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Given that $g(x)$ is a continuously differentiable function of one variable $x$ which satisfies $g(0) = 2$. Assume that the value of the line integral $$I = \int_C xy^2 dx + yg(x)dy$$
is independent of the path C which joins $(0,0)$ to $(2,3)$, find an exact formula for the function $g(x)$, then find the exact value of the line integral $I$.


I have worked out that since the line integral is independent of the path C, it should be a conservative field (is this always true?). Hence going by the criteria of a conservative field I equate the differential of $xy^2$ with respect to $y$, to the differential of $yg(x)$ with respect to x and in the end I have worked out that $g(x) = x^2 + 2$.
Thus I now have the equation $$I = \int_C xy^2 dx + y(x^2 + 2)dy$$

However, I am stuck at the part finding the value of the line integral, in a sense I have no idea what to do with the information of path C's coordinate being from $(0,0)$ to $(2,3)$.

Thank you~

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  • $\begingroup$ Do you know how to find the potential function and are you familiar with the fundamental theorem of line integrals? $\endgroup$ – benguin Nov 17 '16 at 14:37
  • $\begingroup$ Unfortunately, your approach is not correct: in order to deduce that the field $(xy^2, yg(x))$ is conservative, you should be given that its line integrals between any (!) two points in $\Bbb R^2$ are path-independent. This is not what you are given though. $\endgroup$ – Alex M. Nov 17 '16 at 14:54
  • $\begingroup$ @AlexM. Hmm I thought that since the integral was given with respect to path C and they told us that we can assume it is independent of the path which joins $(0,0)$ to $(2,3)$, we can safely deduce that it is a conservative field? $\endgroup$ – stephchia Nov 17 '16 at 15:06
  • $\begingroup$ @Benguin I briefly understand how to find a potential function, but am not sure why it'd be used in this context. $\endgroup$ – stephchia Nov 17 '16 at 15:07
  • $\begingroup$ @stephchia: Unfortunately you can't, but you're not far from being right. I'm typing the solution, give me some time please. $\endgroup$ – Alex M. Nov 17 '16 at 15:12
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You are very, very close to the correct solution. The problem is that you cannot deduce that the field $(xy^2, yg(x))$ is conservative only from the fact that its integral between two points is path-independent. In order to deduce conservativeness, you would have to be told that the line integral between any two points is line-independent - which you are not given. Fortunately, not all is lost, and the information you are given is enough!

First, let us find one particular $g$ among the many that satisfy the given equality. Your approach is correct: let us try our chances and look for a conservative field (but nobody guarantees that we shall be able to find one!). Therefore, let us impose (as you do) that

$$\frac {\partial (xy^2)} {\partial y} = \frac {\partial (y g(x))} {\partial x} ,$$

which gives $2xy = y g'(x)$, whence (since the equality must be true for every $x$ and $y$) $g(x) = x^2 + c$ with $c$ some unknown integration constant. The condition $g(0) = 2$ forces $c=2$, so $g(x) = x^2 + 2$.

The problem is that, in principle, this $g$ that we have obtained is not unique. Let $x \mapsto g(x) + \epsilon (x)$ be another function with the same property. Then $I$ computed with $g$ and $I$ computed with $g + \epsilon$ are equal, so

$$0 = I - I = \int \limits _C xy^2 \ \Bbb d x + y \big( g(x) + \epsilon (x) \big) \ \Bbb d y - \int \limits _C xy^2 \ \Bbb d x + y g(x) \ \Bbb d y = \int \limits _C y \epsilon (x) \ \Bbb d y .$$

This equality says that if $\epsilon$ satisfies the equality above and $\epsilon(0) = 0$, then $x^2 + 2 + \epsilon(x)$ is as good as $x^2 + 2$, meaning that when plugged into $I$ they give the same result. It is impossible to tell them apart from the point of view of $I$ or, in more mathematical terms, the problem of finding $g$ does not have a unique solution.

Finally, in order to explicitly compute $I$ just choose any among these many functions, for instance $g(x) = x^2 + 2$. Next, since $I$ is path-independent, just choose any path between $(0,0)$ and $(2,3)$ and compute $I$. I shall choose a line segment joining the two, given by the formula

$$[0,1] \ni t \mapsto (2t, 3t) \in \Bbb R^2 .$$

Along this curve we get

$$I = \int \limits _0 ^1 (2t) \ (9t^2) \ 2 + (3t) [(2t)^2 + 2] \ 3 \ \Bbb d t = 36 \frac {t^4} 4 \Bigg| _0 ^1 + 36 \frac {t^4} 4 \Bigg| _0 ^1 + 18 \frac {t^2} 2 \Bigg| _0 ^1 = 27 .$$

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  • $\begingroup$ I appreciate you going beyond the mile and explaining why my assumption has to be validated before proceeding! I believe I understand the rationale behind parameterizing the line segment, however I am curious if, given $I$ in the form of an integral, are we able to apply the concept used in finding potential function of a given conservative vector field (usually in the form of $(xi, yj, zk)$ here in this particular problem to solve for $F(B) - F(A)$? I'm not disputing your approach, but in learning this chapter I wish to learn of as many alternative approaches to a problem as possible. $\endgroup$ – stephchia Nov 17 '16 at 15:53
  • $\begingroup$ @stephchia: The approach that I use here is precisely the one used to find to extract a potential function given its force field - it's just that the solution is heavily non-unique. Remember that, in general, any two potentials that differ only by a constant give rise to the same force field, so in general the potential is not unique - but at least the space of potentials is $1$-dimensional. In our case it's worse, the space of solutions being infinite-dimensional. Save for this, the approach is the same. $\endgroup$ – Alex M. Nov 17 '16 at 15:58
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Your line $[A,B]$ parametric equations are

$x=x_A+(x_B-x_A)t=2t,\; dx=2dt$

and $y=y_A+(y_B-y_A)t=3t,\; dy=3dt$

and the line integral is

$$\int_C xy^2dx+y(x^2+2)dy=$$

$$\int_0^1\left((2t)(3t)^2.2+(3t)((2t)^2+2).3\right)dt$$

$$=\int_0^1(36t^3+36t^3+18t)dt$$

$$=9+9+9=27$$

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  • $\begingroup$ Isn't this correct. Why down $\endgroup$ – hamam_Abdallah Nov 17 '16 at 14:44
  • $\begingroup$ The down came from someone else, but I gave it an up now so it's equal! Could you explain how you went about doing the parametrization of the line integral? I understand the rationale but am clueless with regards to the steps. $\endgroup$ – stephchia Nov 17 '16 at 14:55
  • $\begingroup$ The complicated part of the problem is finding $g$. Once you have found it, computing $I$ is easy. But how do you find $g$, given that the OP's approach is not correct? $\endgroup$ – Alex M. Nov 17 '16 at 14:55
  • $\begingroup$ @stephchia And now, is it correct. $\endgroup$ – hamam_Abdallah Nov 17 '16 at 15:34
  • $\begingroup$ @AbdallahHammam Thanks for the clarification! I wish I could give another up but unfortunately whoever down'd it did so for unknown reasons. May I clarify if the purpose of parameterizing the line is to be able to integrate the parameters into our integrals and hence evaluate the result? In the same sense, would it then be possible for us to find a potential function for the integral and evaluate the $F(B) - F(A)$? Unfortunately I'm only familiar with finding potential for vector functions (in the i,j,k format) so I'm not sure if its applicable in this case. $\endgroup$ – stephchia Nov 17 '16 at 15:39

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