# Fundamental Theorem for Line Integrals

• Jun 23rd 2010, 04:08 PM
Cursed
Fundamental Theorem for Line Integrals
(a) Find a function $f$ such that $F=\nabla f$, and (b) use part (a) to evaluate $\oint_C \vec{F}\cdot d \vec{r}$ along the given curve C.

$F(x,y) = $
C: $r(t) = $

_____________________________

MY work:

(a)

$F(x,y)=$

Check if it's conservative or not:
$P=xy^2$ and $Q=x^2y$

$\frac{\partial P}{\partial y} = 2xy$
$\frac{\partial Q}{\partial x} = 2xy$

It is conservative.

$f_x = xy^2$
$f=\frac{1}{2}x^2y^2 +g(y)$
$f_y=x^2y +g'(y) = x^2y$. So $g'(y) = 0$, which means $g(y) = K$

Therefore, $f=\frac{1}{2}xy^2 +K$

Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole $F=\nabla f$ thing.
• Jun 23rd 2010, 04:13 PM
Do you know how to compute the gradient? The f you have doesn't have F as its gradient (but its really close).
• Jun 23rd 2010, 04:28 PM
mr fantastic
Quote:

Originally Posted by Cursed
(a) Find a function $f$ such that $F=\nabla f$, and (b) use part (a) to evaluate $\oint_C \vec{F}\cdot d \vec{r}$ along the given curve C.

$F(x,y) = $
C: $r(t) = $

_____________________________

MY work:

(a)

$F(x,y)=$

Check if it's conservative or not:
$P=xy^2$ and $Q=x^2y$

$\frac{\partial P}{\partial y} = 2xy$
$\frac{\partial Q}{\partial x} = 2xy$

It is conservative.

$f_x = xy^2$
$f=\frac{1}{2}x^2y^2 +g(y)$
$f_y=x^2y +g'(y) = x^2y$. So $g'(y) = 0$, which means $g(y) = K$

Therefore, $f=\frac{1}{2}xy^2 +K$

Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole $F=\nabla f$ thing.

You can make the arbitrary choice K = 0. So $f = \frac{1}{2} x^2 y^2$.

Then the given line integral is simply equal to f(B) - f(A) where A and B are the start and endpoints of the curve you are integrating along.
• Jun 23rd 2010, 04:36 PM
Cursed
whoops. I forgot the constraints:

$\vec{r}(t) = $, where $0 \leq t \leq 1$

I know how to plug in points when you're given them, but what do you do with the vector equation?

If I do $\vec{F}(\vec{r}(u,v))$, I get something really ugly. o.O
• Jun 23rd 2010, 04:41 PM
mr fantastic
Quote:

Originally Posted by Cursed
whoops. I forgot the constraints:

$\vec{r}(t) = $, where $0 \leq t \leq 1$

I know how to plug in points when you're given them, but what do you do with the vector equation?

If I do $\vec{F}(\vec{r}(u,v))$, I get something really ugly. o.O

I have told you what to do. Where are you stuck? Can you not get the coordinates of A and B and substitute into f?
• Jun 23rd 2010, 04:55 PM
Cursed
How do I get the coordinates of A and B if the problem doesn't give them?

The problem just gives a vector function $\vec{F}$ and a curve $C$.

If I'm given the coordinates, I can plug them in. I just don't have any.
• Jun 23rd 2010, 05:02 PM
mr fantastic
Quote:

Originally Posted by Cursed
How do I get the coordinates of A and B if the problem doesn't give them?

The problem just gives a vector function $\vec{F}$ and a curve $C$.

If I'm given the coordinates, I can plug them in. I just don't have any.

You are told $0 \leq t \leq 1$. Therefore A has coordinates corresponding to t = 0 and B has coordinates corresponding to t = 1. And you are told $x = t + \sin \left(\frac{\pi}{2} t\right)$ and $y = t + \cos \left(\frac{\pi}{2} t\right)$. Therefore the coordinates of A and B are ....
• Jun 23rd 2010, 05:06 PM
Cursed
Oh, Okay. I didn't think you could do that. Thanks. :D

My book said to plug $\vec{r}$ into $\vec{f}$ symbolically and then evaluate the integral from A to B, in this case: from t=0 to t=1. o_o
• Jun 24th 2010, 01:26 PM
HallsofIvy
Quote:

Originally Posted by Cursed
(a) Find a function $f$ such that $F=\nabla f$, and (b) use part (a) to evaluate $\oint_C \vec{F}\cdot d \vec{r}$ along the given curve C.

$F(x,y) = $
C: $r(t) = $

_____________________________

MY work:

(a)

$F(x,y)=$

Check if it's conservative or not:
$P=xy^2$ and $Q=x^2y$

$\frac{\partial P}{\partial y} = 2xy$
$\frac{\partial Q}{\partial x} = 2xy$

It is conservative.

$f_x = xy^2$
$f=\frac{1}{2}x^2y^2 +g(y)$
$f_y=x^2y +g'(y) = x^2y$. So $g'(y) = 0$, which means $g(y) = K$

Therefore, $f=\frac{1}{2}xy^2 +K$

This is probably a typo- you dropped the ^2 on the x.

Now, what exactly IS the "given curve c"? You give parametric equations for a curve but no beginning and ending values for t.

If you have such values, calculate the corresponding beginning and ending values for x and y, calculate F(x,y) for those values and subtract, just like you do for integrals on the real numbers. Of course, the constant, K, will cancel out when you do that. That is why mr fantastic said "You can make an arbitrary choice of K= 0".

Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole $F=\nabla f$ thing.[/QUOTE]