Vector Fields & Line Integrals
Vector Fields & Line Integrals
A function $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ that assigns a vector to each point in a region.
$$\int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\|\,dt$$
$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt$$
This computes the work done by $\mathbf{F}$ along curve $C$.
$\mathbf{F}$ is conservative if $\mathbf{F} = \nabla f$ for some potential $f$. Then:
$$\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A)$$
Test: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$.
Evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = \langle y, x \rangle$ and $C$ is $\mathbf{r}(t) = \langle t, t^2 \rangle$, $0 \leq t \leq 1$.
$\mathbf{r}'(t) = \langle 1, 2t \rangle$. $\mathbf{F}(\mathbf{r}(t)) = \langle t^2, t \rangle$.
$\int_0^1 (t^2 \cdot 1 + t \cdot 2t)\,dt = \int_0^1 3t^2\,dt = 1$.
Is $\mathbf{F} = \langle 2xy, x^2 \rangle$ conservative? If so, find the potential.
$P_y = 2x = Q_x$ ✓. $f_x = 2xy \Rightarrow f = x^2y + g(y)$. $f_y = x^2 + g'(y) = x^2 \Rightarrow g'(y) = 0$.
$f(x,y) = x^2y + C$.
Use the potential from Example 2 to evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ from $(0,0)$ to $(1,3)$.
$f(1,3) - f(0,0) = 3 - 0 = 3$.
Practice Problems
Show Answer Key
1. $P_y = 1$, $Q_x = -1$; not conservative
2. $P_y = 0 = Q_x$; $f = x^2+y^2+C$
3. $\|\mathbf{r}'\| = \sqrt{2}$; $\int_0^1 t^2\sqrt{2}\,dt = \sqrt{2}/3$
4. $\mathbf{r}(t) = \langle 3t,4t \rangle$; $\int_0^1(3+8)\,dt = 11$
5. $\int_0^1(t+t)\,dt = 1$
6. Work done by force $\mathbf{F}$ along path $C$
7. $P_y = 3x^2 = Q_x$; yes, conservative
8. $f(2,3)-f(1,1) = 6-1 = 5$
9. $\mathbf{r}(t) = \langle 1-t, t \rangle$, $0 \leq t \leq 1$
10. $\int_0^{2\pi}(\sin^2 t + \cos^2 t)\,dt = 2\pi$
11. $f$ such that $\nabla f = \mathbf{F}$
12. $0$ (for any closed path)