| Practice 3 Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.Problem 1 Please answer the following true or false. If false, explain why or provide a counter example. If true explain why. A. Let f(x,y,z) be a function with continuous second order partial derivatives and let F(x,y,z) be the gradient of f(x,y,z). If S is the ellipsoid x2 4 oriented outward, then  False Using the divergence theorem, we have  If gradf(x,y,z) = F, then divF = fxx + fyy + fzz There theorem would be true if the function was harmonic, however if it not harmonic. All bets are off. For example, if f(x,y,z) = 1/6 (x2 + y2 + z2) and divF = 1/6 (2 + 2 + 2) = 1 Hence the integral represents the volume of the ellipse which is certainly not zero. B. Let F(x,y) be a conservative vector field, then  True, The first line integral traces out the line segment from (0,1) to (1,0) and the second traces out the quarter-circle from (0,1) to (1,0). Notice that in the first integral  r1(t) = (1 - t)i + tj r1'(t) = -i + j and in the second integral r2(q) = (cos q)i + (sin q)j r1'(t) = (-sin q)i +(cos q) j By the fundamental theorem of line integral, the integral is independence of path, hence the two integrals are equal. Problem 2 Show that  for any closed surface S.  Where C is the boundary of the surface S. But since S is a closed surface, it has not boundary. Hence C is a curve of zero length and the right hand integral is zero. Problem 3 A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2). The current can by represented by the vector field F(x,y,z) = (2x + 2z)i + (1 - 3z)j + (2x - 3y + 5)k Find the total work done by the current. The important thing to not here is that  Since F is conservative, we can use the fundamental theorem of line integrals. We seek a potential function f. We have fx = 2x + 2z Integrating with respect to x gives f = x2 + 2xz + C(y,z) Now taking the derivative with respect to y gives fy = Cy(y,z) = 1 - 3z Integrating with respect to y gives C(y,z) = y - 3yz + C(z) so that f = x2 + 2xz + y - 3yz + C(z) Now we take the derivative with respect to z to get 2z - 3y + C'(z) = 2x - 3y + 5 so that C'(z) = 5 Integrate with respect to z to get C(z) = 5z Hence f(x,y,z) = x2 + 2xz + y - 3yz + 5z The fundamental theorem of line integrals gives that the integral is f(1.5,1,2) - f(0.2,0.1,0.3) = [(1.5)2 + 2(1.5)(2) + 1 - 3(1)(2) + 5(2)] - [(0.2)2 + 2(0.2)(0.3) + 0.1 - 3(0.1)(0.3) + 5(0.3)] = 11.58 Problem 4 Evaluate  where F is the vector field  and S is the rectangular solid with vertices (0,0,0), (1,0,0), (1,2,0), (0,2,0), (0,0,3), (1,0,3), (1,2,3), (0,2,3). divF = 1 - 1 + 2 = 2 We have  Since the integrand of the right hand side is just a constant, its value is equal to the constant times the volume of the solid. Since the solid is a rectangular solid with side lengths 1, 2, and 3, we have 2(Volume E) = 2(1)(2)(3) = 12 Problem 5 Find the work done by sailing a ship from the point (2,3) the the point (-1,2) against the wind with velocity field F(x,y) = yi + (3x + 2y)j r(t) = (2 + (-1 - 2)t)i + (3 + (2 - 3))j = (2 - 3t)i + (3 - t)j We have dr = -3i - j F(t) = (3 - t)i + (3(2 - 3t) + 2(3 - t))j = (3 - t)i + (12 - 11t)j The integrand becomes F . dr = -3(3 - t) + (-1)(12 - 11t) = -21 + 14t Now we integrate  Problem 6 Find the flux of F through the surface S where F(x,y,z) = 3zi - 4j + yk and S is the part of the plane x + y + z = 1 in the first octant with upwardly pointing unit normal. We use Stokes' theorem. We have  We can write the surface as z = 1 - x - y Using Stokes theorem we get  |
Saturday, July 24, 2010
Vectors Calaulus 1.35
Vectors Calaulus 1.34
| Stokes' Theorem Stokes' Theorem The divergence theorem is used to find a surface integral over a closed surface and Green's theorem is use to find a line integral that encloses a surface (region) in the xy-plane. The theorem of the day, Stokes' theorem relates the surface integral to a line integral. Since we will be working in three dimensions, we need to discus what it means for a curve to be oriented positively. Let S be a oriented surface with unit normal vector N and let C be the boundary of S. Then C is positively oriented if its orientation follows the right hand rule, that is if you right hand curls around N in the direction of C's orientation, then your thumb will be pointing in the direction of N.  Now we are ready to state Stokes' Theorem. The proof will be left for a more advanced course.
