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Math Help - 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 \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?
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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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  4. #4
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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 \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
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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...

    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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