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MATH 2203 ñFinal Exam (Version 1) Solutions December 7, 2007 S. F. Ellermeyer Name
Instructions. Your work on this exam will be graded according to two criteria: mathe- matical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just write down an ìanswerî with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator but you may not use any books or notes.
- Match each of the twoñdimensional vector Öelds, F, given in parts añe with its graph (given in 1ñ10 on the separately attached sheet).
a) F (x; y) =
y p x^2 + y^2 + 1
i +
x p x^2 + y^2 + 1
j matches Graph 4.
b) F (x; y) = y p x^2 + y^2 + 1
i + x p x^2 + y^2 + 1
j matches Graph 9.
c) F (x; y) = x p x^2 + y^2 + 1
i + y p x^2 + y^2 + 1
j matches Graph 7.
d) F (x; y) = p y^2 y^4 + x^2 + 1
i + p x y^4 + x^2 + 1
j matches Graph 2.
e) F (x; y) = p y x^2 + y^2 + 1
i + p x x^2 + y^2 + 1
j matches Graph 3.
- Let F be the vector Öeld F (x; y) = x sin (y) i + yj and let C be the oriented parabolic curve y = x^2 starting at the point ( 1 ; 1) and ending at the point (2; 4). (The vector Öeld and curve are pictured below.)
- (a) Evaluate the line integral (^) Z
C
F dr.
(b) If some curve, C, beginning at the point ( 1 ; 1) and ending at the point (2; 4) other than the one you used in part a was used, would you expect to get the same answer that you got in part a for
R
C F^ ^ dr? Explain why or why not. Solution: The curve C can be parameterized as
x = x y = x^2 1 x 2.
Thus Z
C
F dr =
Z
C
P (x; y) dx + Q (x; y) dy
Z 2
1
x sin
x^2
dx
cos
x^2
x^4
x=
x= 1
cos (4) + 8
cos (1) +
If some other curve beginning atR ( 1 ; 1) and ending at (2; 4) is used to compute C F^ ^ dr, then we would expect to get a di§erent answer because the vector Öeld F is not conservative.
(b) The curve pictured below is called an epicycloid. It has parametric equations x = 5 cos (t) cos (5t) y = 5 sin (t) sin (5t) 0 t 2 . Use the formula given in part a to Önd the area of the region, D, bounded by the epicycloid.
Solution: Area (D) =
Z
C
x dy y dx
Z 2
0
[(5 cos (t) cos (5t)) (5 cos (t) 5 cos (5t))
(5 sin (t) sin (5t)) ( 5 sin (t) + 5 sin (5t))] dt. Now we simplify the integrand: (5 cos (t) cos (5t)) (5 cos (t) 5 cos (5t)) (5 sin (t) sin (5t)) ( 5 sin (t) + 5 sin (5t)) = 5
5 cos^2 (t) 6 cos (t) cos (5t) + cos^2 (5t)
5 sin^2 (t) + 6 sin (t) sin (5t) sin^2 (5t)
= 5 [5 + 1 6 cos (4t)] = 30 (1 cos (4t)). Thus Area (D) = 15
Z 2
0
(1 cos (4t)) dt = 30.
- Let F be the vector Öeld
F (x; y; z) = x^2 yzi + xy^2 zj + xyz^2 k.
(a) Compute curl(F). (b) Compute div(F). (c) Is F conservative? Why or why not? (A yes or no answer will not su¢ ce here. You must explain why or why not.) Solution:
curl (F) =
i j k @ @x
@ @y
@ @z x^2 yz xy^2 z xyz^2 =
xz^2 xy^2
i
yz^2 x^2 y
j +
y^2 z x^2 z
k and div (F) =
@x
x^2 yz
@y
xy^2 z
@z
xyz^2
= 6xyz.
Note that F is not conservative because curl (F) 6 = 0.
- Let S be the surface of the tetrahedron with vertices (0; 0 ; 0) ; (1; 0 ; 0) ; (0; 1 ; 0) ; and (0; 0 ; 1) with the positive (outward) orientation. This tetrahedron is pictured below. Let F be the vector Öeld F (x; y; z) = yi + (z y) j + xk. Evaluate (^) Z Z
S
F dS.
(0,0,0)
(1,0,0)
(0,1,0)
(0,0,1)
Solution: Let Sxz be the face of the tetrahedron that lies in the xz plane. Let Syz be the face that lies in the xz plane. Let Sxy be the face that lies in the xz plane. Let S 0 be the other face.
Since S 0 is the graph of the function z = 1 x y, we have Z Z
S 0
F dS =
Z Z
D
( P zx Qzy + R) dA
Z Z
D
(z + x) dA
Z Z
D
(1 y) dA
Z 1
0
Z (^1) x
0
(1 y) dy dx
=
In conclusion Z Z
S
F dS =
Z Z
Sxz +Syz +Sxy +S 0
F dS =
