VectorPlot[{y^2-2x*y,3x*y-6*x^2},{x,-5,5},{y,-5,5}]

hapter 16 16.1. 6. Notice that F(x, y) has length 1 and that it is perpendicular to the position vector (x, y) for all x and y (except at the origin). Think about drawing the vectors based on concentric circles of various radii. 19. To draw vector fields in Mathematica, use the following command: VectorPlot[{y^2-2x*y,3x*y-6*x^2},{x,-5,5},{y,-5,5}] For 3-dimensional vector fields use VectorPlot3. 25. First calculate F(x, y) = f(x, y). Then sketch F(x, y). 16.2. 3. First parametrize the right half of the circle x 2 + y 2 = 16 in the form r(t) = (x(t), y(t)) = (R cos t, R sin t) = (R cos t)i + (R sin t)j, a t b, with appropriate a and b and radius R. Then use Formula 3. 7. To parametrize the segments, use Equation 8 and compare with Example 6. As far as notation, you can either use Formula 7 (and the three lines following it) or the box at the very end of the section (and the four lines preceding it). 19. Use efinition 13. ompare with Example 7. 21. ompare with Example 8. 36. The density function is given by λ(x, y, z) = x 2 + y 2 + z 2 and your parametrization is given by r(t) = (x(t), y(t), z(t)) = (t, cos t, sin t); 0 t 2π Moreover, recall that and that, m = x = 1 m λ(x, y, z)ds xλ(x, y, z)ds

ȳ = 1 m z = 1 m yλ(x, y, z)ds zλ(x, y, z)ds where ds = r (t) dt. Use Formula 9 to evaluate these integrals. 40. Parametrize the curve as y(t) = t and x(t) = t 2 + 1. 47. (a) Use F(x, y) = (a, b) with constants a and b and evaluate F(r) dr where is parametrized as x(t) = cos t, y(t) = sin t, 0 t 2π. 51. Focus on the component of the force in the direction of motion. 16.3. 3. Before you start integrating, check if P y? = Q x. If this equation does not hold, then there cannot be a function f(x, y) with F(x, y) = f(x, y). 6. First put this function to the test of Theorem 5. (It will pass this test.) In order to find a scalar field f(x, y) with F(x, y) = f(x, y), start with f x (x, y) = ye x and proceed as in Example 4. 11. Simply find a function f(x, y) with F(x, y) = f(x, y). 13. o not make the mistake of working out the line integral by hand using the given parametrization of the curve. Instead, find a function f(x, y) with F(x, y) = f(x, y) and use the fundamental theorem for line integrals: F(r) dr = f(r(b)) f(r(a)). 15. Follow Example 5. 19. Proceed as in Problem 13. 30. heck the formulas of Problem 29 with P (x, y, z) = y, Q(x, y, z) = x and R(x, y, z) = xyz.

c 36. (a) Writing r = (x, y, z), we have F(x, y, z) = (x, y, z). (x 2 +y 2 +z 2 ) 3/2 That is, in the form F(x, y, z) = (P (x, y, z), Q(x, y, z), R(x, y, z)), we have cx P (x, y, z) = (x 2 + y 2 + z 2 ) 3/2 Q(x, y, z) = R(x, y, z) = cy (x 2 + y 2 + z 2 ) 3/2 cz (x 2 + y 2 + z 2 ) 3/2 c Show that the function f(x, y, z) = has the property that f(x, y, z) = F(x, y, z). Note that this function can x 2 +y 2 +z2 be written as f(r) = c r. Now use Theorem 2 with r(a) = d 1 and r(b) = d 2. (b) Use Part (a) with the given constant c and d 1 = 1.52 10 8 and d 2 = 1.47 10 8. (c) This is similar to Part b: You are given c = ɛqq with ɛ = 8.985 10 9, q = 1, Q = 1.6 10 19. Moreover, you are moving from d 1 = 10 12 to d 2 = d 1 2. 16.4. 3. You are given P (x, y) = xy and Q(x, y) = x 2 y 3. Evaluate both sides of Green s Theorem, where is the triangle with the given vertices and is its boundary curve in counterclockwise direction. 7. The point of this problem is to not do the line integral by hand. Rather, compute it by evaluating the other side of Green s Theorem. 9. See Problem 7. 18. W = F(r) dr = P (x, y) dx + Q(x, y) dy = (Hint: Use polar coordinates.) ( ) Q P da. x y 19. If we apply Green s Theorem to the vector field F(x, y) = ( y, 0), then Q P = 0 ( 1) = 1. So, we can use this to compute x y area by integrating along the boundary curve: ( Q A = 1 da = x P ) da = ( y) dx + 0 dy. y

