic determinant i j k url(f) = x y z P Q R. (iv) Theorem: If f is a function of three variables that has continuous partial derivatives, then url( f) =. This implies that the curl of a conservative vector field is zero. The following theorem states that the converse is true if the vector field is defined on all of R 3. (v) Theorem: If F is a vector field defined on all of R 3 and its components have continuous partial derivatives and url(f) =, then F is conservative. Example. etermine whether the vector field F(x, y, z) = (siny)i + (xcos y)j (sin z)k is conservative or not. Solution: We have url(f) = i j k x y z (sin y) (xcos y) sinz = i + j + (cos y cos y)k = hence, the vector field is conservative.

11 hapter (vi) ivergence. If F = Pi + Qj + Rk is a vector field such that the first order partial derivatives of the components exist, then we define the divergence of F as div(f) = P x + Q y + R z. In other words, div(f) = F. Example: Given F = i + (x + yz)j + (xy z)k, find div(f). Solution: This is straightforward. div(f) = + z 1 2 z. (vii) Theorem: If F = Pi + Qj + Rk is a vector field on R 3 and the components have continuous second-order partial derivatives, then div(url(f)) =. (viii) Vector form of Green s Theorem. Note that Green s Theorem as stated in the class works only for vector fields on R 2 but we can express it in vector form using the curl as follows. Let s start with a vector field F(x, y) = P(x, y)i + Q(x, y)j in R 2. We can express is as a vector field F(x, y, z) = P(x, y)i + Q(x, y)j + k on R 3. Then, Hence, we can write i j k url(f) = x y P(x, y) Q(x, y) F dr = The above gives us a vector form of Green s Theorem. z ( Q = i + j + x P ) k. y (url(f) k)da There is another way to write this theorem in vector form. Let n represent the unit normal vector to the curve. Then, we have F n ds = div(f(x, y))da. 6. Parametric surfaces and their areas (Section 17.6) (i) efinition: Just like a curve in R 3 can be parameterized by a single parameter t, a surface in R 3 can be parameterized by two parameters. A parametric surface is described by a vector function r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k

12 12 Review where (u, v) lie in a planar region. This can also be expressed as Example: The parametric surface given by x = x(u, v), y = y(u, v) z = z(u, v), (u, v). r(u, v) = 2 cos ui + vj + 2 sinuk is a cylinder symmetric with respect to the y axis since x = 2 cos u, y = v and z = 2 sinu implies that x 2 + z 2 = 4 and y = v. Example: Identify the surface with the vector equation r(u, v) = 2 sin ui + 3 cos uj + vk, v 2. Solution: We have x = 2 sin u, y = 3 cos u and z = v. Thus, the surface is given by x y2 9 = 1, z 2. This is the portion of elliptic cylinder lying between the planes z = and z = 2. Example: Find a parametric representation for the lower half of the ellipsoid 2x 2 + 4y 2 + z 2 = 1. Solution: The ellipsoid is given by z = ± 1 2x 2 4y 2. Since the surface is the lower half of the ellipsoid, it is represented by z = 1 2x 3 4y 2. Hence, a parameterization is given by x = u, y = v, z = 1 2u 2 3v 2. (ii) Tangent planes. To find the tangent plane to a parametric surface given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) at a given point P, we find the vectors r u and r v as follows: ( ) ( ) ( ) x y z r u = i + j + k u u u ( ) ( ) ( ) x y z r v = i + j + k v v v Then, n = r u r v is a normal vector to the given plane. This, along with the given point P through which the plane passes enables us to write down an equation for the tangent plane. Example: Find an equation for the tangent plane to the given parametric surface at the point (1,, 1). r(u, v) = u 2 i + 2u sinvj + u cos vk.

13 hapter Solution: We have r u = 2ui + 2 sin vj + cos vk r v = i + 2u cos vj + u sinvk. Thus, n = r u r v = 2ui + 2u 2 sinvj + 4u 2 cos vk At the given point (1,, 1), we must have 1 = u 2, = 2u sinv and 1 = u cos v, from which we get u = 1, v = and thus n = 2i + 4k. An equation of the tangent plane is given by or 2x + 4z 2 =. 2(x 1) + (y ) + 4(z 1) = (iii) Surface area. If a smooth parametric surface S is given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) and S is covered just once as (u, v) ranges through the domain then the surface area of S is A(S) = r u r v da. Example: Fins the surface area of the part of the plane 3x+2y+z = 6 that lies in the first octant. Solution: The portion of the plane in the first octant has vertices (,, 6), (2,, ) and (, 3, ). Let be the projection of this portion on the xy-plane. Then, = {(x, y) x 2, y 3 2 x + 3}. Now, it is easy to write a parameterization for the given surface. We have In vector form this is represented by Hence, we calculate and x = u, y = v, z = 6 3u 2v, (u, v) r(u, v) = ui + vj + (6 3u 2v)k, (u, v). r u = i + j 3k = i 3k r v = j 2k r u r v = 3i + 2j + k

