- #36

- 234

- 79

Thank you ... I have done it with a math editor.LaTeX is preferable, but if your work is too involved, you can post a picture as long as it is clearly legible and right side up. Use the "Attach files" link, lower left.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter MatinSAR
- Start date

Yeah, you replace it with whatever variable you want.the equation of path is x=(2a) sinθ and y=(a) cosθWhat is ##\theta## here? It can't be the argument parameter of the polar representation of (x,y) since we would deduce ##\tan^2(\theta)=\frac 12##. So I assume it is an arbitrary parameter unrelated to polar coordinates. But then, where do the integration bounds come from?I see no diagram.f

- #36

- 234

- 79

Thank you ... I have done it with a math editor.LaTeX is preferable, but if your work is too involved, you can post a picture as long as it is clearly legible and right side up. Use the "Attach files" link, lower left.

- #37

- 234

- 79

What about Theta ?! Did I specify theta correctly?I don't get the factor as 6. Did you handle the signs correctly? The rest looks fine.

I get the factor 6 again ! Let me send a picture ...

- #38

- 234

- 79

I don't get the factor as 6. Did you handle the signs correctly? The rest looks fine.

- #39

Homework Helper

Gold Member

- 2,774

- 1,137

I'm getting what you are getting.

$$ \int 3 x y^2 dy = \int 3 \cdot y^2 \cdot 2 \sqrt{a^2 - y^2} dy $$

Let:

##\frac{y}{a} = \sin \beta##

## \frac{ \sqrt{a^2 - y^2} }{a} = \cos \beta##

It follows that

$$ y^2 = a^2 \sin^2 \beta$$

$$\sqrt{a^2 - y^2} = a \cos \beta$$

$$ dy = a \cos \beta d \beta$$

$$ \implies \int 3 x y^2 dy = 6 a^4 \int \sin^2 \beta \cos^2 \beta d \beta $$

$$ \int 3 x y^2 dy = \int 3 \cdot y^2 \cdot 2 \sqrt{a^2 - y^2} dy $$

Let:

##\frac{y}{a} = \sin \beta##

## \frac{ \sqrt{a^2 - y^2} }{a} = \cos \beta##

It follows that

$$ y^2 = a^2 \sin^2 \beta$$

$$\sqrt{a^2 - y^2} = a \cos \beta$$

$$ dy = a \cos \beta d \beta$$

$$ \implies \int 3 x y^2 dy = 6 a^4 \int \sin^2 \beta \cos^2 \beta d \beta $$

Last edited:

- #40

- 22,843

- 5,476

I also get a factor of 6, although I would never have done the math the way that he did it>I don't get the factor as 6. Did you handle the signs correctly? The rest looks fine.

- #41

- 234

- 79

Was it true ?!I also get a factor of 6, although I would never have done the math the way that he did it>

What was wrong with the way I have done the math

I'm getting what you are getting.

Can you please tell me what is your final answer ?!

- #42

Homework Helper

Gold Member

- 2,774

- 1,137

I didn't do the integral!Was it true ?!

What was wrong with the way I have done the math

Can you please tell me what is your final answer ?!

- #43

Homework Helper

Gold Member

- 2,774

- 1,137

Post your work, it will get checked by someone...

- #44

- 40,285

- 9,282

Yes, my mistake… I should have waited until I had the time to check properly.What about Theta ?! Did I specify theta correctly?

I get the factor 6 again ! Let me send a picture ...

- #45

- 234

- 79

Post your work, it will get checked by someone...

- #46

- 40,285

- 9,282

It is a line integral, not a double integral, so you must arrange that there is only one variable in the integrand. If you want to do it as an integral wrt y then you must first replace all the occurrences of x with what x is as a function of y.

But really, the parametric approach using theta is much easier.

- #47

Homework Helper

Gold Member

- 2,774

- 1,137

Show the work. Explain how you get the last result?

- #48

- 234

- 79

So my answer is wrong , isn't it ?!