Example Let S be the part of the plane z = 4 - x - 2y with upwardly pointing unit normal vector. Use Stokes' theorem to find   Where F = yi + zj - xyk Solution First notice that without Stokes' theorem, we would have to parameterize three different line segments. Instead we can find this with just one double integral. We have  and N dS = i + 2j + k So that Curl F . N dS = 1 + x + 2y - 1 = x 2y We integrate  Curl and Circulation Just as the divergence theorem assisted us in understanding the divergence of a function ata point, Stokes' theorem helps us understand what the Curl of a vector field is. Let P be a point on the surface and Ce be a tiny circle around P on the surface. The  measures the amount of circulation around P. You can see this by noticing that if F flows in the direction of the tangent vector, then F . dr will be positive. If it flows in the opposite direction, then it will be negative. The stronger the force field in the direction of the tangent vector, the greater the circulation. Since the region enclosed by Ce is tiny, the surface integral can be approximated by  or Curl F . N = Circulation per unit area So the curl tell us how much the force field rotates around the point.  We can see that if this is a small piece of the surface containing P, then Curl F . N > 0 |
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Vectors Calaulus 1.33
| Green's Theorem A Little Topology Before stating the big theorem of the day, we first need to present a few topological ideas. Consider a closed curve C in R2 defined by r(t) = x(t)i + y(t)j a < t < b We say C is simple if it does not intersect itself. A curve intersects itself if r(u) = r(v) for two distinct values u and v. A circle is a simple curve while a figure eight is not simple.  A region is called simply connected if it boundary is a single simple closed curve. Another way of thinking about simply connected regions is that their complement (the space minus the region) consists of only one piece. Below are examples of simply connected and non-simply connected regions.  Our final topological definition is orientation. We have seen that if we traverse a curve in the opposite direction, then the line integral will be the negative of the original. We want to have a way to define a positive orientation. We define it as follows. Let R be a simply connected region with boundary curve C. Then C is called positively oriented if facing the direction that the curve is sketched, the region lies to the left of the curve. Otherwise the curve is said to be negatively oriented. One way to remember this is to recall that in the standard unit circle angles are measures counterclockwise, that is traveling around the circle you will see the center on your left.  Green's Theorem We have seen that if a vector fieldF = Mi + Nj has the property that Nx - My = 0 then the line integral over any smooth closed curve is zero. What can we do if the above quantity is nonzero. Green's theorem states that the line integral is equal to the double integral of this quantity over the enclosed region. Precisely, we have
Sketch of the Proof First we can assume that the region is both vertically and horizontally simple. Otherwise we can carefully cut the region into parts so that each of the parts are both vertically simple and horizontally simple. Below is an example of such a cut. Notice that the line where the regions is cut is drawn once upwards and once downwards. Thus the two line integrals over this line will cancel each other out.  We can assume that the region is as in the figure below  We will show that  The proof for the M part is similar. We will compute both sides and show they are the same. First we break the curve into its left and right half. Call the left half C1 and the right half C2. We have  Now we show that the double integral leads to the same expression. We have  And the two expressions are equal. Using Green's Theorem ExampleDetermine the work done by the force field F = (x - xy) i + y2 j when a particle moves counterclockwise along the rectangle with vertices (0,0), (4,0), (4,6), and (0,6). Solution We could do this with a line integral, but this would involve four parameterizations (one for each side of the rectangle). Instead, we use Green's Theorem. We find Nx - My = 0 - (-x) = x The region is just a rectangle, so the limits are the constants. We have  Example Calculate the line integral  Where C is the union of the unit circle centered at the origin oriented negatively and the circle of radius 2 centered at the origin oriented positively. Solution We cannot use Green's Theorem directly, since the region is not simply connected. However, if we think of the region as being the union its left and right half, then we see that the extra cuts cancel each other out.  In this light we can use Green's Theorem on each piece. We have Nx - My = 1 - 0 = 1 Hence the line integral is just the double integral of 1, which is the area of the region. This area is p(22 - 12) = 4p Green's Theorem and Area The example above showed that if Nx - My = 1 then the line integral gives the area of the enclosed region. There are three special vector fields, among many, where this equation holds. We state the following theorem which you should be easily able to prove using Green's Theorem.
