Line Integral gives no work done? For the following question,

$$
\mathbf{F}=\langle-y, x\rangle
$$
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.
 A: 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 $F(x,y)= \langle-y, x\rangle= \langle 0,t \rangle$ 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 $0$.
A: Here's how physicists often do it:
$$\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (F_x \hat{i} + F_y \hat{y})\cdot (\hat{i}\,dx + \hat{j}\,dy) = \int_C F_x\,dx + F_y\,dy$$
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
$$\int_{C_2} F_x\,dx + F_y\,dy = \int_{y=0}^{y=1} x\,dy = \int_{y=0}^{y=1} dy=1.$$
A: Your second integral is wrong. Note that $x=1$ on this portion of the path.
