Line integral along the curve

**essedra** · February 28th 2011, 01:59 PM

Given that A= yai+xaj, P1(1,1,-1) and P2(8,2,-1) evaluate integral from P1 to P2 (A.dl) along the curve x=y^3.

**NOX Andrew** · February 28th 2011, 03:08 PM

Is the question $\displaystyle \int_C \left(ay\mathbf i + ax\mathbf j\right) \, ds$ , where $\displaystyle C$ is the curve $\displaystyle x = y^3$ from $\displaystyle P_1(1,1,-1)$ to $\displaystyle P_2(8,2,-1)$ , $\displaystyle a$ is a constant, and $\displaystyle \mathbf i$ and $\displaystyle \mathbf j$ are unit vectors?

**essedra** · February 28th 2011, 03:36 PM

Yes, this was the expression I tried to write, can you help me how to solve it, sir? Thanks in advance.

**NOX Andrew** · February 28th 2011, 04:21 PM

I don't believe you can integrate a vector through a line like that. Are you sure you aren't trying to find the work integral?

**HallsofIvy** · March 1st 2011, 02:36 AM

Of course you can. But the problem would be more corectly written $\int (ay\vec{i}+ ax\vec{j})\cdot d\vec{s}$ where $d\vec{s}= \vec{i}dx+ \vec{j}dy$ .

In this problem, $x= y^3$ so $dx= 3y^2dy$ . That means that $ay\vec{i}+ ax\vec{j}= ay\vec{i}+ ay^3\vec{j}$ and $d\vec{s}= (3y^2\vec{i}+ \vec{j})dy$ the dot product gives you an integral of a function of y with respect to y.

Finding "the work integral", given a force vector, is a physics problem which is a special application of "integration on a path". And that is what this problem is asking.

If $\vec{F}(x,y)= f(x,y)\vec{i}+ g(x,y)\vec{j}$ and the path is given by the parametric equations, $x= u(t)$ , $y= v(t)$ , then $d\vec{s}= (u'(t)\vec{i}+ v'(t)\vec{j}) dt$

$\int \vec{F}\cdot d\vec{s}= \int \left(f(x,y)\vec{i}+ g(x,y)\vec{j}\right)\left(\vec{i}dx+ \vec{j}dy}\right)$
or the path integral $\int f(x,y)dx+ g(x,y)dy$ which, in turn, is the same as
$\int \left(f(x(t),y(t))u'(t)+ g(x(t),y(t))v'(t)\right)dt$

**chisigma** · March 1st 2011, 03:12 AM

Originally Posted by essedra

Given that A= yai+xaj, P1(1,1,-1) and P2(8,2,-1) evaluate integral from P1 to P2 (A.dl) along the curve x=y^3.

The vector field can be condidered as two-dimensional and is...

$f_{x}= a\ y$

$f_{y}= a\ x$ (1)

But is also...

$\displaystyle \frac{\partial f_{x}}{\partial y} = \frac{\partial f_{y}}{\partial x}$ (2)

... so that the vector field is conservative and that means that the integral $\displaystyle \int_{P_{1}}^{P_{2}} (f_{x}\ \overrightarrow{i} + f_{y}\ \overrightarrow{j})\ ds$ is independent from the path connecting $P_{1}$ and $P_{2}$ ...

Kind regards

$\chi$ $\sigma$

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