on an open connected region D, then F is conservative on D. (c) If curl F=curl G on R 3, then C F dr = C G dr for all closed path C.

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1 . (5%) Determine the statement is true ( ) or false ( ). 微甲 -4 班期末考解答和評分標準 (a) If f(x, y) is continuous on the rectangle R = {(x, y) a x b, c y d} except for finitely many points, then f(x, y) is integrable on R and R f(x, y)da = (b) If F(x, y) = P (x, y)i + Q(x, y)j and P y = Q x c d a b f(x, y)dxdy = (c) If curl F=curl G on R, then C F dr = C G dr for all closed path C. a b d f(x, y)dydx. c on an open connected region D, then F is conservative on D. (d) If F and G are vector fields and curl F=curl G, div F=div G, then F G is a constant vector field. (e) Let B be a rigid body rotating about the z-axis with constant angular speed ω. If v(x, y, z) is the velocity at point (x, y, z) B, then curl v is parallel to k. Answer. ( 每小題各 分 ) (a) (b) (c) (d) (e). (%) Write the integral y x Answer. ( 每小題錯一格扣 分, 錯兩格以上全錯 ) f(x, y, z)dzdydx in 5 other orders. (a) y y f(x, y, z)dzdxdy (b) z y f(x, y, z)dxdydz (c) y y f(x, y, z)dxdzdy (d) x z x f(x, y, z)dydzdx (e) ( z) z x f(x, y, z)dydxdz Page of 9

2 . (5%) Evaluate the integrals. (a) (b) 4 tan y y cos x tan(cos x)dxdy. x + y dxdy + y x + y dxdy + x x + y dydx. (a) 4 tan y cos x tan (cos x)dxdy = = = 4 4 tan x cos x tan (cos x)dydx (pt) sin x tan (cos x)dx (Let u = cos x, du = sin xdx) tan udu = ln (cos u) (pt) = ln (cos ) ln (cos ) (pt) (b) y x x + y dxdy + y x + y dxdy + = D x + y da, where D is bounded by = and =. By polar coordinate, we have D x + y da = = = cos θ+sin θ cos θ+sin θ r cos θ + r sin θ r (cos θ + sin θ)drdθ (cos θ + sin θ) dθ = (sin θ cos θ) (pt) = (pt) rdrdθ (4pt) x + y dydx Page of 9

3 4. (%) Let S be the surface x + y + z = a, x, y, z (a > ), and let C be the boundary of S. Find the centroid of C. For a quarter circle of radius a (named C ) on a plane, its centroid can be found to be at ( a, a ) by either way: () Parametrize the curve: Parametrize C by r(t) = a cos t, a sin t, t [, ]. r (t) = a sin t, a cos t = a. Arc length s = 4 a = a. x s = C x ds = x = a a = a By the symmetry of the arc, y = x = a. x(t) r (t) dt = a cos(t)adt = a () Pappus s Theorem: Knowing that the surface area of a hemisphere of radius a is a and the arc length of a quarter circle of radius a is a, if the quarter circle is in the first quadrant and is rotated about the x-axis, Pappus s Theorem gives By the symmetry of the arc, x = y = a. (7 points up to this point.) A = y s a = y a y = a The curve C is composed of quarter circles C, C, and C on the xy-, yz-, and xz-planes, respectively. By the above discussion, their centroids are ( a, a, ), (, a, a ), and ( a,, a ), respectively. Since they have equal masses, the centroid of C is the average of them, namely ( 4a, 4a, 4a ). ( points) (Note: if you misunderstood the problem but correctly calculated the centroid of the surface S to be at ( a, a, a ), you still get 4 points. But no points will be given if you calculated the centroid of the part of the volume inside the sphere in the first octant.) Page of 9

