# Math Help - Integral curvilinear

1. ## Integral curvilinear

Prove that integral curvilinear:

$\int_C yzdx + xzdy + yx^2dz$

is dependent on the path

2. Originally Posted by Apprentice123
Prove that integral curvilinear:

$\int_C yzdx + xzdy + yx^2dz$

is dependent on the path

You can choose two different paths to show the dependence of path.

3. Originally Posted by Apprentice123
Prove that integral curvilinear:

$\int_C yzdx + xzdy + yx^2dz$

is dependent on the path

The line integral for conservative vector fields are independent of path. Note that we can describe the vector field here as $\bold{F} = \left< yz, xz, yx^2 \right>$. show that this vector field is conservative, that is, find a function $f(x,y,z)$ such that $\nabla f = \bold{F}$. that will show independents of path.

it is not enough to pick two random paths and show the line integral is the same.

4. Remember if Curl F = 0 if and only if the line integral is independent of path

There are several equivalent statemnts

1Curl F = 0 (Fis irrotational
3.The line Integral is independent of path
4.F is conservative

5. $rotF = (x^2-y)i - (2xy -y)j +(z-z)k$ is different of $0$, not is conservative, then depends the path

6. Originally Posted by Calculus26
Remember if Curl F = 0 if and only if the line integral is independent of path

There are several equivalent statemnts

1Curl F = 0 (Fis irrotational
3.The line Integral is independent of path
4.F is conservative
ah yes, i always forget that curlF = 0 condition. it is perhaps easier than finding the potential function here.

7. Originally Posted by Apprentice123
$rotF = (x^2-y)i - (2xy -y)j +(z-z)k$ is different of $0$, not is conservative, then depends the path
well, that is not what the curl is, but the correct curl is non-zero still. so this is not independent of path

8. What is curl ?
Is Rotational ?

9. You may have learned it as rotation

Del x F

10. Originally Posted by Apprentice123
What is curl ?
Is Rotational ?

i suppose it is the same for you. curl is defined as follows:

let $\bold{F} = P \bold{i} + Q \bold{j} + R \bold{k}$

then $\text{curl}\bold{F} = \nabla \times \bold{F} = \left(\frac {\partial R}{\partial y} - \frac {\partial Q}{\partial z} \right) \bold{i} + \left(\frac {\partial P}{\partial z} - \frac {\partial R}{\partial x} \right) \bold{j} + \left(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right) \bold{k}$

11. I learned how rotational

Thanks, yours are helping me a lot