MVC
Chapter 6 Quiz
NAME:
Instructions: SHOW ALL WORK !!
We don’t really need a calculator, do we?

2t 3 
Let the curve C be given by:  x(t ), y(t ), z (t )    t , t 2 ,
 with 1  t  2 . Set up each
3 

integral below as a single integral in terms of t.
#1.
a.
ds 

xyz ds
 x(t ) 2   y(t ) 2   z(t ) 2 dt 
  
xyz ds  (t ) t
1
b.
C
C
2
C



C
2
1  (2t ) 2  (2t 2 ) 2 dt 
 2t 3 
2
2 2

 1  (2t )  (2t ) dt 
 3 
2

1
2 6
t 1  4t 2  4t 4 dt
3
F  d s given Fx, y, z = zi + yj + xk.
F  d s   zdx  ydy  xdz
C

2
2t 3

dt  t 2 (2tdt )  t (2t 2 dt )
3
1
2
 2t 3

 
 4t 3  dt
3

1
#2.
Find a parameterization for each of the following curves:
a. The line segment from (1, 2, 4) to (  2, 2, 0)
 x  3t  1

 y  2 0  t 1
 z  4t  4

b. The curve that starts at the point (0, 0, 0) and goes to the point (1, 0,1) along the circular
paraboloid z  x 2  y 2 .
 x  t cos  t 

 y  t sin  t  0  t  1

z  t2

MVC
Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve y = x3, and
then returning to the origin along the straight line y = x.
#3.
a. Parameterize the path (in two pieces, most likely) to express the line integral
2
 2 x y dx  xy dy as an integral or integrals in the single variable t. DON’T integrate.
C
 2x
2
y dx  xy dy 
C
 2x
2
C1
y dx  xy dy 
 2x
2
y dx  xy dy
 C2
1
1
0
0
  2t 2t 3 dt  tt 3 3t 2 dt   2t 2t dt  tt dt
1
  (2t 5  3t 6  2t 3  t 2 ) dt
0
b. Apply Green’s Theorem to the integral in part a to obtain a double integral, making sure
to provide appropriate limits of integration. DON’T integrate.
1 x
 N M 
2
2
x
y
dx

xy
dy

C
0 3  x  y 
x
1 x
 N M 
 


x y 
0 x3 
1 x
    y  2 x 2 dydx
0 x3
MVC
#3. (continued) Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve
y = x3, and then returning to the origin along the straight line y = x.
c. Given the vector field F( x, y)  2 x 2 y i  xy j , write the integral(s) in the single variable t
you would need to evaluate to find the outward flux of F across the curve C. DON’T
integrate.
  N dx  M dy    xy dx  2 x
C
2
y dy
C

  xy dx  2 x
2
y dy 
  xy dx  2 x
2
y dy
 C2
C1
1
1
  tt dt  2t t 3t dt   tt dt  2t 2t dt
3
2 3
2
0
0
1
  (t 4  6t 7  t 2  2t 3 )dt
0
d. Apply the divergence form of Green’s theorem to obtain a double integral that would
calculate this flux. DON’T integrate.
1 x
  N dx  Mdy    div( F )dA
C
0 x3
1 x
 M N 


 dA
x y 
0 x3 
1 x
    4 xy  x dA
0 x3
MVC
#4 Use Green’s Theorem (with a thoughtful choice of F ( x, y )  M ( x, y )iˆ  N ( x, y ) ˆj ) to set up a line
integral that gives the area of the region inside the curve C : x(t )  t sin t , y (t )  3cos t;0  t  2 .
Write the line integral in terms of t. [Note that C is simple, smooth, closed curve].
Let F ( x, y )  M ( x, y )iˆ  N ( x, y ) ˆj  xjˆ; i.e. M ( x, y )  0 and N ( x, y )  x . Then,
 N M 
Area( D)  1dA  

 dA   M ( x, y)dx  N ( x, y)dy 
y 
D
D  x
C
2
 xdy   (t sin t )(3sin t )dt
C
0
#5. Set up a double integral that represents the area of the parametrized surface:
X  (2 s  3t  1, t , s  t ); 1  s  1,0  t  1
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x2  y 2
#6. Find a parameterization for the portion of the surface z 
that lies above the region
x2
1
1
4
in the xy-plane bounded by the curves y  x, y  3x, y  , and y  .
3
x
x
MVC