Line Integral Evaluation

  • #1
NewtonApple
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Homework Statement


Evaluate the line integral [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
(a) a straight line from (0,1) to (1,2);
(b) the parabola x=t, y=t2 + 1;
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

Homework Equations


Equation of line: y = mx + c

The Attempt at a Solution


(a) [/B]To convert given integral into one variable I used Equation of line
y=mx + c
where slope m = [itex]\frac {y_2 - y_1}{x_2 - x_1} = \frac{2-1}{1-0}=1[/itex]
and y intercept c = 1
which gives us y = x+1
dy = dx

Thus given integral

[itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]

becomes one variable x dependent

[itex] = ∫^{1}_{0} (x^2-x-1)dx + (({x+1})^2+x)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+2x+1+x)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+3x+1)dx[/itex]
[itex] = ∫^{1}_{0} (x^2-x-1+x^2+3x+1)dx[/itex]
[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex] ------------------ (1)

solving
[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]
applying limits gives
[itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

(b) need some hints for this part

(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

from equation (1) above
[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex]

solving
[itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]

from (0,1) to (1,1)
we use only x: 0 to 1
we get [itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

from (1,1) to (1,2)
we use only x: 1 to 1
we get [itex]\frac{2}{3} - \frac{2}{3} + 1 - 1 = 0 [/itex]

hence adding both parts gives [itex] \frac{5}{3} [/itex]


Please tell me am I on the right track?
 
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Answers and Replies

  • #2
ehild
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Homework Statement


Evaluate the line integral [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
(a) a straight line from (0,1) to (1,2);
Check the upper boundary, it must be (1,2). With that boundary, your answer to (a) is correct.

(b) the parabola x=t, y=t2 + 1;

What is t at the lower boundary x=0, y=1, and what is it at x=1, y=2?
Write x, dx, and y, dy in the integrand in terms of t, and integrate with respect to t.

(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

from equation (1) above
[itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex]

You can not use eq (1). Separate the integral from(0,1) to (1,1) along the straight line connecting these points (a horizontal path) and then from (1,1) to (1,2) (a vertical path).
 
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  • #3
ARaslan
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a)
A vector parallel to the line segment from (0,1) to (1,2) is <1,1>. Therefore the vector equation for the line segment is r(t) = <0,1> + t<1,1> = <t, t +1>. So the parameter of the line integral is :
x =t
y = t+1
0 <= t <= 1.
dy/dt = 1
dx/dt = 1

so the integral becomes: ∫(t^2 - t - 1) + (t+1)^2 +t dt from 0 to 1
= ∫2t^2 + 2t dt from 0 to to 1 which evaluates to 5/3.

follow the same procedure for the rest of the section.
1)parameter
2)find dy/dt and dx/dt
3)plug in the x's and y's in term of t
4)plug in dy = dy/dt * dt and dx = dx/dt *dt
5) integrate from 0 to 1

note: You will have to divide C into to problems identical to A.
 
  • #4
NewtonApple
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Check the upper boundary, it must be (1,2). With that boundary, your answer to (a) is correct.

yes, you are right. It's a typo, upper bound should be (1,2).
 
  • #5
ARaslan
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The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
 
  • #6
NewtonApple
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The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.

The problem is from Mathematical Methods for Physicists by Tai L. Chow, see the attached file.
 

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  • #7
ARaslan
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The problem is from Mathematical Methods for Physicists by Tai L. Chow, see the attached file.

Yes, those limits given are the exact same limits given by the line segments. My guess is that those limits are given for problem b. For problems A and C, ignore the given limits and focus on the limits given by the line segment
 
  • #8
haruspex
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The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
No, it makes sense, but it in (a) and (c) it's redundant because the bounds are also given separately. You could equally well complain, instead, that the curve specifications should be:
(a) a straight line segment;
(b) a segment of the parabola x=t, y=t2 + 1;
(c) two straight line segments meeting at (1,1)
Does it really matter?
 
  • #9
NewtonApple
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So for part (b)
Evaluate the line integral [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
(b) the parabola x=t, y=t2 + 1

x= t
dx = dt

[itex]y= t^2 +1[/itex]
dy = 2tdt

when x=0 then t=0
when x=1 then t=1
when y=1 then t = imaginary
when y=2 then t= 1

Hence given integral
[itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]
in terms of t is
[itex](t^2 -t^2 -1)dt + [(t+1)^2 + t]2tdt[/itex]
[itex](-1)dt + [(t^2 + 2t +1) + t]2tdt[/itex]
[itex]∫^{1}_{0} [-1 + 2t^3 + 6t^2 +2t]dt[/itex]
[itex]-t + \frac{2}{4}t^4 + \frac{6}{3}t^3 + \frac{2}{2}t [/itex]
after applying lower and upper bound we get
[itex]-1 +\frac{1}{2} + 2 + 1[/itex]
[itex] \frac{1}{2} + 2 = \frac{5}{2}[/itex]

Which doesn't seem right.
 
  • #11
NewtonApple
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When y=1 then t = 0

No. Try that again.

You dropped an exponent.

thx! yes and it should be
[itex] (t^2 -t^2 -1)dt + [(t^2 + 1)^2 +1]2dt[/itex]

After solving and applying bounds we get [itex] \frac{5}{3}[/itex]
 
  • #12
haruspex
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When y=1 then t = 0



thx! yes and it should be
[itex] (t^2 -t^2 -1)dt + [(t^2 + 1)^2 +1]2dt[/itex]

After solving and applying bounds we get [itex] \frac{5}{3}[/itex]
I get a slightly higher value. Please post your working.
 
