My Math Forum Evaluate the line integral vector filed (curve)

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December 30th, 2016, 07:38 PM   #11
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Quote:
 Originally Posted by zylo Actually, by inspection, u=$\displaystyle (x^{2}+z^{2})y$. Then $\displaystyle \int f\cdot dr = \oint_{1}^{2} du$
Apparently that wasn't obvious:

$\displaystyle \int_{1}^{2}du = \int_{P_{1}}^{P_{2}}du = u(P_{2})-u(P_{1})$

$\displaystyle \int_{P_{1}}^{P_{2}}du= \Delta u_{1}+\Delta u_{2}+.....+\Delta u_{n}$, n $\displaystyle \rightarrow \infty$, along arbitrary continuous path.

$\displaystyle \int_{P_{1}}^{P_{2}}du = (u_{2}-u_{1}) + (u_{3}-u_{2}) + ......+ (u_{n}-u_{n-1}) = (u_{n}-u_{1}) = u(P_{2})-u(P_{1})$

Potential is not the solution context of the problem, it is an example of the solution.

December 30th, 2016, 08:31 PM   #12
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Quote:
 Originally Posted by xl5899 If we don't want to carry out the test to check whether it's conservative or not, we still can use the traditional method, which is using ∫Fdr, right?
Since you haven't been given a curve, you need to carry out the check to be sure that any curve will do. But once you know that any curve will do, you also know that you just need to evaluate the potential function at the end-points. Actually calculating the integral over any particular path is a complete waste of effort. It's only worth doing to demonstrate that the integral is indeed independent of the path.

December 31st, 2016, 06:45 AM   #13
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Why do you keep referring to it as a potential function, OP doesn't. The mathematics is "independence of path," which no calculus text introduces as a potential function. Google it. Potential is an application, just like entropy, where independence of path of $\displaystyle \int$ dQ/T establishes the entropy function S.

Quote:
 Originally Posted by zylo 1) Ck that $\displaystyle \int f\cdot dr$ is independent of path. 2) Evaluuate $\displaystyle \int f\cdot dr$ for simplest path. $\displaystyle \int f\cdot dr$ = Xdx+Ydy+Zdz is independent of path if: $\displaystyle f=\triangledown u$ so that $\displaystyle \int f\cdot dr = \int \triangledown u\cdot dr=\int du$ f = $\displaystyle \triangledown u$ if $\displaystyle \frac{\partial Z}{\partial y}=\frac{\partial Y}{\partial z}$, $\displaystyle \frac{\partial X}{\partial z}=\frac{\partial Z}{\partial x}$, $\displaystyle \frac{\partial Y}{\partial x}=\frac{\partial X}{\partial y}$

December 31st, 2016, 06:51 AM   #14
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Quote:
 Originally Posted by zylo which no calculus text introduces as a potential function
By which you presumably mean that yours doesn't.

December 31st, 2016, 08:55 AM   #15
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Heat and Entropy, and path independence

Quote:
 Originally Posted by zylo $\displaystyle \int_{1}^{2}du = \int_{P_{1}}^{P_{2}}du = u(P_{2})-u(P_{1})$ $\displaystyle \int_{P_{1}}^{P_{2}}du= \Delta u_{1}+\Delta u_{2}+.....+\Delta u_{n}$, n $\displaystyle \rightarrow \infty$, along arbitrary continuous path. $\displaystyle \int_{P_{1}}^{P_{2}}du = (u_{2}-u_{1}) + (u_{3}-u_{2}) + ......+ (u_{n}-u_{n-1}) = (u_{n}-u_{1}) = u(P_{2})-u(P_{1})$
Let Q be "heat." Why isn't $\displaystyle \int_{1}^{2}dQ = Q(P_{2})-Q(P_{1})$? Because Q is not a function of (thermodynamic) coordinates.
$\displaystyle \int_{1}^{2}dQ$ depends on work interchange with the system.
In general, $\displaystyle \oint$ dQ = W for a cycle, heat in equals work out.
If W=0, $\displaystyle \oint$ dQ = 0 and $\displaystyle \int$ dQ is independent of path so that Q = Q(P) and $\displaystyle \int_{1}^{2}dQ = Q(P_{2})-Q(P_{1})$.

In general, $\displaystyle \int dQ/T$ is independent of path* and so defines a function S (entropy) of coordinates, so that between any two thermodynamic states, $\displaystyle \int_{1}^{2}dS = S(P_{2})-S(P_{1})$.

