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.
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?
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
$\displaystyle \int \left(f(x(t),y(t))u'(t)+ g(x(t),y(t))v'(t)\right)dt$
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
$\displaystyle \chi$ $\displaystyle \sigma$