# Thread:

1. ## 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?
Thanks in advance

Edit: Apologies for the spelling mistake in the thread title. Can't edit it, it seems.

2. Can I ask to check on the sign of one of your terms in the integral

i.e. is it $<br /> \int \limits_{c} (10x^4 {\color{red}{+}}\, 2xy^3)dx + 3x^2y^2dy$

3. Ah, I did write it down incorrectly, but it's not that either. Correct version is:

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

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

$<br /> \int (10x^4 - 2xy^3)dx - 3x^2y^2dy<br />$
Note that $2xy^3dx + 3 x^2 y^2 dy = d(x^2y^3)$.

5. 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

$<br /> \int 10x^4 dx - d(x^2y^3)<br />$
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)?

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

$<br /> \int 10x^4 dx - d(x^2y^3)<br />$
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

$<br /> \int \limits_c 10x^4 dx - d(x^2y^3) = <br /> \int \limits_c d(2x^5 - x^2y^3) = \left. 2x^5 - x^2y^3 \right|_{(0,0)}^{(2,1)}<br />$

7. Ooh.. I totally didn't see that. Thanks!
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