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Thread: Line integral along the curve

  1. #1
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    Question Line integral along the curve

    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.
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  2. #2
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    Is the question $\displaystyle \displaystyle \int_C \left(ay\mathbf i + ax\mathbf j\right) \, ds$, where $\displaystyle \displaystyle C$ is the curve $\displaystyle \displaystyle x = y^3$ from $\displaystyle \displaystyle P_1(1,1,-1)$ to $\displaystyle \displaystyle P_2(8,2,-1)$, $\displaystyle \displaystyle a$ is a constant, and $\displaystyle \displaystyle \mathbf i$ and $\displaystyle \displaystyle \mathbf j$ are unit vectors?
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  3. #3
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    Yes, this was the expression I tried to write, can you help me how to solve it, sir? Thanks in advance.
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    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?
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  5. #5
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    Of course you can. But the problem would be more corectly written $\displaystyle \int (ay\vec{i}+ ax\vec{j})\cdot d\vec{s}$ where $\displaystyle d\vec{s}= \vec{i}dx+ \vec{j}dy$.

    In this problem, $\displaystyle x= y^3$ so $\displaystyle dx= 3y^2dy$. That means that $\displaystyle ay\vec{i}+ ax\vec{j}= ay\vec{i}+ ay^3\vec{j}$ and $\displaystyle d\vec{s}= (3y^2\vec{i}+ \vec{j})dy$ 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 $\displaystyle \vec{F}(x,y)= f(x,y)\vec{i}+ g(x,y)\vec{j}$ and the path is given by the parametric equations, $\displaystyle x= u(t)$, $\displaystyle y= v(t)$, then $\displaystyle d\vec{s}= (u'(t)\vec{i}+ v'(t)\vec{j}) dt$

    $\displaystyle \int \vec{F}\cdot d\vec{s}= \int \left(f(x,y)\vec{i}+ g(x,y)\vec{j}\right)\left(\vec{i}dx+ \vec{j}dy}\right)$
    or the path integral $\displaystyle \int f(x,y)dx+ g(x,y)dy$ which, in turn, is the same as
    $\displaystyle \int \left(f(x(t),y(t))u'(t)+ g(x(t),y(t))v'(t)\right)dt$
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by essedra View Post
    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...

    $\displaystyle f_{x}= a\ y$

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

    But is also...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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