# Finding parametrizattion for a Line Integral

• May 29th 2010, 09:21 AM
AbAeterno
Finding parametrizattion for a Line Integral
I've been asked to evaluate the line integral

$\int (10x^4 - 2xy^3)dx - 3x^2y^2dy$

Over the curve

$x^4 - 6xy^3 - 4y^2 = 0$

Between the points (0,0) and (2,1).

What I'm trying to do is finding a parametrization x(t) from the given curve and then put that into the integral so I can solve it.

However I can't seem to make it work. I tried solving x for y, so I can say x=t and y is a function of t. Or the other way round would work as well. But the crossterm $6xy^3$ gets in the way. And the powers of both x and y aren't right to use the ABC formula.

Could anyone help me out on what to do here?

Edit: Apologies for the spelling mistake in the thread title. Can't edit it, it seems.
• May 29th 2010, 10:05 AM
Jester
Can I ask to check on the sign of one of your terms in the integral

i.e. is it $
\int \limits_{c} (10x^4 {\color{red}{+}}\, 2xy^3)dx + 3x^2y^2dy$
• May 29th 2010, 10:07 AM
AbAeterno
Ah, I did write it down incorrectly, but it's not that either. Correct version is:

$
\int (10x^4 - 2xy^3)dx - 3x^2y^2dy
$
• May 29th 2010, 10:12 AM
Jester
Quote:

Originally Posted by AbAeterno
Ah, I did write it down incorrectly, but it's not that either. Correct version is:

$
\int (10x^4 - 2xy^3)dx - 3x^2y^2dy
$

Note that $2xy^3dx + 3 x^2 y^2 dy = d(x^2y^3)$.
• May 29th 2010, 10:29 AM
AbAeterno
Quote:

Originally Posted by Danny
Note that $2xy^3dx + 3 x^2 y^2 dy = d(x^2y^3)$.

Okay. So I can write that down in the integral and get

$
\int 10x^4 dx - d(x^2y^3)
$

But then what? Still doesnt help me find a parametrisation as far I can see. Or is that not what I'm supposed to do?

Edit: oh wait.. so now I don't need to find a x(t) and y(t) anymore, but an x(t) and x^2y^3(t)?
• May 29th 2010, 10:36 AM
Jester
Quote:

Originally Posted by AbAeterno
Okay. So I can write that down in the integral and get

$
\int 10x^4 dx - d(x^2y^3)
$

But then what? Still doesnt help me find a parametrisation as far I can see. Or is that not what I'm supposed to do?

Since the vector field is conservative, the parameterization doesn't matter. So

$
\int \limits_c 10x^4 dx - d(x^2y^3) =
\int \limits_c d(2x^5 - x^2y^3) = \left. 2x^5 - x^2y^3 \right|_{(0,0)}^{(2,1)}
$
• May 29th 2010, 10:47 AM
AbAeterno
Ooh.. I totally didn't see that. Thanks!