Sections minutes. 5 to 10 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed.

Transcription

1 MTH 34 Review for Exam 4 ections minutes. 5 to 1 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed. Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem.

2 Line integrals (16.1) Integrate the function f (x, y) = x 3 /y along the plane curve given by y = x / for x, ], from the point (, ) to (, ). olution: We have to compute I = f ds, by that we mean f ds = t1 t f ( x(t), y(t) ) r (t) dt, where r(t) = x(t), y(t) for t t, t 1 ] is a parametrization of the path. In this case the path is given by the parabola y = x /, so a simple parametrization is to use x = t, that is, r(t) = t, t, t, ] r (t) = 1, t. Line integrals (16.1) Integrate the function f (x, y) = x 3 /y along the plane curve given by y = x / for x, ], from the point (, ) to (, ). olution: r(t) = f ds = f ds = t1 t, t for t, ], and r (t) = 1, t. t f ( x(t), y(t) ) r (t) dt = f ds = We conclude that t t t / dt, t 1 + t dt, u = 1 + t, du = t dt. 5 1 u 1/ du = 3 u3/ 5 1 = 3 ( 5 3/ 1 ). f ds = 3( ).

3 Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. Vector fields, work, circulation, flux (plane) (16.) Find the work done by the force F = yz, zx, xy in a moving particle along the curve r(t) = t 3, t, t for t, ]. olution: The formula for the work done by a force on a particle moving along given by r(t) for t t, t 1 ] is W = F dr = t1 t F(t) r (t) dt. In this case r (t) = 3t, t, 1 for t, ]. We now need to evaluate F along the curve, that is, F(t) = F ( x(t), y(t) ) = t 3, t 4, t 5.

4 Vector fields, work, circulation, flux (plane) (16.) Find the work done by the force F = yz, zx, xy in a moving particle along the curve r(t) = t 3, t, t for t, ]. olution: F(t) = t 3, t 4, t 5 and. r (t) = 3t, t, 1 for t, ]. The Work done by the force on the particle is W = t1 t F(t) r (t) dt = t 3, t 4, t 5 3t, t, 1 dt W = ( 3t 5 + t 5 t 5) dt = 4t 5 dt = 4 6 t6 We conclude that W = 7 /3. = 3 6. Vector fields, work, circulation, flux (plane) (16.) Find the flow of the velocity field F = xy, y, yz from the point (,, ) to the point (1, 1, 1) along the curve of intersection of the cylinder y = x with the plane z = x. olution: The flow (also called circulation) of the field F along a curve parametrized by r(t) for t t, t 1 ] is given by F dr = t1 t F(t) r (t) dt. We use t = x as the parameter of the curve r, so we obtain r(t) = t, t, t, t, 1] r (t) = 1, t, 1. F(t) = t(t ), (t ), t (t) F(t) = t 3, t 4, t 3.

5 Vector fields, work, circulation, flux (plane) (16.) Find the flow of the velocity field F = xy, y, yz from the point (,, ) to the point (1, 1, 1) along the curve of intersection of the cylinder y = x with the plane z = x. olution: r (t) = 1, t, 1 for t, 1] and F(t) = t 3, t 4, t 3. t1 1 F dr = F(t) r (t) dt = t 3, t 4, t 3 1, t, 1 dt, t F dr = We conclude that 1 ( t 3 + t 5 t 3) dt = 1 F dr = 1 3. t 5 dt = 6 t6 1. Vector fields, work, circulation, flux (plane) (16.) Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t, π]. olution: The flux (also normal flow) of the field F = F x, F y across a loop parametrized by r(t) = x(t), y(t) for t t, t 1 ] is given by t1 F n ds = Fx y (t) F y x (t) ] dt. t Recall that n = 1 r (t) y (t), x (t) and ds = r (t) dt, therefore F n ds = ( F x, F y 1 ) r (t) y (t), x (t) r (t) dt, so we obtain F n ds = F x y (t) F y x (t) ] dt.

6 Vector fields, work, circulation, flux (plane) (16.) Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t, π]. t1 olution: F n ds = Fx y (t) F y x (t) ] dt. We evaluate F along the loop, t F(t) = a cos(t), a cos(t) sin(t) ], and compute r (t) = a sin(t), a cos(t). Therefore, F n ds = F n ds = π π a cos(t)a cos(t) a ( cos(t) sin(t) ) ( a) sin(t) ] dt a cos (t) + a sin(t) cos(t) a sin (t) ] dt Vector fields, work, circulation, flux (plane) (16.) Find the flux of the field F = x, (x y) across loop given by the circle r(t) = a cos(t), a sin(t) for t, π]. olution: F n ds = ince π π a cos (t) + a sin(t) cos(t) a sin (t) ] dt. π F n ds = a ] 1 + sin(t) cos(t) dt, F n ds = a π sin(t) dt =, we obtain sin(t) ] dt. F n ds = πa.

