Computing $\int_c\vec{\nabla} F \cdot \vec{t}\,ds$, any clever way to eyeball the answer?

Question

I want to compute the integral$$\int_c \vec{\nabla} F \cdot \vec{t}\,ds,$$where $F = x - y$, $\vec{t}$ is the unit tangent vector, and $c$ is the following "circle" with counterclockwise rotation, i.e. a path along the edges of a square with corners $(1, 1)$, $(-1,1)$, $(-1,-1)$, and $(1,-1)$.

This is a Math GRE practice question, and so I'm wondering what is the quickest way to evaluate this without doing a full setup. How can we eyeball the answer?

Answer 1

The question as it stands does not make any sense: $$\int_c \vec F\cdot \vec dr=\int_c\vec F\cdot \vec Tds$$ Which, in either, case implies the Function of Force $\vec F$ is a vector function given by: $$\vec F=x\vec i- y\vec j$$ and not by a scalar:$$F=x-y$$

As for the solution of the integral, let us parameterize $x$ and $y$ such that: $$x=t;0\leq t\leq1$$ $$y=t;0\leq t\leq1$$

Now, notice for any integral: $\int f(u_i)du_i$ $\forall i\in \mathbb N$, and bounds $a\leq u_k\leq b$ for some index $k\in \mathbb N$ $$\int_a^cf(u_i)du_i+\int _c^bf(u_i)du_i=\int_a^b f(u_i)du_i$$

where $a\leq b \leq c$.

Also note, $$\int_{-c}\vec F\cdot \vec Tds=-\int_c\vec F\cdot \vec Tds$$

Given those facts, then integral then becomes:

$$\int_0^1t+\int_0^1t-\int_0^1t-\int_0^1t=0$$

Which, of course, makes sense because the force $\vec F$ is a conservative force since its curl is zero.

Knowing that, always check if the curl of a given force vector is zero. Furthermore, if it is zero, then see if the path is circular. That is, it ends where it began. If it is circular, then the work done is zero, which implies $$\int_c\vec F\cdot \vec Tds=0$$

Answer 2

Short answer: $\nabla F$ is a conservative field. By the gradient theorem, the closed loop integral is zero.

Evaluating it is quick, though. You ought to practice it, if this looked tough. Just draw a picture. It's quick, since

$$\nabla F = \hat{x} - \hat{y}$$

doesn't depend on $x,y$.

$$\begin{align} \int_{1 square} \nabla{F}\cdot d\vec{s} &= \int_1^{-1}\nabla{F}\big|_{y=1}\cdot (-\hat{x})dx + \int_1^{-1}\nabla{F}\big|_{x=-1}\cdot (-\hat{y})dy \\&\quad\quad + \int_{-1}^{1}\nabla{F}\big|_{y=-1}\cdot\hat{x}dx + \int_{-1}^{1}\nabla{F}\big|_{x=1}\cdot \hat{y}dx \\ &= -\int_1^{-1}dx + \int_1^{-1}dy + \int_{-1}^{1}dx - \int_{-1}^{1}dx \\ &= 2-2+2-2=0 \end{align}$$