Never really grasped this until this thread, which put things together for me. Have to thank OP.

*As shown by taking a system through a cycle while interchanging heat with heat reservoirs using Carnot cycles. Also demonstrated mathematically by Caratheodory.

Last edited by zylo; December 31st, 2016 at 08:57 AM.

December 31st, 2016, 11:22 AM   #16
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Quote:
 Originally Posted by zylo Why do you keep referring to it as a potential function, OP doesn't. The mathematics is "independence of path," which no calculus text introduces as a potential function. Google it. Potential is an application, just like entropy, where independence of path of $\displaystyle \int$ dQ/T establishes the entropy function S.
my Calc 1-3 book, Ellis and Gulick, does in fact introduce the topic as independence of path and state that if F is a gradient of some function it doesn't give a name to, then the line integral is path independent.

In the very next sub-section, it's not even given a separate section number, the authors discuss conservation of energy and potential functions and how this function they didn't give a name to in the previous section is called a potential function.

 December 31st, 2016, 12:02 PM #17 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 So we have for a system cycle, heat in = work out, $\displaystyle \oint dQ = \oint dW$. Suppose heat is zero so that $\displaystyle \oint dW=0$. Then $\displaystyle \int_{1}^{2}dW$ is independent of path and in this case $\displaystyle \int_{1}^{2}dW =W(P_{2})-W(P_{1})$ and W is called the potential. For example, if you slide a block with friction, $\displaystyle \int_{1}^{2}dW = \int_{1}^{2} f\cdot dr$ is dependent on path and $\displaystyle \oint dW$ = heat generated $\displaystyle \neq$ 0. $\displaystyle \int_{1}^{2}dW = \int_{1}^{2}dQ+ \int_{1}^{2}dU$ (first law of thermo) but $\displaystyle \oint dW = \oint dQ$ because internal energy U is a state function and Q isn't. NOTE: $\displaystyle \oint$ dW = 0 iff $\displaystyle \int$ dW is independent of path because $\displaystyle \oint dW = \int_{1}^{2}dW+\int_{2}^{1}dW = \int_{1}^{2}dW-\int_{1}^{2}dW = 0$ (two different paths make up the cycle).
December 31st, 2016, 12:08 PM   #18
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Apostol II introduces the concept of independence of path in section 10.10, gives and proves the result $$\int_{\mathbf a}^{\mathbf b} \nabla \phi \cdot d\mathbf \alpha = \phi(\mathbf b) - \phi(\mathbf a)$$ in section 10.11 as the second fundamental theorem of calculus for line integrals. Section 10.12 opens with the sentence
Quote:
 If a vector field $\mathbf f$ is the gradient of a scalar field $\phi$, then $\phi$ is called a potential function for $\mathbf f$.
The notion of conservative fields was mentioned six pages previously in general terms.

Last edited by v8archie; December 31st, 2016 at 12:18 PM.

 January 2nd, 2017, 08:22 AM #19 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 Independence of Path and Exact Differentials To clean things up, the basic ideas are: Definition of line integrals: $\displaystyle \int_{C}H(x,y,z)ds=\sum H(x_{k},y_{k},z_{k})\Delta s_{k}$ $\displaystyle \int_{C}H(x,y,z)dx=\sum H(x_{k},y_{k},z_{k})\Delta x_{k}$ $\displaystyle \int_{C}H(x,y,z)dy=\sum H(x_{k},y_{k},z_{k})\Delta y_{k}$ $\displaystyle \int_{C}H(x,y,z)dz=\sum H(x_{k},y_{k},z_{k})\Delta z_{k}$ If x,y,z are independent variables, $\displaystyle \int dx=x, \int dy=y, \int dz=z$. If dH is a dependent variable given by $\displaystyle dH \equiv H_{1}(x,y,z)dx+H_{2}(x,y,z)dy+H_{3}(x,y,z)dz$, then $\displaystyle \int_{C} dH$ is a line integral dependent on path, unless $\displaystyle \int_{C}$ dH is independent of path $\displaystyle (\int_{C} dH = H)$ iff $\displaystyle dH=\frac{\partial H}{\partial x}dx+\frac{\partial H}{\partial y}dy+\frac{\partial H}{\partial z}dz$, which says dH is an exact differential and H=H(x,y,z) exists.

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# check whether the line integral ∫ (1,2,3) (2,0,0) (xdx ydy zdz) is independent of path or not

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