# Calculus

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What is this in relation to?

Gravity is conservative, friction is non conservative.

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Yes I want to know about conservative and non conservative fields. Also I want to know about zero and nonzero curl.

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So how did your query arise?

You need to put more in here to get more out.

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My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?

Edited by SF Shawn

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It is linked to the line only. But the the "integrand" itself may be parametric. Line is the path.

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My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?

I am puzzled by this question. Any text that introduces path integrals will tell you the conditions under which this is true. If a path integral, $\int_p^q f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z) dy$ is independent of the path, then we could define a function F(x,y,z) by "F(x,y,z) is the integral from (0, 0, 0) to (x, y, z)". But then it would follow that $\frac{\partial F}{\partial x}= f$, $\frac{\partial F}{\partial y}= g$, $\frac{\partial F}{\partial z}= h$. But then we would have to have $\frac{\partial^2 F}{\partial x\partial y}= f_y= g_x= \frac{\partial F}{\partial y\partial x}$, $\frac{\partial^2 F}{\partial x\partial z}= f_z= h_x= \frac{\partial F}{\partial z\partial x}$, and $\frac{\partial^2 F}{\partial y\partial z}= g_z= h_y= \frac{\partial F}{\partial z\partial y}$. (A differential for which that is true is called an exact differential.)

As an easy example, if we integrate $\int y^2dx+ xdy+ 2dz$ on the straight line from (0, 0, 0), we can take as parametric equations x= y= z= t so the integral becomes $\int_0^1 t^2dt+ tdt+ 2dt= \int_0^1 (t^2+ t+ 2)dt= \left[\frac{1}{3}t^3+ \frac{1}{2}t^2+ 2t\right]_0^1= \frac{1}{3}+ \frac{1}{2}+ 2= \frac{17}{6}$.

But if we integrate that same integrand from (0, 0, 0) to (1, 1, 1) by taking the path (0, 0, 0) to (1, 0, 0), then to (1, 1, 0), then to (1, 1, 1) we get:

On (0, 0, 0) to (1, 0, 0) take x= t, y= 0, z= 0. The integral becomes $\int_0^1 0 dt= 0$.

On (1, 0, 1) to (1, 1, 0) take x= 1, y= t. z= 0. The integral becomes $\int_0^1 1 dt= 1$.

On (1, 1, 0) to (1, 1, 1) take x= 1, y= 1. z= t. The integral becomes $\int_0^1 2dt= 2$.

The integral from (0, 0, 0) to (1, 1, 1), along that path is the sum 0+ 1+ 2= 3.

Here, of course, $\frac{\partial y^2}{\partial y}= 2y\ne 1= \frac{\partial x}{\partial x}$

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