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Quiz 6 Solution: Line Integral of Conservative Vector Field on Parabolic Curve - Prof. Sea, Quizzes of Advanced Calculus

The solution to quiz 6 in math 2203, where students are required to evaluate the line integral of a conservative vector field f(x, y) = 32j over the parabolic curve y = x² from (0, 0) to (1, 1). The solution utilizes the fundamental theorem of line integrals and the definition of line integrals.

Typology: Quizzes

2010/2011

Uploaded on 06/03/2011

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MATH 2203 –Quiz 6 (Version 1) Solution
November 20, 2009
NAME________________________________
Let Fbe the vector eld
F(x; y) = 32j
and let Cbe the parabolic curve y=x2beginning at the point (0;0) and
ending at the point (1;1).
Evaluate the line integral
ZC
Fdr.
You must include all details of your work.
Solution: The easiest way to do this is to use the Fundamental Theorem
of Line Integrals: The vector eld F(x; y) = 32jis conservative and has
potential function f(x; y) = 32y. Thus
ZC
Fdr=ZC
rfdr=f(1;1) f(0;0) = 32 0 = 32.
Now we show how to do the integral using the de…nition of line integrals:
The curve Ccan be parameterized as
r(t) = ti+t2j
0t1
from which we see that
r0(t) = i+ 2tj
and
F(r(t)) r0(t) = (32j)(i+ 2tj) = 64t.
Thus ZC
Fdr=ZC
F(r(t)) r0(t)dt =Z1
0
64t dt =32.
1

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MATH 2203 ñQuiz 6 (Version 1) Solution November 20, 2009

NAME________________________________

Let F be the vector Öeld

F (x; y) = 32 j

and let C be the parabolic curve y = x^2 beginning at the point (0; 0) and ending at the point (1; 1). Evaluate the line integral Z

C

F  dr.

You must include all details of your work. Solution: The easiest way to do this is to use the Fundamental Theorem of Line Integrals: The vector Öeld F (x; y) = 32 j is conservative and has potential function f (x; y) = 32 y. Thus Z

C

F  dr =

Z

C

rf  dr = f (1; 1) f (0; 0) = 32 0 = 32.

Now we show how to do the integral using the deÖnition of line integrals: The curve C can be parameterized as

r (t) = ti + t^2 j 0  t  1

from which we see that r^0 (t) = i + 2tj

and F (r (t))  r^0 (t) = ( 32 j)  (i + 2tj) = 64 t.

Thus (^) Z

C

F  dr =

Z

C

F (r (t))  r^0 (t) dt =

Z 1

0

64 t dt = 32.