Example Use the third part of the area formula to find the area of the ellipse x2 y2 4 9 Solution To compute the line integral, we parameterize the curve r(t) = 2 cos t i + 3 sin t j r'(t) = -2 sin t i + 3 cos t j We have  |
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Vectors Calaulus 1.32
| Surface Integrals Surface Integrals for Parametric Surfaces In the last section, we learned how to find the surface area for parametric surfaces. We cut the region in the uv-plane into tiny rectangles and added up the area of the corresponding tiny parallelograms in the xy-plane. The area of these parallelograms was  If we think of the surface as having varying density f(x,y,z), then the mass of this parallelogram will be  and adding up all these masses and taking the limit as the rectangle sizes approach zero, gives the definition of the surface integral.
As with finding the surface area the integral typically results in an impossible integral. Example Find  where S is the surface r(u,v) = ui + u2j + (u+ v)k 0 < u < 2 1 < v < 4 and f(x,y,z) = x + 2z Solution We find ru = i + (2u)j + k rv = k and take the cross product  We have f(x(u,v),y(u,v),z(u,v)) = x(u,v) +2z(u,v) = u +2(u + v) = 3u + v We find  Although this integral is possible, its solution is quite involved. You can verify that the surface integral evaluates to approximately 525.27. Surface Integrals for Surfaces that are Functions of Two Variables We have seen before that if z = g(x,y) is a surface such that g has continuous first order partial derivatives, then the parameterization r(u,v) = ui + vj + g(u,v)k has the property that  This leads to the formula for surface integrals.
Example Find where S is the part of the paraboloid z = x2 + y2 that lies inside the cylinder x2 + y2 = 1 and f(x,y,z) = z Solution We have  and f(x,y,z) = z = x2 + y2 At this point, you should be thinking, "This looks like a job for polar coordinates." And we get  Let u = 1 + 4r2 du = 8r dr r2 = 1/4 u - 1/4 and the substitution gives us  Oriented Surfaces and Flux we look to see if the region is on the right or left. Unfortunately, this definition does not work will for surfaces in three dimensions. The idea of right and left is not well defined. In fact not all surfaces can be oriented. We say that a surface is orientable if a unit normal vector can be defined on the surface such that it varies continuously over the surface. Below is an example of a non-orientable surface (called the Mobius Strip)  You can see that there is no front or back of this surface. Recall that a unit normal vector to a surface can be given by  There is another choice for the normal vector to the surface, namely the vector in the opposite direction, -N. By this point, you may have noticed the similarity between the formulas for the unit normal vector and the surface integral. This idea leads us to the definition of the Flux Integral Consider a fluid flowing through a surface S. The Flux of the fluid across S measures the amount of fluid passing through the surface per unit time. If the fluid flow is represented by the vector field F, then for a small piece with area DS of the surface the flux will equal to DFlux = F . N DS Adding up all these together and taking a limit, we get
Notice that the denominator of N and the formula for dS both involve ||ru x rv||. Canceling, we get NdS = ru x rv dvdu for a surface that is defined by the function z = g(x,y), we get the nice formula NdS = -gx(x,y)i - gy(x,y)j + k (oriented upward) or NdS = gx(x,y)i + gy(x,y)j - k (oriented downward) Example Find the flux of F(x,y,z) = xi + 2yj + zk across the part of the surface z = x + y2 with upward pointing normal that lies within the box 0 < x < 3 2 < y < 5 Solution We compute NdS = -i - 2yj + k dydx and F . N dS = -x - 4y2 + x + y2 = -3y2 The flux integral is  |
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