A nice description of the cycloid can be found in Example 7 of Section 10.1. Figure 13 in Section 10.1 depicts several arches of the curve. As explained there, and as given in the problem, one single arch of the cycloid can be parametrized by r(t) = (x(t), y(t)) = (t sin t, 1 cos t) with 0 t 2π. aution: The arch of the cycloid is parametrized clockwise. So, you will have to switch the sign of your line integral! Also, do not neglect the part of the curve which loops back along the x-axis. 21. (a) Parametrize the line segment : r(t) = (x(t), y(t)) from the point A(x 1, y 1 ) to the point B(x 2, y 2 ) by x(t) = x 1 +t(x 2 x 1 ) and y(t) = y 1 + t(y 2 y 1 ) with 0 t 1. Take the vector field F(x, y) = (P (x, y), Q(x, y)) = ( y, x) and compute by hand ( y) dx+x dy = P (x, y) dx+q(x, y) dy = F(r) dr = (b) Once you connect all these line segments to go around a polygon, you can use Green s Theorem to compute the area, as you did in Problem 19 (where you used a different vector field, though). Since Q P = 1 ( 1) = 2, we get x y A = 1 da = 1 ( Q 2 x P ) da = 1 ( y) dx+x dy, y 2 where goes once around the polygon, that is, is the concatenation of the segments 1, 2, 3,, n connecting the vertices (x 1, y 1 ) to (x 2, y 2 ), (x 2, y 2 ) to (x 3, y 3 ), (x 3, y 3 ) to (x 4, y 4 ),, (x n, y n ) to (x 1, y 1 ), respectively. The formula follows now from adding up the corresponding results from Part (a). (c) Use the formula of Part (b).

22. Recall that x = 1 A x da. If we use Green s Theorem with P (x, y) = 0 and Q(x, y) = x 2, then Q P = 2x. So, x y x = 1 x da = 1 ( Q A 2A x P ) da y = 1 0 dx + x 2 dy. 2A The formula for ȳ is found similarly. 25. Proceed similar to Problem 22. This time, start with I y = x 2 ρ da. 16.5. 10. We have F = (P, Q, R) with R = 0. See if you can tell whether the partial derivatives P x, P y, Q x, Q y are positive, negative or equal to zero. From that information, decide whether div F is positive, negative or equal zero, and in which direction the vector curl F points, or if it is equal to the zero vector. 14. Every conservative field is irrotational : if F = f, then curl F = curl ( f) = 0. Is this true for your field? 17. First verify that curl F = 0. Then Theorem 4 (which we will prove later using Stokes Theorem) guarantees that F is conservative. Now find a potential function as in Section 16.3. 19. Every curl is incompressible : div(curl G) = 0. oes this check out? 33. First of all, 2 g simply means 2 g = g = div g. Use Problem 25 to work out the right hand side of Green s Theorem: F n ds = div F da, with F = f g.

16.6. 10. The following command draws a surface in Mathematica and allows you to stretch and rotate it using the mouse. Try it, it looks quite pretty: ParametricPlot3[ {u,sin[u+v],sin[v]}, {u,-pi,pi},{v,-pi,pi}] 21. Solve for x and put u = y, v = z. 24. Find r(u, v) = (x(u, v), y(u, v), z(u, v)), by taking u = y and v = θ in x = 3 cos θ and z = 3 sin θ. o not neglect to state the domain intervals for u and v. 29. The answer is in the back of your book. 32. This is the surface depicted in Figures 4 and 5 of Section 16.7. 40. Use efinition 6. 44. Treat the surface as a graph z = f(x, y) and use Formula 9. 47. Again, this is a graph y = f(x, z) = x 2 + z 2 with x 2 + z 2 16. So, use Formula 9, with y, x and z instead of z, x and y, respectively. Polar coordinates might come handy at the end. 48. You will run into an integral for which you need the trigonometric substitution u = tan θ and with du = sec 2 θ dθ and 1+u 2 = sec 2 θ. Then see Example 8 of Section 7.2. 49. Use efinition 6. 16.7. 10. Solve for z = g(x, y) = 4 2x 2y, consider r(x, y) = (x, y, g(x, y)) and use the domain given by 0 x 2, 0 y 2 x, in Formula 4. 17. Parametrize the hemisphere using spherical coordinates with radius ρ = 2. That is, r(θ, φ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ) where 0 θ 2π and 0 φ π/2. When calculating r θ r φ, compare with Example 10 of Section 16.6. ompute the integral with Formula 2.

