Review Questions for Test 3 Hints and Answers

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1 eview Questions for Test 3 Hints and Answers A. Some eview Questions on Vector Fields and Operations. A. (a) The sketch is left to the reader, but the vector field appears to swirl in a clockwise direction, and the arrows get longer farther away from the origin. (b) The vectors are vectors pointing away from the origin, with length equal to the distance of the tail to the origin. A2. (a) divf, 2k. (b) divf 2,. A3. (a) First, x y z M N P (P y N z )i + (M z P y )j + (N x M y )k. Thus, div(curlf) P yx N zx + M zy P xy + N xz M yz assuming the second partials are continuous because then the mixed partials are equal. (b) First, curl( f) x y z (f zy f yz )i + (f yz f zy )j + (f yx f xy )k i + j + k f x f y f z as desired because the second mixed partials of f are equal. A4. Yes, as long as (x, y) (, ) because M y 2kxy (x 2 + y 2 ) N 2 x. A. First, the curl is x y z y 2 2xyz xy 2 (2xy 2xy)i (y 2 )j + (2yz 2y)k y 2 j + (2yz 2y)k.

2 Because the components of F have continuous partial derivatives and curlf, the vector field is not conservative. A6. f(x, y, z) xy 2 z + B. Evaluation and Applications of Line Integrals B. A parametric representation of the curve is x t, y 2t, t 2. Then ds + 4 dt dt. Therefore, the surface area is obtained by the line integral of the height function over the curve as follows. f(x, y)ds 2 4 t 2 (2t) 2 dt 4t [32 ()] B2. Parameterize the curve as x t, y t 2 where 3 t 3. Then A 3 t + 4t 2 dt 2 3 t( + 4t 2 ) /2 dt ( + 4t2 ) 3/2 6 [373/2 ]. 3 B3. (a) There are six spirals in the helix. (b) Wire starts at (,, ) and ends at (,, 2). (c) First, we find ds r 36π2 + (t) dt dt. The length is 2π 4π 4π 36π2 + L ds t 2π 2 36π 2 +. (d) The number of charged particles along the wire is 4π [ N δ(x, y, z)ds 6 t2 36π2 + dt 6 t 3 36π 4π 2 2π π 3 4π ] π B4. First, the curve can be written in parametric form as x t, y 4t t 2 for t 4 to. Therefore, the line integral is given by y dx + x 2 dy (4t t 2 ) dt + t 2 (4 2t) dt 4 4 3t 2 + 4t 2t 3 dt t 3 + 2t 2 2 t4 2 ( )

3 B. Parameterize the curve as x t 2, y t for t Then 2 2 x dx + xy dy t 2 2t dt + t 2 t dt 3t 3 dt. 2 2 B6. The work done is W F dr (t 3 i + t 2 j t 3 k)(i + 2tj + k)dt 2t 3 dt 2. B7. Parameterize the curve as x t, y t 2, t 2, then the line integral becomes t 3 dt + (t + t 2 )(2t) dt 2 3t 3 + 2t 2 dt B8. (a) hanges the sign in the interval. (b) Yes, F dr (F T)ds, so the function f F T.. Path Independence and The Fundamental Theorem of Line Integrals. (a) Because M, N and P have continuous partial derivatives, it suffices to show that, which we now do. x y z ( 3 ( 3))i + (2 2)j + ( )k. y + 2z x 3z 2x 3y (b) potential function for F is f(x, y, z) xy + 2xz 3yz, using the Fundamental Theorem of Line integrals F dr xy + 2xz 3yz (3,2,) (,,) ( ) ( + + ) (a) Because M, N and P have continuous partial derivatives, it suffices to show that, which we now do. x y z ( )i + ( )j + ( e x sin y e x sin y)k. e x cos y e x sin y 2

4 (b) A potential function for F is f(x, y, z) e x cos y + 2z, using the Fundamental Theorem of Line integrals (,π,3) F dr e x cos y + 2z e () + 6 ( + 2) 4 e. (, π 2,) Thus the integral along any piecewise smooth curve is 4 e. 3. (a) It suffices to check that the curl is. x y z f(x) g(y) h(z) i + j + k. Therefore, F is conservative. (b) Yes, let F, G and H represent antiderivatives of f, g and h. Then p(x, y, z) F (x) + G(y) + H(z) is a potential function for V. Hence V is conservative by the definition. Then by the fundamental theorem of line integrals V dr p(a 2, b 2, c 2 ) p(a, b, c ) F (a 2 ) + G(b 2 ) + H(c 2 ) (F (a ) + G(b ) + H(c )) a2 a f(t) dt + b2 b g(t) dt + c2 c h(t) dt. (c) First, a potential function for F is f(x, y, z) x 2 +xy 2z 2 z. Thus using the fundamental theorem of line integrals, we evaluate F dr x 2 + xy 2z 2 z (2,,) (3,, 2) (4 2) ( ). 4. No. The above would have to be true for all piecewise smooth curves. For example, try F x 2 j. Then F is not conservative, yet F dr, where is a circle of radius centered at the origin. However, F dr for the square with vertices (, ), (, ), (, ), (, ). You may wish to verify these line integrals either directly, or by using Green s theorem. (b) Yes, because the vector field F f(x)i + g(y)j is conservative.. Notice that this is a path independent line integral by 4(b). However, it is hard to find a potential function because of the difficulty in finding an antiderivative for e y2. Therefore, we choose a different path, namely, x t, y, for t to t. The the line integral becomes t 2 dt 2 3.

5 6. In general, think about the following concepts. What is the definition of a conservative vector field? (see p. for answer). What are tests in terms of partial derivatives or curl to test whether certain vector fields are conservative? (see Theorem 4. and 4.2). What are the connections between conservative vector fields, path independence and line integrals of the vector field over closed curves? (see Theorem 4.7). D. Green s Theorem D. Using Green s Theorem we calculate (x + y) dx + (3y + y 2 x 2 ) dy ( 2 2x 4 3 x3 x 2 2 x (3y + y2 x 2 ) (x + y) y ( 2x ) dy dx ) da ( 4x 2 2x)dx D2. (a) Using Green s Theorem we have N dy M dx 2 2 ( N x + M y The last integral is equal to the area of the region. (b) Using the formula in (a) we obtain A 2 x dy y dx π 2π 2π ) da ( + ) da da. 2 (a cos t)(b cos t) dt (b sin t)( a sin t) dt ab(cos 2 t + sin 2 t) dt ab dt ab 2π πab. 2 D3. (a) Using Green s theorem x dy dx + x dy (b) Using the formula from (a) A 3 2π 2π x (x) y ()da da Area of. 2π (cos 3 t)(3 sin 2 t)(cos t) dt 3 cos 4 t( cos 2 t) dt 3 [ 3π 4 π 8 cos 4 t sin 2 t dt ] 3π 8.

6 D4. (i) The counterclockwise circulation is computed as F dr M dx + N dy yx 2xy y dx y ( N x M ) da y 2x 2 x dx 6. x 2x dy dx (ii) The flux is computed as M dy N dx ( M x + N ) da y yx y + y 2 dx y x x + x 2 dx y dy dx D. (a) Green s theorem implies the integral over a counterclockwise oriented curve is 3 (Area of ) because N x M 3. In this case, is a triangle with area 2. Thus the y integral is 7 in a counterclockwise direction, and 7 in a clockwise direction. (b) The integral will be (Area of ). This confirms what the previous section said about integrals of conservative vector fields over closed curves.