we can conclude that ϕ(x, y, z) = sin (xz) + e yz + const. If ϕ is written as a vector but the above three calculations are right, you lose 3pts.

Transcription

1 5 微甲 6- 班期末考解答和評分標準. (%) Let F = z cos(z)i + ze yz j + ( cos(z) + ye yz )k. (a) (8%) Find a scalar function ϕ(, y, z) such that ϕ = F. (b) (%) Evaluate C F dr, where C is the curve r(t) = (cos(t ), ln(t + ), tan (t)), t. (a) Because ϕ (, y, z) = z cos (z) ϕ(, y, z) = sin (z) + g (y, z) ϕ y (, y, z) = ze yz ϕ(, y, z) = e yz + g (, z) ϕ z (, y, z) = cos (z) + ye yz ϕ(, y, z) = sin (z) + e yz + g (, y), we can conclude that ϕ(, y, z) = sin (z) + e yz + const. If ϕ is written as a vector but the above three calculations are right, you lose pts. (b) Because F is conservative, C F dr = ϕ(r()) ϕ(r()) = ϕ(, ln, /) ϕ(,, ) = + /.. (%) Let C be the polar curve defined by r = cos θ in the first quadrant. Evaluate C y ds. Let = r cos θ, y = r sin θ (pt) C yds = r sin θ r + ( dr dθ ) dθ (pt) = r sin θ sin θ cos θ + ( cos θ ) dθ = r sin θ cos θ dθ (pt) = sin θdθ = cos θ = (pt). (%) Sketch the region of integration and evaluate the integral 8 cos ( + y + ) dyd. Page of 7

2 ( points) 8 cos( + y = = = ) dy d y + cos( ) d dy y + [(y + ) sin( y + )] (y + ) [sin(y )] dy y dy ( points) Let u = y +, dv = sin(y )dy, and by applying Integration by parts, (y + ) [sin(y )] dy = [ (y + )cos(y )] + [sin(y )] ( points) = cos() + + sin() ( points) = + sin() cos() Note:. If you sketch the wrong part of the region, you will gain ( points) for sketching part.. If you write ( points). 8 y instead of the correct answer, you will gain. If you write ( point). 8 y instead of the correct answer, you will gain. If you miss the minus in Integration by parts, you will gain ( point).. (%) Evaluate R e +y y da, where R is the region in the y-plane bounded by the four lines y =, y =, Page of 7

3 y = and y =. Method : (, y) Let u = + y ; v = y, then we can get J = (u, v) = 5 Therefore, e +y y da = R v v e u v J dudv = v v Method : (, y) Let u = y ; v = y, then we can get J = (u, v) = 5 Therefore, e +y y da = R Method : u e u v u J dvdu = u Let u = y ; v = y), then we can get J = (, y (u, v) = Therefore, e +y y da = R e v+ v J dudv = e u v 5 dudv = 5 ve u v u=v u=v = 5 e v u e 5 dvdu = (e e). u (v ) e v+ v u (v ) dudv = (e e). 給分標準 : J 算錯扣 分積分上下界找錯扣 分前面兩項找對答案算錯扣 分. v(e e) = (e e). 5. (6%) Let E be the space region bounded by the surfaces S = {(, y, z) z = + + y, z }, S = {(, y, z) z = y, z }, and V(, y, z) = yi + j + zk. z S (a) (6%) Evaluate divv dv, where Ω is the solid region enclosed by S S. Ω S y (b) (%) State the Divergence Theorem and evaluate the total outward flu S S V ds. (c) (6%) Compute the upward flu of V across S. Solution 5(a). Ω divv dv = Ω dv = = 5 6 r +r rdzdrdθ Note:. The shape of region is not hemisphere.. divv =, points. Volume of region =, points Page of 7

