Is This Function a Gradient? Testing for Line Integrals

[SigmaCrisis]

Calculating line integrals...

Ok, the problem is:
h(x,y) = 3x (x^2 + y^4)^1/2 i + 6y^3 (x^2 + y^4)^1/2 j;
over the arc: y = -(1 - x^2)^1/2 from (-1,0) to (1,0).

In my notes, I had written: if h is a gradient, then the INTEGRAL of g*dr over curve C depends only on the endpoints. Also, if the curve C is closed AND h is a gradient, then the integral of g*dr over curve c is 0.

So, my question is, when testing to see if a given function like the one above, should you just test to see if:

partial derivative of the i component with respect to y EQUALS the partial derivative of the j component with respect to x?

If not equal, then it is not a gradient, right?
Thanks.

 

[arildno]

Ok, the problem is:
h(x,y) = 3x (x^2 + y^4)^1/2 i + 6y^3 (x^2 + y^4)^1/2 j;
over the arc: y = -(1 - x^2)^1/2 from (-1,0) to (1,0).
In my notes, I had written: if h is a gradient, then the INTEGRAL of g*dr over curve C depends only on the endpoints. Also, if the curve C is closed AND h is a gradient, then the integral of g*dr over curve c is 0.

True enough, but you have placed your focus on a SIMPLIFYING SPECIAL CASE, rather than on the general case (which is what you SHOULD focus on).
So, how are line integrals IN GENERAL computed?

EDIT:
As far as your question is concerned, yes that is how you could test for whether h is a gradient field.

 

[SigmaCrisis]

I couldn't parametrize the arc...and I thought that perhaps taking the integrals by considering only the endpoints would be easier. In this example, I parametrized the straight line connecting the endpoints, r(u) = (2u-1)i + (0)j. This however produces the wrong answer.

To answer the question on how to calculate the line integral:

[INTEGRAL on curve C] h*dr, where r is the parametrization of the curve. Right?

 

[arildno]

Try the parametrization:
x=\cos\theta, y=\sin\theta, \pi\leq\theta\leq2\pi

I'm fairly sure it'll work out if you fiddle about with it for a while.