7 Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. onservative fields, potential functions (16.3) Is the field F = y sin(z), x sin(z), xy cos(z) conservative? If yes, then find the potential function. olution: We need to check the equations y F z = z F y, x F z = z F x, x F y = y F x. y F z = x cos(z) = z F y, x F z = y cos(z) = z F x, x F y = sin(z) = y F x. Therefore, F is a conservative field, that means there exists a scalar field f such that F = f. The equations for f are x f = y sin(z), y f = x sin(z), z f = xy cos(z).

8 onservative fields, potential functions (16.3) Is the field F = y sin(z), x sin(z), xy cos(z) conservative? If yes, then find the potential function. olution: x f = y sin(z), y f = x sin(z), z f = xy cos(z). Integrating in x the first equation we get f (x, y, z) = xy sin(z) + g(y, z). Introduce this expression in the second equation above, y f = x sin(z) + y g = x sin(z) y g(y, z) =, so g(y, z) = h(z). That is, f (x, y, z) = xy sin(z) + h(z). Introduce this expression into the last equation above, z f = xy cos(z) + h (z) = xy cos(z) h (z) = h(z) = c. We conclude that f (x, y, z) = xy sin(z) + c. onservative fields, potential functions (16.3) ompute I = y sin(z) dx + x sin(z) dy + xy cos(z) dz, where given by r(t) = cos(πt), 1 + t 5, cos (πt)π/ for t, 1]. olution: We know that the field F = y sin(z), x sin(z), xy cos(z) conservative, so there exists f such that F = f, or equivalently df = y sin(z) dx + x sin(z) dy + xy cos(z) dz. We have computed f already, f = xy sin(z) + c. ince F is conservative, the integral I is path independent, and I = (1,,π/) (1,1,π/) y sin(z) dx + x sin(z) dy + xy cos(z) dz ] I = f (1,, π/) f (1, 1, π/) = sin(π/) sin(π/) I = 1.

9 onservative fields, potential functions (16.3) how that the differential form in the integral below is exact, 3x dx + z y dy + z ln(y) dz ], y >. olution: We need to show that the field F = is conservative. It is, since, 3x, z y, z ln(y) y F z = z y = zf y, x F z = = z F x, x F y = = y F x. Therefore, exists a scalar field f such that F = f, or equivalently, df = 3x dx + z y dy + z ln(y) dz. onservative fields, potential functions (16.3) (1, 1,) ompute I = x cos(z) dx + z dy + (y x sin(z)) dz. (,,) olution: The integral is specified by the path end points. That suggests that the vector field is a gradient field. F = x cos(z), z, y x sin(z)] = f = x f, y f, z f. x f = x cos(z) f = x cos(z) + g(y, z). y f = z = y g g = yz + h(z) f = x cos(z) + yz + h(z). z f = y x sin(z) = x sin(z) + y + h h = ince f = x cos(z) + yz + c, we obtain I = (1, 1,) (,,) f dr = f (1, 1, ) f (,, ) I = 1.

10 Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. The Green Theorem in a plane (16.4) Use the Green Theorem in the plane to evaluate the line integral ] given by (6y + x) dx + (y + x) dy on the circle defined by (x 1) + (y 3) = 4. olution: F dr = curl F dx dy; and curl F = ( ) x F y y F x. Here F = (6y + x), (y + x). ince x F y = and y F x = 6, hence curl F = 6 = 4. Green s Theorem implies ] (6y + x) dx + (y + x) dy = F dr = ( 4) dx dy. ince the area of the disk = {(x 1) + (y 3) 4} is π( ), F dr = 4 dx dy = 4(4π) F dr = 16π.

11 The Green Theorem in a plane (16.4) Use the Green Theorem in the plane to find the flux of F = (x y ) i + (x + y) j through the ellipse 9x + 4y = 36. olution: Recall: F n ds = div F dx dy. Recall: div F = x F x + y F y. Here is simpler to compute the right-hand side than the left-hand side. div F = =. Green s Theorem implies F n ds = () dx dy. = A(R). R ince R is the ellipse x /4 + y /9 = 1, its area is A(R) = ()(3)π. We conclude F n ds = 1π. Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem.