4 5. (%) Let C be the curve of intersection of x + y + z = 4, x + y = x, z, oriented C to be counterclockwise when viewed from above. Evaluate C y dx + z dy + x dz. ˆ Solution : Using line integral to solve this problem directly. r(θ) =< + cos θ, sin θ, sin θ >, θ [, ]. ( points) The original equation = [ sin θ + sin θ cos θ + ( + cos θ) cos θ ] dθ ( points) By symmetry, the first and the third term will be zero in the end. Therefore, the above equation will change as follows: 4 sin θ dθ = 4 cos θ cos θ dθ = (4 points) ˆ Solution : Using Stokes Theorem to solve this problem. F =< y, z, x > F =< z, x, y > ( points) r(x, y) =< x, y, 4 x y > x r x =<,, > 4 x y r y =<,, r x r y =< y 4 x y > x 4 x y, y 4 x y, > ( points) By Stokes Theorem, c F dr = S ( F ) ds = D < 4 x y, x, y > < = D ( x xy y) da 4 x y x 4 x y, ( points) y 4 x y, > da By symmetry, the second and the third term will be zero in the end. Therefore, the above equation will change as follows: r cos θrdrdθ cos θ = 6 = = = 8 cos θ ( r ) cos θ rdrdθ cos 4 θ dθ cos4 θ dθ cos θ ( ) dθ (cos θ cos θ + ) dθ By symmetry, the second term will be zero in the end. Therefore, the above equation will change as follows: = 8 =. ( points) Page 4 of 9

5 6. (%) Let F = (x y) y (x +y ) i + (x y) x (x +y ) j. (a) erify that F is conservative on the right half plane x >. Find a potential function of F on the right half plane. (b) Evaluate C F dr where C is the ellipse x 4 + (y ) =. (c) Evaluate C F dr where C is the curve with polar equation r = e θ, 9 4 θ 9 4. y x (a) (6%) y f = (x y) x x f = tan ( y (x +y ) (x +y ) x ) + g(x) x f = xy y + + g (x) = (x y) y + g (x) = (x y) y (x +y ) (x +y ) (x +y ) (x +y ) g (x) = g is constant x f = tan ( y (x +y ) x ) or + x tan ( y (x +y ) x ) = y tan ( y (x +y ) x ) (6%) Other point: P y = Q x = x4 4x y+4xy y 4 (%); P (x +y ) y = Q x implies f is conservative (%) (b) (4%) Since {y > } is simple connected, F is conservative on {y > }. On the other hand, C is closed curve on y > (%); therefore, C F dr = (%) (c) (%) method the integral on C is equal to the integral on unit circle times two and integral on the tail. D F dr = (cos θ sin θ) dθ = (4%), where D is unit circle. The integral on tail is independent of path, which equals to f(cos 9 4 e 9 4, sin 9 4 e 9 4 ) f(cos 9 4 e 9 4, sin 9 4 e 9 4 ) = (%), where f is potential function of F Therefore, C F dr = 4 = 9 (%) method C F dr = γ F dr + γ F dr, where γ is the θ part of C, γ is θ < part of C, in which x(θ) and y(θ) is differentiable. 9 4 C F dr = (cos θ sin θ) dθ + = (cos θ sin θ) dθ = (cos θ sin θ) dθ (4%) (6%) (the answer worth point) Page 5 of 9

6 7. (%) Evaluate C (y +sin x)dx+(z +cos 4 y)dy +(x +tan 5 z)dz where C is the curve r(t) = sin t i+cos t j+sin t k, t. [Hint: C lies on the surface z = xy.] First observe that r(t) = sin t i + cos t j + sin t k is negative oriented. Thus by Stoke s theorem: C (y + sin x) dx + (z + cos 4 y) dy + (x + tan 5 z) dz = S F ds (%) where S is the surface z = xy bounded by D = {x + y } i j k F = x y z = z i x j k (%) y + sin x z + cos 4 y x + tan 5 z n = ( z x, z z, )/ ( y x, z, ) = ( y, x, )/ ( y, x, ) (%) y F ds = ( z, x, ) ( y, x, ) da S D = 4yz + 6x da D = 8x y + 6x da D = (8r cos θ sin θ + 6r cos θ )r dr dθ = (4%) (y + sin x) dx + (z + cos 4 y) dy + (x + tan 5 z) dz = C Page 6 of 9

