Quiz #6
Math 223-01 (Salomone)
August 3, 2007
Show all your work!
Name: Solution
| {z }
Prob.1
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| {z }
Prob.2
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| {z }
Prob.3
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Problem 1. (8 points) For each of the following vector field /curve combinations, simply circle the techniques
that you could use to evaluate RCF·dr. There may be more than one answer. No partial credit will be
assessed on this problem.
(a) (2 points) F(x,y,z)=hy,z,xiand Cis the triangle from (0,0,0) to (1,0,2) to (−1,−1,0) and back to (0,0,0).
Path Independence Green’s Theorem Stokes’ Theorem None of the Above
Note that Cis a closed curve, and ∇ × F=h−1,−1,−1i,0so Fis not path-independent.
(b) (2 points) F(x,y)=hx2−2xy +y2,x2−2x y −y2iand Cis the line segment from (−3,3) to (6,2).
Path Independence Green’s Theorem Stokes’ Theorem None of the Above
Here, Cis not a closed curve, and ∇ ∗ F=2x−2y−(2y−2x)=−4x−4y,0, so it is not path independent.
(c) (2 points) F(x,y,z)=hx yz,xyz,xyziand Cis the curve parameterized by r(t)=(cos t,sin t,1+sin2t),
0≤t≤2π.
Path Independence Green’s Theorem Stokes’ Theorem None of the Above
Since r(0) =(1,0,1) =r(2π), this is a closed curve. However, ∇ × F=hxz −xy,x y −yz,yz −xzi,0so it is not path independent.
(d) (2 points) F(x,y)=*x
√x2+y2−1,y
√x2+y2−1+and Cis the circle x2+y2=4, traversed counterclockwise.
Path Independence Green’s Theorem Stokes’ Theorem None of the Above
This vector field is undefined on the unit disk x2+y2≤1; since the path Cencircles this hole, all bets are off. Even though F
happens to pass the screening test and Cis a closed curve, none of these techniques is guaranteed to give the right answer.
Problem 2. (8 points) A particle moves along a rectangular path from (−2,0) to (3,0) to (3,1) to (−2,1) and
back to (−2,0). How much work does the force field F(x,y)=hsin2x+2x y2,yy+4x2yido along this path?
Since the path involved is a closed curve in R2, we may use Green’s theorem. The region Rinside this
curve is the rectangle −2≤x≤3, 0 ≤y≤1. Check that the curve Cis counterclockwise, so no extra
negative signs are necessary. By Green’s theorem, we have
IC
F·dr=x
R ∂Q
∂x−∂P
∂y!dA.
We compute
∂Q
∂x−∂P
∂y=∂
∂x(yy+4x2y)−∂
∂y(sin2x+2xy2)=8xy −4x y =4xy.
Thus the integral on the right-hand side of Green’s theorem is
x
R
4xy dA =Z1
0Z3
−2
4xy dxdy =Z1
0
2x2y
x=3
x=−2dy
=Z1
0
10y dy =5y2
1
0=5.
