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Test 2, MATH 550
Instructions: Write your name legibly. Name: There are 100 points. Show all work to get partial credit. You can not use calculators capable of symbolic computations, calculating deriva- tives or integrals.
(1) Let C be the polygonal path from (1, 0 , 0) to (1, 1 , 0) and then from (1, 1 , 0) to (1, 1 , 2) and let F = (2xy + yz)i + (x^2 + xz + z^2 )j + (xy + 2yz)k.
a. (8pts) Compute
C F·dR^ by parametrizing^ C(use 2 pieces). Solution:
C F·dR^ =^
0 F^2 (1, y,^ 0)^ dy+
∫ 1 0 F^3 (1,^1 , z)^ dz^ = 0 1 dy^ +^
0 (1 + 2z)^ dz^ = 1 + 6 = 7. b. (8 pts )Compute that the curl ∇ × F = 0 by explicitly showing all parts. Solution: ∇ × F = ((x + 2z) − (x + 2z))i − (y − y)j + ((2x + z) − (2x + z))k = 0. c. (8pts )Find a potential of F. Solution: φ(x, y, z) = x^2 y + xyz + z^2 y.
d. (6 pts) Recompute
C F^ ·^ dR^ by using this potential. Solution:
C F^ ·^ dR^ =^ φ(1,^1 ,^ 2)^ −^ φ(1,^0 ,^ 0) = (1 + 2 + 4)^ − 0 = 7. (2) Let S be the open top box bounded by x = 0, x = 1, y = 0, y = 1, z = 0, and z = 2. Let F = −yi + xj + (z − x)k. Let n denote the unit outward normal of the box.
a. (10 points) Compute ∇ × F. Solution: ∇ × F = 0i − (−1)j + (1 + 1)k = j + 2k. b. (10 pts) Compute
S ∇ ×^ F^ ·^ n^ dS^ by computing this in- tegral over each of the five faces of S. Solution: n = i on front, so ∫ ∫
front
∇ × F · n dS =
front
0 dS = 0.
Similarly ∫ ∫
back
∇ × F dS =
back
0 dS = 0.
On the bottom n = −k, so ∫ ∫
bottom
∇ × F · n dS =
bottom
− 2 dS = − 2.
On the lefts side n = −j ,so ∫ ∫
left
∇ × F · n dS =
left
− 1 dS = − 2.
On the right side n = j, so ∫ ∫
right
∇ × F · n dS =
right
1 dS = 2.
Hence
S ∇ ×^ F^ ·^ n^ dS^ =^ −2 + (−2) + 2 =^ −2.
(4) Let S be the surface of the cylinder x^2 + y^2 = 4 with 0 ≤ z ≤ 3 and let F = xi + (y + z)j − 2 zk.
a. (15 pts) Compute directly
S F^ ·^ n^ dS, where^ n^ denotes the unit outward normal. Solution: The unit outward normal is n = xi+ 2 yj, so F · n = x
(^2) +y (^2) +zy 2 =^
4+zy 2 on^ S. Also^ dS^ = 2dθdz^ in cylindrical coordinates, so ∫ ∫
S
F · n dS =
∫ (^2) π
0
0
4 + 2z sin θ dzdθ = 24π.
b. (5 pts) Compute ∇ · F. Solution: ∇ · F = 1 + 1 − 2 = 0. c. (5 pts) If S 1 denotes the closed surface bounded by z = 0, z = 3, and S, compute the value of
S 1 F^ ·^ n^ dS^ by using the divergence theorem. Solution: By the divergence theorem
∫∫∫^ S^1 F^ ·^ n^ dS^ = D 0 dV^ = 0. d. (5 pts) Use the information from c. to recompute
S F^ ·
n dS from part a (Compute over the top and bottom of the cylindrical region). Solution: By part c. we know that
S F^ ·^ n^ dS^ = −
top F^ ·^ n^ dS^ −^
bottom F^ ·^ n^ dS.^ Now^ n^ =^ k^ on top, so F · k = − 2 z = −6 on top. Hence
top F^ ·^ n^ dS^ = − 6 × area of the top = − 6 × 4 π = − 24 π. On the bottom n = −k, so F · n = 2z = 0, so
bottom F^ ·^ n^ dS^ = 0. Hence
S F^ ·^ n^ dS^ =^ −(−^24 π)^ −^ 0 = 24π, which agrees with part b..
