(10 points) 1. Give a parameterization in vector form of the plane containing the points (1, 2, 3), (1, 3, 5),
and (2, 1, 4).
!
(12 points) 2. Let F! = 2x!i + 6y 2!j + 12z 3!k. Calculate C F! · d!r, where C is the path from (4, 0, 0) along the
quarter circle in the xy-plane to (0, 4, 0), followed by the path to (0, 0, 0) along the y-axis, and
then along the z-axis to (0, 0, 5).
!
(10 points) 3. Calculate C ((x2 − y)!i + (y 2 + x)!j) · d!r, where C is the path along the graph of the parabola
y = x2 + 1 from (0, 1) to (1, 2).
!
(12 points) 4. Use Green’s Theorem to calculate C ((2x2 + 3y)!i + (2x + 3y 2 )!j) · d!r, where C is the path
around the triangle with vertices (2, 0), (0, 3), (−2, 0), oriented counterclockwise.
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(14 points) 5. Let S be the open cylindrical can with a bottom, but no top, given by the union of the cylinder
x2 +y 2 = 9, 0 ≤ z ≤ 2, and the disk x2 +y 2 ≤ 9, z = 0. If S is oriented outward and downward,
find
!
!
curl(−y!i + x!j + z!k) · dA.
S
(Hint: The boundary of S is the circle x2 + y 2 = 9 in the plane z = 2.)
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(14 points) 6. Use the Divergence Theorem to find the flux of F! through the closed cylinder of radius 2,
centered on the z-axis, with 3 ≤ z ≤ 7, where F! = (x + 3eyz )!i + (ln(x2 z 2 + 1) + y)!j + z!k.
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(14 points) 7. Calculate directly the flux of the vector field F! = x!i + y!j through the part of the surface
z = 25 − (x2 + y 2 ) above the disk of radius 5 centered at the origin in the xy-plane, oriented
upward.
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!
(14 points) 8. Use Stokes’ Theorem to find C (−z!j + y!k) · d!r, where C is a circle of radius 2 in the plane
x + y + z = 3, centered at (1, 1, 1), and oriented clockwise when viewed from the origin.
(Hint: A unit normal vector for the plane x + y + z = 3 is !n =
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√1 (!i
3
+ !j + !k).)