Prove that integral curvilinear:

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

is dependent on the path

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- Apr 26th 2009, 06:47 AMApprentice123Integral curvilinearProve that integral curvilinear:

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

is dependent on the path

- Apr 26th 2009, 08:55 AMcurvature
- Apr 26th 2009, 11:37 AMJhevon
The line integral for conservative vector fields are independent of path. Note that we can describe the vector field here as $\displaystyle \bold{F} = \left< yz, xz, yx^2 \right>$. show that this vector field is conservative, that is, find a function $\displaystyle f(x,y,z)$ such that $\displaystyle \nabla f = \bold{F}$. that will show independents of path.

it is not enough to pick two random paths and show the line integral is the same. - Apr 26th 2009, 11:40 AMCalculus26
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

2.F= gradf

3.The line Integral is independent of path

4.F is conservative - Apr 26th 2009, 11:51 AMApprentice123
$\displaystyle rotF = (x^2-y)i - (2xy -y)j +(z-z)k$ is different of $\displaystyle 0$, not is conservative, then depends the path

- Apr 26th 2009, 11:51 AMJhevon
- Apr 26th 2009, 11:55 AMJhevon
- Apr 26th 2009, 12:01 PMApprentice123What is curl ?

Is Rotational ?

- Apr 26th 2009, 12:03 PMCalculus26
You may have learned it as rotation

Del x F - Apr 26th 2009, 12:04 PMJhevon
i suppose it is the same for you. curl is defined as follows:

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

then $\displaystyle \text{curl}\bold{F} = \nabla \times \bold{F} = \left(\frac {\partial R}{\partial y} - \frac {\partial Q}{\partial z} \right) \bold{i} + \left(\frac {\partial P}{\partial z} - \frac {\partial R}{\partial x} \right) \bold{j} + \left(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} \right) \bold{k}$ - Apr 26th 2009, 12:07 PMApprentice123I learned how rotational

Thanks, yours are helping me a lot