Line integral of a curve

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1. Dec 27, 2017

Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Line integral of a curve

$I = \int_{ }^{ } yz dx + \int_{ }^{ } zx dy + \int_{ }^{ } xy dz$ with proper limits.

$I = \int_{\frac { \pi }4}^{ \frac { 3 \pi} 4} abc ( \cos^2 t - \sin^2 t ) dt = -abc$

|I| = abc

So, the answer is option (a).

Is this correct?

2. Dec 27, 2017

kuruman

Except for the negative sign the answer correct. You just can't strip the negative sign because it doesn't match one of the choices. What if $-abc$ were a sixth choice? A better way to do this would be to see if $\vec{F}$ can be derived from some scalar function U, such that $\vec{F}=-\vec{\nabla}U$ (Hint: It can). Think of $\vec{F}$ as a conservative force and $U$ as the potential from which it is derived. Then the integral depends on the end points, i.e. $I=U(t_2)-U(t_1)$.

3. Dec 27, 2017

Pushoam

I had the impression that value of I = |I| (which is wrong according to your above comment). So, I removed the negative sign.

I did the calculation again and I got -abc.
$I = \int_{\frac { \pi }4}^{ \frac { 3 \pi} 4} abc ( \cos^2 t - \sin^2 t ) dt =$
$\frac { \sin{(2t)}} 2 |_ {\frac { \pi }4} ^{ \frac { 3 \pi} 4} abc = -abc$

I do not see what is wrong in the above calculation.

Taking $\vec F = - \nabla U$

I have to calculate $U = - \int \vec F \cdot d \vec x$ $= -abc = \frac { \sin{ (2t)}} 2$

Then, I have to calculate $\int \vec F \cdot d \vec x$ for t going from $\frac { \pi} 4 ~ to ~ \frac { 3 \pi }4$ .

This gives $U(\frac { \pi} 4) – U(\frac {3 \pi} 4)$ , which is again –abc.

I am not getting what mistake I am doing here, but I am coming towards the same answer.

4. Dec 27, 2017

Staff: Mentor

This is what I get, as well. It's not unheard of for posted answers to have typos.
The line integral calculation should be pretty straightforward. I don't see why you would need to do this.

5. Dec 27, 2017

kuruman

I also get this answer after discovering an excess of a negative sign.

6. Dec 28, 2017

Staff: Mentor

So I'm thinking that there's a typo in the answer.

7. Dec 28, 2017

Pushoam

So, I = -abc
And value of I = |I| = abc
Is this correct?

8. Dec 28, 2017

kuruman

I think so too. I did it two ways, using OP's method and using the scalar function but introduced an extraneous negative sign in the latter which got me going for a short while.
The value of the integral is I = - abc as we all agree by now. Why do you feel you should take the absolute value?

9. Dec 28, 2017

Staff: Mentor

I agree that I = -abc, but $I \ne |I|$. You can't just arbitrarily take the absolute value.

10. Dec 28, 2017

Staff: Mentor

I would say to force the computed answer to agree with the posted answer. This is not a good reason to do so.

11. Dec 28, 2017

Pushoam

Actually, I am having problem with the word " value".
I thought value of I = |I|...(1)
But,(1) is wrong.
value of I = the integral which I get after the calculation
absolute value of I = |I|

12. Dec 28, 2017

kuruman

Values can be positive or negative. Absolute values are positive only. The question does not specify "absolute".

13. Dec 28, 2017

Staff: Mentor

No. The value of I is whatever it is -- positive, negative, or zero. I = |I| if and only if I >= 0.