Question

Use the fundamental theory of line integrals to calculate

Answer 1

\[\int\limits_{C}^{}F.dr = \]

If:

\[F=(x ^{\frac{ 2 }{ 3 }},e ^{7y})\]

and C is the unit circle oriented clockwise.


I got that:

\[f(x,y)=\frac{ 3x ^{\frac{ 3 }{ 5 }} }{ 5 }+\frac{ e ^{7y} }{ 7}+C\]

and my final answer to be:
\[\frac{ 26-5e ^{14\pi} }{ 35}\]

But I have no idea if this is correct or not. Any help would be appreciated.

Answer 2

Okay, well how about this. The next problem is the same equation but C is now a quarter of the unit circle in the first quadrant traced counter clockwise from (1,0) to (0,1). Now I got :
\[\frac{ 1-e ^{7} }{ 7}+\frac{ 3 }{ 5 }\]

But the book says it is:

\[\frac{ e ^{7}-1 }{ 7}-\frac{ 3 }{ 5 }\]

Which is the opposite. Why is that like that if you are going counter clockwise?

Answer 3

Are there any singularities in your circle?

Answer 4

I don't know what you mean.

The circle is just a basic unit circle in the first and 1/4th of that in the first quadrant for the second.

Answer 5

Oooooh. I get it. The reason that the answer was negative is because I was doing f(P)-f(Q). But the FTC is f(Q)-f(P) with P being the starting point and Q being the ending.

Answer 6

Well, you do have to go the right direcdtion.

Why do you care that 1/4 is in the 1st Quadrant? Did you share the entire problem statement?

Answer 7

Well yes, yes I did.

I've been trying several of these problems and I keep having issues with negatives or positives. The first point is P and the second is Q but when I do Q - P I get the wrong answer.

Answer 8

How did you parameterize the Unit Circle?

Answer 9

I didn't. I just: f(1,0)-f(0,1).

Answer 10

Okay, there's that 1st Quadrant again. What about the problem statement suggests that you should restrict your investigation to the 1st Quadrant?

Are you SURE your line integral is path independent? If it is, why did the problem statement specify the unit circle in a clockwise direction? Seems suspicious.

Answer 11

Okay, well how about this. The next problem is the same equation but C is now a quarter of the unit circle in the first quadrant traced counter clockwise from (1,0) to (0,1). Now I got :

There are two questions: The first is the whole circle and has no answer in the back of the book and the second is a quarter circle (with the same F) but answer is in the back of the book.

Answer 12

Parameterize your path. \[r(t) = <\cos(t),\sin(t)>\] for \[0\le t\le \frac{\pi}{2}\]

Answer 13

Are we to go down the y-axis and then back to where we started, or just the arc?

Answer 14

But that isn't the point. I can do it that way, but the idea of this section is to use the Fundamental Theorem of Calculus for Line Integrals.

Answer 15

Okay, then can we agree that the first one is just zero, since the beginning and ending points are the same?

Answer 16

Oh. Yes, because it would be closed. So yes, the first one is zero.

Answer 17

Excellent, now let's work on your f(x,y). Try that x-part again. You had thirds in that exponent. How did you end up with fifths?

Answer 18

Integrated in terms of x:

\[x ^{\frac{ 2 }{ 3 }}\]
and got
\[\frac{ 3x ^{\frac{ 5 }{ 3 }} }{ 5 }\]
Because 2/3 +1 = 5/3 and then you flip that and put it out front to deal with the exponent.

Answer 19

I would agree if your exponent would not have changed to 3/5 instead of 5/3.

Answer 20

What? Why would the exponent be 3/5? Oh. I'm sorry. I typed that in wrong into equation. That is my mistake; only on OpenStudy.

Answer 21

That's where you lost me. Now that we agree that's just a typo, we should be close to done.

And you evaluated f(0,1) - f(1,0) and you were done.

Answer 22

And it is from (0,1) to (1,0) because it is going counter clockwise and you need to do the end point minus the start point correct?

Thank you very much. Hopefully I can translate this across problems.

Answer 23

A little patience and maybe one or two fewer typos and you'll get it. Good work!