dzdydx. [Hint: switch dzdydx into dxdydz.] z(2 z) dzdydx. z(2 z) 2 )dz z(2 z) 2 sin(πz)dz = 1 x = u + 3v (2 points) y = v (2 points) (u, v) = 1 2

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1 Page of 6. (%) Evaluate x z( z) 微甲 8- 班期末考解答和評分標準 dzddx. [Hint: switch dzddx into dxddz.] We need to compute x z( z) dzddx. The region of integration is bounded b [, ] [, ] [, ] {(x,, z) 3 x +, z }. (4% If ou derived the correct region or drawed a correct graph indicating the correct region) Therefore, x did not provide our reasons) Hence x f(x,, z)dzddx z( z) dzddx z z f(x,, z)dxddz. (4%; ou won t get full points if ou z( z) z( z) dxddz z ( )ddz z (z z( z) )dz dz π cos(πz) π. (4%). (%) Evaluate e 4x +x dxd where is the region satisfing x and. [Hint: rewrite 4x x + as u + v b suitable changing of coordinates.] Let Then let u x 3 and v then The Jacobian is 4x x + u + v ( points) { x u + 3v v (x, ) (u, v) ( points) ( points) e 4x +x dxd u π 4 π 6 e (u +v ) dvdu (6 points) e r rdrdθ 3. (%) Evaluate x x x + (x + + z ) dzddx.

2 Page of 6 The domain is defined b x + + z <, z > x +, < < x, z >, x >. (4%) hange the variable b x r sin φ cos θ r sin φ sin θ z r cos φ And the domain is defined b < r <, < φ < π 4, < θ < π. (4%) π 4 π r 3 sin φ dr dθ dφ (%) π ( 4 ) π 4 ( ) (%) 4. (%) Find the work done b the force field F (xz + z + ) i + (x z 4x + x) j + (x + xz + z) k in moving a particle along the curve parameterized b r(t) t i + sin t j + πt k, t. Let φ(x,, z) x z + xz x + x + z, then φ (xz + z + ) i + (x z 4x) j + (x + x + z) k φ(x,, z) F dr r() r() F φ + x j (4%) ( φ + x j) dr φ(x,, z) φ(x,, z) xd (, π, π) (,, ) r() r() + xd (4%) π π + π + π π t t t F dr π 4 (4%) 5. (%) Evaluate the line integral F dr counterclockwise around the region bounded b x and x, where F(x, ) sin(x 3 ) i + x + j. B Green Theorem and Fubini Theorem

3 Page 3 of 6 F dr x x + sin(x3 )da x x + ddx(6%) x + dxd(3%) x + d x + d ln( + ) ln.(3%) 6. (%) Evaluate the integral S F ˆNdS, where F (x + ) i + ( + e z ) j + (x sin + z) k, S is the surface of the region D inside the sphere x + + z 4 and the clinder x +, ˆN is the unit outward normal of S. B Divergence Theorem F ˆNdS S divfdv D 4dV 4V (4%) 4 D π 4 r 6π 3 (8 3 3).(%) 4 r rdzdrdθ(6%) 7. (%) Find the flux of F z i xz j + (x + ) k upward through the surface where u, v π. r(u, v) e u cos v i + e u sin v j + u k, r u (eu cosv, e u sinv, ) r v ( eu sinv, e u cosv, ) r u r v ( eu cosv, e u sinv, e u ) ( points)