14 14 Review Thus, Finally, by the above formula, A(S) = r u r v = = dA = x 2 +3 dydx = (iv) If the surface S is given by z = f(x, y), then we can write a parameterization as x = u, y = v and z = f(u, v). That is, r(u, v) = ui + vj + f(u, v)k. Then, Hence, and r u = i + j + f u k r v = i + j + f v k. r u r v = f u i f v j + k r u r v = 1 + (f u ) 2 + (f v ) 2. This means that, for a surface S given by z = f(x, y), we have A(S) = 1 + (fu ) 2 + (f v ) 2 da. Example: Find the surface area of the surface z = xy that lies within the cylinder x 2 + y 2 = 1. Solution: Since the surface is the portion of z = xy inside the cylinder x 2 + y 2 = 1, we know that its projection on the xy plane is given by the disk of radius 1 centered at the origin. We thus have z = uv, (u, v). Here, let f(u, v) = uv so f u = v and f v = u. Using the formula above, we get A(S) = 1 + v 2 + u 2 da switching to polar coordinates, we get which you can easily calculate. A(S) = 2π 1 (1 + r 2 )rdrdθ

15 hapter Surface integrals (Section 17.7) (i) The idea. Previously, we have taken double integrals over rectangles and other general planar regions. We can take integrals over parametric surfaces as well. Here, instead of integrating with respect to the da, we take the integral with respect to an infinitesimal patch on S. (ii) efinition. Given a parametric surface S described by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) and let f be a function of three variables whose domain contains S. Then, the surface integral of f over S is given by f(x, y, z)ds = f(r(u, v)) r u r v da. S Example: Find the surface integral S yzds where S is the part of the plane x + y + z = 1 lying in the first octant. Solution: Let f(x, y, z) = yz. We write a parameterization for S first. We ve already done a similar problem earlier. The portion of the plane in the first octant has vertices (1,, ), (, 1, ) and (,, 1). Then, the projection onto the xy-plan is the region given by = {(x, y) = x 1, y x + 1}. A parameterization for the surface is x = u, y = v, z = 1 u v, (u, v), or r(u, v) = ui + vj + (1 u v)k, (u, v). We calculate Also, Hence, f(r(u, v)) = v(1 u v) = v uv v 2. r u = i + j k r v = i + j k. r u r v = i + j + k and r u r v = 3. The surface integral is given by S yzds = (v uv v 2 ) 3dA = 1 u+1 (v uv v 2 )dvdu and you can easily find the double integral on the right-hand side to get the final answer.

16 16 Review (iii) Oriented surfaces. If it is possible to choose a unit normal vector n at every point (x, y, z) on a given surface S in a continuous manner, then S is called an oriented surface. An example of a non-orientable surface is the Mobius strip. (iv) Surface integrals of vector fields. If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is defined by F ds = F n ds. S Using the definition of the surface integral on the right-hand side, and the fact that n = ru rv r, we u r v also have F ds = F (r u r v )da. S S Example: Evaluate the surface integral S F ds for the given vector field F and oriented surface S. S F(x, y, z) = xyi + yzj + zxk where S is the part of the paraboloid z = 4 x 2 y 2 that lies above the square x 1, y 1. Solution: We first write a parameterization for S as follows x = u, y = v, z = 4 u 2 v 2, (u, v) where = {(u, v) u 1, v 1}. In vector form, this is given by Next, we compute and thus Furthermore, and Hence, r(u, v) = ui + vj + (4 u 2 v 2 )k, (u, v). r u = i + j 2uk r v = i + j 2vk r u r v = 2ui + 2vj + k. F(r(u, v)) = uvi + v(4 u 2 v 2 )j + u(4 u 2 v 2 )k F(r(u, v)) (r u r v ) = 2u 2 + 2v 2 (4 u 2 v 2 ) + u(4 u 2 v 2 ). S F ds = 1 1 ( 2u 2 + 2v 2 (4 u 2 v 2 ) + u(4 u 2 v 2 ) ) dudv and you can easily find the double integral on the right-hand side.

17 hapter Stokes Theorem (Section 17.8) (i) Stokes Theorem can be thought of a a generalization of Green s Theorem to R 3. It relates the surface integral over a given surface to a line integral over the boundary curve of the surface. (ii) A positively oriented surface S induces a positive orientation of the boundary curve. This means that if you traverse the curve in the positive direction around with your head pointing in the direction of the unit normal, then the surface will always be on your left. (iii) Statement. Let S be an oriented, piecewise-smooth surface bounded by a simple, closed, piecewisesmooth boundary curve with positive orientation on an open region in R 3 that contains S. Then F dr = url(f) ds. S

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