  • #13
ARaslan
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No, it makes sense, but it in (a) and (c) it's redundant because the bounds are also given separately. You could equally well complain, instead, that the curve specifications should be:
(a) a straight line segment;
(b) a segment of the parabola x=t, y=t2 + 1;
(c) two straight line segments meeting at (1,1)
Does it really matter?

Well technically it does not; however a line integral is defined as an integral which is taken around a curve. By definition line integrals must be calculated along a certain path, so it does not make sense to give coordinates in the integral itself.
 
  • #14
NewtonApple
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I get a slightly higher value. Please post your working.


Ok! solving again

Evaluate the line integral [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
(b) the parabola x=t, y=t2 + 1

x= t
dx = dt

[itex]y= t^2 +1[/itex]
dy = 2tdt

when x=0 then t=0
when x=1 then t=1
when y=1 then t = 0
when y=2 then t= 1

Hence given integral
[itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]
in terms of t is
[itex](t^2 -t^2 -1)dt + [(t^2+1)^2 + t]2tdt[/itex]
[itex](-1)dt + [(t^4 + 2t^2 +1) + t]2tdt[/itex]
[itex][-1 + t^4 + 2t^2 +1 + t]2tdt[/itex]
[itex]∫^{1}_{0} [2t^5 + 4t^3 +2t^2]dt[/itex]
[itex] \frac{2}{6}t^6 + \frac{4}{4}t^4 + \frac{2}{3}t^3 [/itex]
after applying lower and upper bound we get
[itex]\frac{1}{3} + 1 + \frac{2}{3} [/itex]
[itex] \frac{1+3+2}{3} = \frac{5}{3}[/itex]
 
  • #15
ehild
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after applying lower and upper bound we get
[itex]\frac{1}{3} + 1 + \frac{2}{3} [/itex]
[itex] \frac{1+3+2}{3} = \frac{5}{3}[/itex]
1+3+2=6 :D
 
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  • #16
NewtonApple
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1+3+2=6 :D
:oops: now I know my problem, I don't know simple arithmetic :).

Jokes apart. Integral suppose to be path independent so why two different answers for part (a) and (b)?
 
  • #17
haruspex
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You also went wrong in this step
$$(-1)dt + [(t^4 + 2t^2 +1) + t]2tdt$$
$$[-1 + t^4 + 2t^2 +1 + t]2tdt$$
but it turned out to make no numerical difference.
 
  • #20
NewtonApple
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You also went wrong in this step

but it turned out to make no numerical difference.

Part(b) Correction

[itex]=\left(-1+2t^{5}+4t^{3}+2t^{2}+2t\right)dt[/itex]

[itex]=-t+\frac{2t^{6}}{6}+\frac{4t^{4}}{4}+\frac{2t^{3}}{3}+\frac{2t^{2}}{2}[/itex]

[itex]=-t+\frac{t^{6}}{3}+t^{4}+\frac{2t^{3}}{3}+t^{2}[/itex]

applying bounds from 0 to 1

[itex]=-1+\frac{1}{3}+1+\frac{2}{3}+1[/itex]

[itex]=\frac{1}{3}+\frac{2}{3}+1=\frac{1+2+3}{3}=\frac{6}{3}=2[/itex]
 
  • #22
NewtonApple
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Part (c)
Evaluate the line integral [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along

(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

Solution:
Two paths. Let's call it Path 1 and Path 2.

(i) Path 1: (x1=0, x2=1, y1=1, y2=1)

Equation of line is y=1; dy=0

Integral becomes

[itex]\intop_{0}^{1}(x^{2}-1)dx=\frac{x^{3}}{3}-x=\frac{1}{3}-1=\frac{1-3}{3}=\frac{-2}{3}[/itex]

(ii) Path 2: (x1=1, x2=1, y1=1, y2=2)

Equation of line is x=1; dx=0

Integral becomes

[itex]\intop_{1}^{2}(y^{2}+1)dy=\frac{y^{3}}{3}-y=\frac{2^{3}}{3}-\frac{1^{3}}{3}+2-1=\frac{8}{3}-\frac{1}{3}+1=\frac{8-1+3}{3}=\frac{11-1}{3}=\frac{10}{3}[/itex]


Adding (i) and (ii)

[itex]\frac{-2}{3}+\frac{10}{3}=\frac{10-2}{3}=\frac{8}{3}[/itex]
 
  • #23
NewtonApple
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How do I insert LaTeX? I don't see any tool bar. Right now I'm doing it manually.
 
  • #24
ehild
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(ii) Path 2: (x1=1, x2=1, y1=1, y2=2)

Equation of line is x=1; dx=0

Integral becomes

[itex]\
\intop_{1}^{2}(y^{2}+1)dy=\frac{y^{3}}{3}-y=\frac{2^{3}}{3}-\frac{1^{3}}{3}+2-1=\frac{8}{3}-\frac{1}{3}+1=\frac{8-1+3}{3}=\frac{11-1}{3}=\frac{10}{3}
[/itex]


Adding (i) and (ii)

[itex]\frac{-2}{3}+\frac{10}{3}=\frac{10-2}{3}=\frac{8}{3}[/itex]

It is correct, but you made a sign mistake in the integral, it should be ##\frac{y^{3}}{3}+y##
 
  • #25
haruspex
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It is correct, but you made a sign mistake in the integral, it should be ##\frac{y^{3}}{3}+y##
Must have been a typo in posting the working. It was correct at the next step.
 
  • #26
NewtonApple
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It is correct, but you made a sign mistake in the integral, it should be ##\frac{y^{3}}{3}+y##

Must have been a typo in posting the working. It was correct at the next step.

Thx! for pointing it out - yes, it's a typo.
 

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