7 8. (%) Evaluate S xds where S is the part of the cone z = (x + y ) that lies below the plane z = + x. Step. Find the projection onto the xy-plane of the curve of intersection of the cone z = (x + y ) and the plane z = + x. { z = (x + y ) z = + x (x + y ) = (x + ) ( x ) + y = (pt) Step. If we regard x and y as parameters, then we can write the parametric equations of S as x = x y = y z = (x + y ) (pt) where ( y ) x + ( y ), y (pt) and the vector equation is r(x, y) = xi + yj + (x + y )k Step. Find r x r y. r x = i + j + r y = i + j + x k x +y y k x +y r x r y = x i + y j + k (pts) r x r y = (pt) Step4. Evaluate S xds. S xds = D x r x r y dxdy (pts) = = + ( y ) ( y ) = 6 = 6 (pts) y dy xdxdy Page 7 of 9

8 9. (%) Let S be the surface of the solid bounded by x + y + z = and z. Find the total flux of F(x, y, z) = x i + y j + z k across S. (Method I) Let = {(x, y, z) R x + y + z, z }, then by Divergence Theorem, Flux of F = F ds = divf ds = ( + z) ds (%) S From the symmetry of, we have Therefore, divf ds = z ds = x ds = y ds =. sec φ ρ cos φρ sin φdρdφdθ = 4 ρ cos φρ sin φdρdφdθ = sec φ cos φ sin φ 6 tan φ sec φ dφ = 4 4 ( sin φ ρ 4 cos φ sin φ sec φdφdθ tan φ) = 9 64 S F ds = z ds = 9 64 = 9 (7 %) (Method II) Let S = {(x, y, z) R x + y + z =, z } and S = {(x, y, z) R x + y 4, z = } Flux of F = S = S F ds = (x, y, z ) ds + (x, y, z ) ds S S (x, y, z ) (x, y, z) ds + (x, y, z ) (,, ) ds. S = S x + y + z ds S z ds = S x + y + z ds 4 Area(S ). (%) From the symmetry of S, we have x ds = y ds =. S S Thus, we only need to calculate z ds, then by Spherical coordinate S S z ds = cos φ sin φ dφdθ = ( 4 cos4 φ) = [( ) 4 ] = 5. S F ds = z ds 4 Area(S ) = 5 4 ( ) = 9 (7 %) S Page 8 of 9

9 . (%) Solve the differential equation y + y = x e x + tan x, x (, ). Complementary equation: y + y =. Auxiliary equation: r + = r = ±i y c = c sin x + c cos x. ( points) For the particular solution: () For y + y = x e x it s a better idea to use the method of undetermined coefficients: Let y p = (Ax + Bx + C)e x, y p = [Ax + (A + B)x + (B + C)]e x, y p = [Ax + (4A + B)x + (A + B + C)]e x. y p + y p = [Ax + (4A + B)x + (A + B + C)]e x x e x A =, B =, C = y p = ( x x + )ex. ( points for the formulation, points for solving the coefficients.) () For y + y = tan x we use the method of variation of parameters: Let y p = u sin x + u cos x, y p = (u sin x + u cos x) + u cos x u sin x. Setting u sin x + u cos x = (equation ), we have y p = u cos x u sin x u sin x u cos x. y p + y p = u cos x u sin x tan x (equation ). Solving the system of equations and, we have { u sin x + u cos x = u cos x u sin x = tan x { u (x) = sin x, u (x) = tan x sin x = sin x cos x = cos x = cos x sec x cos x { u (x) = cos x, u (x) = sin x ln sec x + tan x = sin x ln(sec x + tan x) for x (, ) y p (x) = u sin x + u cos x = (cos x) ln(sec x + tan x) ( points for the system of equations, points for solving and integrating them.) Combining the above results, we have the general solution y(x) = c sin x + c cos x + ( x x + )ex (cos x) ln(sec x + tan x) Page 9 of 9