23. The surface is a graph z = g(x, y) = 4 x 2 y 2 over the given domain. So, use Formula 10. Follow Example 5 below the formula. 27. This surface S has two parts to it, call them S 1 and S 2. So, you have to do two integrals and add the answers. (a) S 1 is a graph, but the variables are switched. We have a function y = g(x, z) = x 2 + z 2 over the domain x 2 + z 2 1. Adjust Formula 10 appropriately and use the downward flux from the graph towards its domain: F ds = S 1 x 2 +z 2 1 R g z P g + Q dz dx =... x Then use polar coordinates, z = r cos θ and x = r sin θ:... = 2π 1 0 0 (r 2 + 2r 2 cos 2 θ)r dr dθ =... = π (b) S 2 is the disk x 2 + z 2 1 with y = 1. This surface can be parametrized by r(x, z) = (x, 1, z) with x 2 + z 2 1. Be very careful: r x r z =... = j. However, you need the flux through S 2 in the other direction. So, use Formula 9 F ds = F n da. S 2 with n = (r x r z ). (Why could you have guessed the overall answer?) 29. This surface has six sides. eal with them like you dealt with S 2 in Problem 27. The outward fluxes through the individual faces should come out: 4, 8, 12, 4, 8, 12; with a total of 48. 40. This surface is a graph z = g(x, y) = x 2 + y 2 with domain 1 x 2 + y 2 16: m = ρ(x, y, z) ds. S Use Formula 4. Finally, use polar coordinates to evaluate the integral.

41. (a) I z = (x 2 + y 2 )ρ(x, y, z) ds S 16.8. 7. alculate curl F =... = ( 2z, 2x, 2y). Then use (P, Q, R) = ( 2z, 2x, 2y) with z = g(x, y) = 1 x y in Formula 10 of Section 16.7, where F has been replaced by curl F, to compute the right hand side of Stokes Theorem. 17. You want to use Stokes Theorem to evaluate this line integral. (See Problem 7.) 18. Notice that the curve : r(t) = (sin t, cos t, sin 2t) is running clockwise, when viewed from above, because the sine and the cosine in the first two coordinates are switched. Follow the hint in the text: the curve lies on the surface z = g(x, y) = 2xy, because 2xy = 2 sin t cos t = sin 2t = z. Therefore, consider F dr = curl F n ds = S where S is the part of the surface which is bounded by. Notice that the domain for the graph z = g(x, y) in the xy-plane is bounded by the curve (sin t, cos t), i.e., is the unit disk. At the very end of your calculation, switch to polar coordinates. (Answer: π) 16.9. 4. You are asked to calculate both sides of the ivergence Theorem. The surface integral splits into three parts: the cylinder has a top (x = 2), a bottom (x = 0) and a cylindrical side (y 2 + z 2 = 9). For the top and the bottom, the outward unit normals are given by n = i and n = i, respectively, leading to F n = x 2 = 4 for the top and F n = x 2 = 0 for the bottom surface. (So, the flux can easily be calculated using the known disk area.) For the cylindrical side, use r(x, θ) = (x, 3 cos θ, 3 sin θ) with 0 x 2 and 0 θ 2π. Then use Formula 9 of Section 16.7. 5. You are asked to work out the triple integral side of the divergence theorem. 9. Start with figuring div F. 11. Work out the triple integral in cylindrical coordinates.

14. Observe that F(x, y, z) = ((x 2 + y 2 + z 2 )x, (x 2 + y 2 + z 2 )y, (x 2 + y 2 + z 2 )z). 24. Notice that on the sphere, the outward unit normal is given by n = (x, y, z) (x, y, z) = (x, y, z) 1 = (x, y, z). So, in order to apply the ivergence Theorem, you need to find a vector field F = (P, Q, R) with F n = P x + Qy + Rz = 2x + 2y + z 2. Once you have this field F, you will see that div F = 1. Hence F n ds = div F dv = 1 dv = volume of the sphere. S E E