# Homework Help: Expressing F(x,y) as a function of x and y when it is a function of u and v

1. Jul 10, 2007

### jacquelinem_00

1. The problem statement, all variables and given/known data
F(x,y) is defined by the integral

(x,y)
S [-2uv^2sin(u^2v)]du + [cos(u^2v) - u^2vsin(u^2v)]dv
(2, pi)

Express F(x,y) as a function of x and y, eliminating the integral sign.

**Plus, I added the question as a bmp file, because I don't know how to write it all pretty like you guys do!!

*** Also, the S is the integral sign :D

2. Relevant equations

I'm guessing I should let x = uv^2 and y = u^2v.

3. The attempt at a solution
Well, I know there is some substitution involved, but I honestly have no idea where to start with this question. Any hints would be greatly appreciated!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### question 4.bmp
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2. Jul 11, 2007

### olgranpappy

Are you missing a bracket somewhere in your integrand? As it's written the integral doesn't seem to make sense... I guess I will wait for the .bmp file to get approved before I can help.

3. Jul 11, 2007

### HallsofIvy

That's a path integral in the uv-plane but it does not appear to be independent of the path:
(-2uv^2 sin(u^2v))v= -4uv sin(u^2v)- 4u^2v^3 cos(u^2v)
(c0s(u^2v)-u^2v sin(u^2v))u= -2uv sin(u^2v)- 2uv sin(u^2v)- 2u^3v^2 cos(u^2v)= -4uv sin(u^2v)- 2u^3v^2 cos(u^2v)
are not equal.

4. Jul 11, 2007

### olgranpappy

Ah, I see now. And it is indep of path, both
$$\frac{\partial f_u}{dv}=\frac{\partial f_v}{du}=-4uv\sin-2u^3v^2\cos$$.

5. Jul 11, 2007

### olgranpappy

so, anyways, Jacqueline, I think it would be easiest to choose a path from (2,pi) to (x,y) that first runs straight up the "v-axis" from pi to y (at fixed u=2). I.e., use du=0 and plug in u=2 to the one integrand which multiplies dv. Now look at the resulting integrand. Is it a total derivative? if so then we are in luck and don't have to work very hard. What do you find?

Next, integrate straight across the "u-axis" from 2 to x (at fixed v=y). I.e., use dv=0 and plug in v=y to the one integrand which multiplies du. Now in this integrand there is a (one dimensional) change of variables that should be obvious. what do you find? Make the change of variables and complete the integral.

Drawing a picture of the u-v plane would probably help you too. label the points (2,pi) and (x,y) and draw the path that you are taking... in fact, you can use any path you like, but I think the problem was probably designed with the boring path I just described in mind. Cheers,

6. Jul 12, 2007

### jacquelinem_00

Thank you so much Adam! You're great!!!!

7. Jul 12, 2007

### olgranpappy

You're welcome.

8. Jul 22, 2007

### jacquelinem_00

I think I thanked you prematurely. From reading your response I thought I knew what I was doing, until I tried to finish the question, and realized I was still stuck.

I took your advice and did everything. I fixed u=2 and set du=0. For the first part, I found that my integral was

-(y)
|
| [cos(u^2v) - u^2vsin(u^2v)]dv
|
- (pi)

= ycos(4y) - picos(4*pi)

Now this is supposed to be a total derivative? Yikes, I have no idea what it even MEANS!

Then I set dv=0 and I had

- (x)
|
| [-2uy^2sin(u^2y)] du
|
- (2)

=ycos(x^2y) - ycos(4y)

What am I even finding here? I cannot wrap my mind around these concepts! I really hate to ask you guys again, but I think all this sunshine this summer has fried my brain!! Please help!! Thank you!!

9. Jul 22, 2007

### olgranpappy

what I meant was that the integrand
$$cos(u^2v)-u^2vsin(u^2v)=\frac{d}{dv}(vcos(u^2v))$$
is a total derivative... but, whatever. it doesn't matter as long as you can do the integral... which you did do, and you did correctly. So the expression you found

$$ycos(4y)-\pi cos(4\pi)=ycos(4y)-\pi$$
is right and it is the first part of your answer. (Oh, btw, I also used cos(4pi)=1)

Right. That is the correct value of the second contribution to the total value of the integral... so the total value of the integral is the sum of the two parts you found. I.e., the answer is
$$(ycos(4y)-\pi) + (ycos(x^2y)-ycos(4y)) = ycos(x^2y)-\pi$$

You are welcome again.

Last edited: Jul 22, 2007