4 Page 4 of 6 flux π π (ue u sinv, ue u cosv, e u ) ( e u cosv, e u sinv, e u )dudv e 4u dudv π 4 (e4 ) ( points) 8. (%) Evaluate F dr around the curve r(t) cos t i + sin t j + (sin t + cos t) k, t π, where F (e x 3 ) i + (e + x 3 ) j + (e z + x + ) k. Method : B Stokes theorem, F dr F Nds. ( pt) { x +, We also have : z x +. { x + z, Let : x + Hence is the boundar of. Its parametric equation is (x, ) xi+j+(x+)k,. x +. N is the upper normal of. Nds x dx d ( i j + k)dx d i j k F det x z e x 3 e + x 3 e z + x + (3 pts) i j + (3x + 3 )k (3 pts) F dr F Nds x + x + π r π 3 r4 4 ( i j + (3x + 3 )k ) ( i j + k)dx d (3x + 3 )dx d 3r r dr dθ 3π. (5 pts) Method : dr(t) ( sin t i + cos t j + (cos t sin t)k ) dt ( pts) ( F(r(t)) dr(t) ( e cos t + sin 3 t) sin t + (e sin t + cos 3 t) cos t ) +(e sin t+cos t + cos t + sin t)(cos t sin t) dt ( pts) e cos t d cos t + sin 4 t dt + e sin t d sin t + cos 4 t dt +e sin t+cos t d(sin t + cos t) + (cos t sin t)dt ( pts)

5 Page 5 of 6 π t F(r(t)) dr(t) ( e cos t + e sin t + e sin t+cos t) π π t + sin 4 t + cos 4 t + cos t sin t dt + π t π π t π + 8 π + π sin 4 t dt +, ( ) cos t dt since cos t + cos t dt π t + cos 4t d4t, since t π t sin n t dt πn π t cos s ds cos n t dt 3π (7 pts) 9. (%) Find the general solution of the differential equation e x + 4e x cos x. First, we consider the homogeneus equation + 3 +, tr e rx, we obtain the auxiliar equation r + 3r +, so (r + )(r + ), thus we get r,. So the general solution to the homogeneous equation is h e x + e x,, are constants Since the differential equation is nonhomogeneus, the solution to this equation is given b h + p, where p is the particular solution to this equation. To compute p. Method I we need to solve the following two equations e x () e x cos x () For (), since e x is alread a solution to this differertial equation, we tr Axe x, and we get A, so the particular solution to () is given b p xe x For (), we tr Be x cos x + e x sin x, and we get B,, so the particular solution to () is given b p e x cos x + e x sin x. Since this is a linear differential equation, we have the particular solution to e x + 4e x cos x, sa p, is the sum of the particular solution of () and (), i.e., p p + p xe x e x cos x + e x sin x So, the solution to this equation is given b h + p e x + e x + xe x e x cos x + e x sin x You can also tr p Axe x + Be x cos x + e x sin x directl to solve A, B, Method II The second method use variation of paramrters. For the homogeneus equation + 3 +, from the above calculation, we have there are two linearl independent solutions, sa e x, e x, and we search the particular solution of the form p u (x) (x) + u (x) (x). For simplicit, we choose p to satisf the differential equation u + u, and b this assumption, for this p need to satisf this differential equation e x + 4e x cos x, we must have u + u e x + 4e x cos x, so we have following two equations, u + u u e x + u e x (3) u + u u e x u e x e x + 4e x cos x (4) (3) + (4) u e x e x + 4e x cos x u + 4 cos x,so, u x + 4 sin x + 3 for som constant 3. (3) + (4) u e x e x + 4e x cos x u e x 4e x cos x.

6 Page 6 of 6 Using integration b parts, we get e x cos xdx e x sin x e x sin xdx e x sin x + e x cos x e x cos xdx, so e x cos xdx (ex sin x + e x cos x). Thus, u e x (e x sin x + e x cos x) + 4, where 4 is a constant. So, p u (x) (x) + u (x) (x) u (x)e x + u (x)e x (x + 4 sin x + 3 )e x + ( e x (e x sin x + e x cos x) + 4 )e x xe x e x + e x sin x e x cos x + 3 e x + 4 e x Thus the general solution is h + p e x + e x + xe x e x + e x sin x e x cos x + 3 e x + 4 e x 5 e x + 6 e x + xe x + e x sin x e x cos x. for some constant 5, 6.. (%) Solve the differential equation in the interval x >. x 3 + x Let x r (%),then we substitute it. We have x r [(r )r(r ) + r ]. [r ] 3 r,, (7%) Then b checking detail (%) the solution is c x + c x ln x + c 3 x(ln x) (%).