# Line Integral gives no work done?

Mathematics Asked on January 5, 2022

For the following question,

$$mathbf{F}=langle-y, xrangle$$
For this field:

Compute the line integral along the path that goes from (0,0) to (1,1) by first going along the $$x$$ -axis to (1,0) and then going up one unit to (1,1) .

I got an answer of $$0$$, by doing: But the answer key concludes that the answer is $$1$$:

To compute $$int_{C} mathbf{F} cdot d mathbf{r}$$ we break the curve into two pieces, then add the line integrals along each piece.
First, fix $$y=0$$ (so $$d y=0$$ ) and let $$x$$ range from 0 to 1 .
$$int_{x=0}^{x=1} mathbf{F} cdot d mathbf{r}=int_{x=0}^{x=1}-y d x+x d y=int_{0}^{1} 0 d x=0$$
Next, fix $$x=1$$ (so $$d x=0$$ ) and let $$y$$ range from 0 to 1:
$$int_{y=0}^{y=1} mathbf{F} cdot d mathbf{r}=int_{y=0}^{y=1}-y d x+1 d y=1$$
We conclude that $$int_{C} mathbf{F} cdot d mathbf{r}=1$$

I understand the solution from the answer key, but I don’t get why my solution doesn’t work. Please assist.

Here's how physicists often do it:

$$int_C mathbf{F}cdot dmathbf{r} = int_C (F_x hat{i} + F_y hat{y})cdot (hat{i},dx + hat{j},dy) = int_C F_x,dx + F_y,dy$$

On the first part $$x$$ goes $$0 to 1$$ and $$y=0$$ is constant so $$dy = 0$$ and hence $$int_{C_1} F_x,dx + F_y,dy = int_{x=0}^{x=1} -y,dx = 0.$$

Similarly, on the second part $$y$$ goes $$0 to 1$$ and $$x=1$$ is constant so $$dx = 0$$ and hence $$int_{C_2} F_x,dx + F_y,dy = int_{y=0}^{y=1} x,dy = int_{y=0}^{y=1} dy=1.$$

Answered by mechanodroid on January 5, 2022

Your error is in the second integral. Along the path from $$(0, 0)$$ to $$(1, 0)$$ we can take as parameterization $$x= t$$ (from $$0$$ to $$1$$), $$y= 0$$ for all $$t$$. So $$F(x,y)= langle-y, xrangle= langle 0,t rangle$$ and the vector differential is $$langle dt, 0 rangle$$ so the integral is $$int_0^1 langle 0, t rangle cdot langle dt, 0 rangle = int_0^t 0= 0$$ That is what you correctly have.

Along the path from $$(1, 0)$$ to $$(1, 1)$$ we can take as parameterization $$x= 1$$ for all $$t$$, $$y= t$$ (from $$0$$ to $$1$$). So $$F(x,y)= langle -y, x rangle= langle -t, 1 rangle$$ (not $$langle-t, 0 rangle$$ because $$x= 1$$) and the vector differential is $$langle 0, dy rangle$$ so the integral is $$int_0^1 langle -t, 1 rangle cdot langle 0, dy rangle = int_0^1 dy= 1$$

So the complete integral is 1. Again, your error is that on the second line, from $$(1, 0)$$ to $$(1, 1)$$ as $$x$$ is always $$1$$, not $$0$$.

Answered by user247327 on January 5, 2022

Your second integral is wrong. Note that $$x=1$$ on this portion of the path.

Answered by Ted Shifrin on January 5, 2022

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