# Homework Help: Conservative and irrotational vectors

1. Jan 25, 2010

### Bevyclare

1. Please can someone assist me in solving the attached vector problem.

I have made several attempt to no avail

3.

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2. Jan 25, 2010

### vela

Staff Emeritus
You've almost got part (i) finished. You just have to keep going. You got to this point:

$$J=\int_C (ye^{xy}dx+(xe^{xy}+1)dy$$

Then you introduced a parameter u and apparently wrote x=u and y=u3 since C is the curve $y=x^3$. This gave you dx=du and dy=3u2du, so you got to here:

$$J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)$$

So far, so good. You just need to substitute for y in terms of u in the first term and figure out what u1 and u2 are. You were told the contour runs from x=1 to x=6. Since you know how x and u are related, you can easily see what u1 and u2 equal.

All the terms can easily be integrated either directly or with a simple substitution

3. Jan 26, 2010

### Bevyclare

Vela,

I still cannot solve the problem.

4. Jan 26, 2010

### HallsofIvy

Where are you haveing trouble?
$$J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)$$
and vela pointed out that, since $y= u^3$, that is the same as
$$J=\int_{u_1}^{u_2} (u^3e^{u^4}du+3u^3e^{u^4}du+3u^2du)$$
so it is really just
$$J=\int_{u_1}^{u_2} (4u^3e^{u^4}du+3u^2du)$$

You are told that x runs from 1 to 6. When x= 1 what is u? When x= 6, what is u?

If you still can't do it, please show your work so we can see exactly where you have a problem.

5. Jan 27, 2010

### Bevyclare

Hello everyone

Please find attached the output of my integration and substitution.

However, I am not convinced that this is correct.

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6. Jan 29, 2010

### Bevyclare

Please let someone assist me by reviewing my attached solution.

I don't think I'm correct.

Please I am counting on you.