But really, the parametric approach using theta is much easier.

Thank you ... I will try to do it using parametric approach and I will send a picture of the work ...

Thank you ... I will try to do it using parametrich approach and I will send a picture of the work ...Show the work. Explain how you get the last result?

- #49

- 40,285

- 9,282

I'm confused. Isn’t that what you already did in post #32? Except that you did not perform the integral.I will try to do it using parametric approach

- #50

- 22,843

- 5,476

The integration is very simple. It just involves the use of trigonometric identities.$$(\sin{\theta}\cos{\theta})^2=\frac{\sin^2{2\theta}}{4}=\frac{1-\cos{4\theta}}{8}$$

Last edited:

- #51

Homework Helper

Gold Member

2022 Award

- 1,889

- 1,784

Hi @MatinSAR. Can I point out a fundamental mistake you are making? I don’t think anyone has mentioned it yet (apologies if they already have).

The final expression for W can

You made the same mistake in your Post #1 solution, where your final expression for W contained x and y.

The expression for W can contains only given constants (here only the 'a') and pure numbers like 5, π and √2 (just picking random values as examples).

For example, if you determine the formula for the circumference of a circle, x²+y²=a², it wouldn’t make sense to have x and/or y in the final formula.

- #52

- 234

- 79

Yes. It was unfinished.I'm confused. Isn’t that what you already did in post #32? Except that you did not perform the integral.

Thank you.

Thank you ... I didn't know about this.

The final expression for W cannotcontain x or y (or θ). These are variables which change as you move along the path. They must disappear when the definite integral is evaluated.

You made the same mistake in your Post #1 solution, where your final expression for W contained x and y.

The expression for W can contains only given constants (here only the 'a') and pure numbers like 5, π and √2 (just picking random values as examples).

For example, if you determine the formula for the circumference of a circle, x²+y²=a², it wouldn’t make sense to have x and/or y in the final formula.

- #53

- 234

- 79

The force is conserative so If I choose another path the final answer shouldn't change, Is it true ?!

- #54

- 40,285

- 9,282

Is it? What do you get if you integrate from 0 to 2π?The force is conserative

I agree with your answer.

- #55

- 234

- 79

It was mentioned in question.Is it?

Thank you.I agree with your answer.

- #56

- 234

- 79

.

.

.

.

.

I hope I haven't forgotten anyone..

Thank you for your help and time.

- #57

- 40,285

- 9,282

Try the path along the axes. Isn't F always zero there?It was mentioned in question.

- #58

- 234

- 79

No it's not.Try the path along the axes. Isn't F always zero there?

- #59

Homework Helper

Gold Member

2022 Award

- 1,889

- 1,784

Oh yes it is! (If you familiar with British pantomime.)No it's not.

Can I expand on what @haruspex said in Post #57?

Referring to the Post #21 diagram, suppose you take the following route:

- from A(0,a), move along the y-axis to the origin (0,0);

- from the origin (0,0) move along the x-axis to C(2a,0).

##\vec F = x^2y~\hat i + xy^2 \hat~j##

While moving along the y-axis (x=0); both components of ##\vec F## are zero.

Similarly while moving along the x-axis.

The total work done along this path is therefore zero. This is different to your calculated value for the original route from A to C.

So you can see (without doing any maths) that the force is

- #60

- 234

- 79

Great !Referring to the Post #21 diagram, suppose you take the following route:

- from A(0,a), move along the y-axis to the origin (0,0);

- from the origin (0,0) move along the x-axis to C(2a,0).

So it's not conserative ...

Thank you.

Share:

- Replies
- 57

- Views
- 2K

- Replies
- 4

- Views
- 578

- Replies
- 11

- Views
- 514

- Replies
- 25

- Views
- 533

- Replies
- 6

- Views
- 779

- Replies
- 29

- Views
- 731

- Replies
- 4

- Views
- 529

- Replies
- 9

- Views
- 465

- Replies
- 94

- Views
- 2K

- Replies
- 17

- Views
- 448