12 urface area (16.5) et up the integral for the area of the surface cut from the parabolic cylinder z = 4 y /4 by the planes x =, x = 1, z =. olution: x 1 z 4 R 4 y We must compute: A() = dσ. Recall dσ = f dx dy, with k R. f k Recall: f (x, y, z) = y + 4z 16. f =, y, 4 f = y = 4 + y. ince R =, 1] 4, 4], its normal vector is k and f k = 4. Then, 4 + y A() = y dx dy A() = dy dx. R Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem.

13 urface integrals (16.6) Integrate the function g(x, y, z) = x 4 + y over the surface cut from the parabolic cylinder z = 4 y /4 by the planes x =, x = 1 and z =. olution: x 1 z 4 R 4 y We must compute: I = g dσ. Recall dσ = f dx dy, with k R f k and in this case f (x, y, z) = y + 4z 16. f =, y, 4 f = y = 4 + y. ince R =, 1] 4, 4], its normal vector is k and f k = 4. Then, ( g dσ = ) 4 + y x 4 + y dx dy. 4 R urface integrals (16.6) Integrate the function g(x, y, z) = x 4 + y over the surface cut from the parabolic cylinder z = 4 y /4 by the planes x =, x = 1 and z =. ( olution: g dσ = ) 4 + y x 4 + y dx dy. R 4 g dσ = 1 x(4 + y ) dx dy = x(4 + y ) dx dy g dσ = g dσ = 1 ( R 4 ] (4+y 1 ] ) dy x dx = 1 (4y + y 3 3 ) 1 ( = ) 3 ) 4 4 ( x ) 1 g dσ = 56 3.

14 Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem. The tokes Theorem (16.7) Use tokes Theorem to find the flux of F outward through the surface, where F = y, x, x and = {x + y = a, z, h]} {x + y a, z = h}. olution: Recall: ( F) n dσ = F dr. The surface is the cylinder walls and its cover at z = h. Therefore, the curve is the circle x + y = a at z =. That circle can be parametrized (counterclockwise) as r(t) = a cos(t), a sin(t) for t, π]. π ( F) n dσ = F dr = F(t) r (t) dt, where F(t) = a sin(t), a cos(t), a cos (t) and r (t) = a sin(t), a cos(t),.

15 The tokes Theorem (16.7) Use tokes Theorem to find the flux of F outward through the surface, where F = y, x, x and = {x + y = a, z, h]} {x + y a, z = h}. olution: F(t) = a sin(t), a cos(t), a cos (t) and r (t) = a sin(t), a cos(t),. Hence ( F) n dσ = We conclude that ( F) n dσ = π π F(t) r (t) dt, ( a sin (t) + a cos (t) ) π dt = a dt. ( F) n dσ = πa. Review for Exam 4 (16.1) Line integrals. (16.) Vector fields, work, circulation, flux (plane). (16.3) onservative fields, potential functions. (16.4) The Green Theorem in a plane. (16.5) urface area. (16.6) urface integrals. (16.7) The tokes Theorem. (16.8) The Divergence Theorem.

16 The Divergence Theorem (16.8) Use the Divergence Theorem to find the outward flux of the field F = x, xy, 3xz across the boundary of the region D = {x + y + z 4, x, y, z }. olution: Recall: F n dσ = ( F)dv. F = x F x + y F y + z F z = x x + 3x F = 3x. F n dσ = D D ( F)dv = It is convenient to use spherical coordinates: F n dσ = π/ π/ D 3x dx dy dz. 3ρ sin(φ) cos(φ) ] ρ sin(φ) dρ dφ dθ. The Divergence Theorem (16.8) Use the Divergence Theorem to find the outward flux of the field F = x, xy, 3xz across the boundary of the region D = {x + y + z 4, x, y, z }. olution: F n dσ = π/ π/ π/ F n dσ = F n dσ = sin(θ) 3ρ sin(φ) cos(φ) ] ρ sin(φ) dρ dφ dθ. ] π/ ] cos(θ) dθ sin ] (φ) dφ 3ρ 3 dρ ] 1 π/ ( ) ] 3 1 cos(φ) dφ 4 ρ4 ] π/ F n dσ = (1) 1 ( π ) (1) F n dσ = 3π.