What is the line integral of a curve?

[Pushoam]

Homework Statement

Homework Equations

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?

 

[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. $I=U(t_2)-U(t_1)$.

 

[Pushoam]

You just can't strip the negative sign because it doesn't match one of the choices. What if −abc were a sixth choice?

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.

 

[Mark44]

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

This is what I get, as well. It's not unheard of for posted answers to have typos.

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

Taking $\vec F = - \nabla U$

The line integral calculation should be pretty straightforward. I don't see why you would need to do this.

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.

 

[kuruman]

This is what I get, as well. It's not unheard of for posted answers to have typos.

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

 

[Mark44]

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

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

 

[Pushoam]

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

 

[kuruman]

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

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.

And value of I = |I| = abc
Is this correct?

The value of the integral is I = - abc as we all agree by now. Why do you feel you should take the absolute value?

 

[Mark44]

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

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

 

[Mark44]

Why do you feel you should take the absolute value?

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

 

[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|

 

[kuruman]

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

 

[Mark44]

Actually, I am having problem with the word " value".
I thought value of I = |I|...(1)

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