4 . Volume of region =, points Solution 5(b). Divergence Theorem: Ω: simple solid region S:boudary surface of Ω with positive outward orientation. V = (P, Q, R): vector field with P, Q y, R z continuous on an open region that contains Ω. Then we have Ω V dv = S V ds V ds = 5 S S 6 Note:. Ω, point. P, Q y, R z, point. Ω V dv = S V ds, point. S S V ds = 5 6, point Solution 5(c). Method : Direct calculation r(u, v) = (u cos v, u sin v, u ), r u = (cos v, sin v, u) r v = ( u sin v, u cos v, ) r u r v = (u cos v, u sin v, u) V r u r v = u u u, v S V ds = u u dudv = Method Let S = {(, y, z) + y, z = }. Then V n = on S where n = k which implies This leads to S V ds = S V ds = Ω V dv S V ds = where volume of region has been calculated in part (a). Note: S V ds = Ω V dv is WRONG. Method S V ds = Ω V dv S V ds = 5 6 where the surface integral of S is as follows: = r(u, v) = (u cos v, u sin v, + u), r u = (cos v, sin v, ) r v = ( u sin v, u cos v, ) r v r u = (u cos v, u sin v, u) V r u r v = u u u, v Page of 7

5 Note: Be careful of the sign of the normal S V ds = u u dudv = 6. (%) Evaluate C ( y + y)d (y )dy, where C is described by the polar equation r = cos θ oriented counterclockwise. Let D = { r cos θ, θ } be the region bound by the curve C. Apply Green s Theorem, C ( y + y) d (y ) dy = D = D ( y + ) ( y + ) da ( y y ) da. (6 points) Use polar coordinate = r cos θ, y = r sin θ, D ( y y ) da = = = ( r sin θ) r dr dθ (5 points) (r r= cos θ 5 r5 sin θ ) dθ r= ( cos θ) 5 sin θ( cos θ)5 dθ = + + [ ( cos θ)6 ] θ= θ= =. ( points) 7. (%) Find the area of the sphere + y + z = lying inside the cylinder ( ) + y =. The area = + z ( ) +y + zyda [ points] (z = y, z = y, z y y = y ) [ points] = ( ) +y y da [ points] (Use the polar coordinate to calculate) = cos θ rdrdθ r [ points] = = 8 = 8 cos θ ( r ) dr dθ [ points] ( r ) r= cos θ r= dθ [ points] sin θdθ [ points] = 8( ) [ points] 8. (%) Compute S curlf ds, where F = e z i+( +z )j+(y+cos z)k and where S = {(, y, z) + y 9 + z = and + z } oriented so that the boundary is counterclockwise when viewed from above. Page 5 of 7

6 y y z 9 S z z Method : curlf = F = z, e z, ( pts) Let T be the intersection of the ellipsoid with the plane + z =. Note that the boundary of S and T are the same. On the plane + z = which is z =. Parametrize T by γ(, y) = (, y, γ = (,, ), γ y = (,, ) γ γ y = ( Let D be the projection of T onto y plane. Now, by Stoke s Theorem 5 By symmetry, D ddy =. So ) with the condition + y 9 ( + y,, ) ( pts) S curlf ds = T curlf ds + y 9 Thus S curlf ds =. 9 + z ). = +, e, (,, )ddy D = + y ddy (6 pts) + 5 ddy = Area(D) = = Method : C is the intersection of + y 9 + z = and + z =. We have z = and + y =. So we can parametrize C as 9 ( pts) On C, we have γ(θ) = ( cos(θ) cos(θ), sin(θ), ), θ ( pts) F = e cos (θ), 5 cos(θ) cos (θ), sin(θ) + cos( ) ( pt) γ (θ) = sin(θ) sin(θ), cos(θ), ( pt) F γ (θ) = sin(θ)e cos (θ) + 5 cos (θ) + sin (θ) + sin(θ) cos( cos(θ) ) Then F dr = ( sin(θ)e cos (θ) + 5 cos (θ) + sin (θ) + sin(θ) cos( cos(θ) ) )dθ ( pts) Page 6 of 7

7 Note that f(θ) = sin(θ)e cos (θ) cos(θ) sin(θ) cos( ) + is an odd function. Thus f(θ)dθ = f(θ)dθ =. We also have cos (θ)dθ =. So C F dr = sin (θ) dθ = ( cos(θ)) = = ( pts) Thus, by Stoke s Theorem S curlf ds = ( pts). Page 7 of 7