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.
Best Answer
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 $0$.