Every integral along a curve starts the same way: describe the curve by a vector function $\mathbf{r}(t)$, $a \le t \le b$. For line integrals of vector fields, the direction matters too: a curve from $A$ to $B$ is a different oriented curve from the same points traced from $B$ to $A$. So a parametrization has two jobs: trace the right points, and trace them the right way.
Key idea
- Segment from $A$ to $B$: $\mathbf{r}(t) = A + t(B - A)$, ${}0 \le t \le 1$.
- Circle with center $(h, k)$ and radius $R$, counterclockwise from $(h + R, k)$: $\mathbf{r}(t) = \langle h + R\cos t, k + R\sin t \rangle$, ${}0 \le t \le 2\pi$. Clockwise: $\langle h + R\cos t, k - R\sin t \rangle$.
- Graph $y = g(x)$ from $x = a$ to $x = b$: $\mathbf{r}(t) = \langle t, g(t) \rangle$, $a \le t \le b$.
- Helix around the $z$-axis: $\langle R\cos t, R\sin t, ct \rangle$.
- Reversed curve $-C$: if $\mathbf{r}(t)$, $a \le t \le b$, traces $C$, then $\mathbf{r}(a + b - t)$ traces the same points backward.
A curve made of several smooth pieces, such as the boundary of a triangle, is parametrized one piece at a time, each piece in its direction of travel. We write $C = C_1 + C_2 + C_3$, and an integral over $C$ is the sum of the integrals over the pieces.
Why it works
Why the reversal works: $s = a + b - t$ runs from $b$ down to $a$ as $t$ runs from $a$ to $b$, so $\mathbf{r}(s)$ visits the points in the opposite order. Its velocity is $-\mathbf{r}'(s)$.
Worked examples
Worked example: A segment
- $B - A = \langle 3, 4 \rangle$.
- $\mathbf{r}(t) = \langle 1 + 3t, -2 + 4t \rangle$, ${}0 \le t \le 1$.
- Check: $\mathbf{r}(0) = (1, -2)$ and $\mathbf{r}(1) = (4, 2)$. The speed is $|\langle 3, 4 \rangle| = 5$, the length of the segment.
Answer: $\langle 1 + 3t, -2 + 4t \rangle$
Worked example: A clockwise circle
- Counterclockwise from $(5, -1)$ would be $\langle 2 + 3\cos t, -1 + 3\sin t \rangle$.
- Flip the sine to go clockwise: $\mathbf{r}(t) = \langle 2 + 3\cos t, -1 - 3\sin t \rangle$, ${}0 \le t \le 2\pi$.
- Check: at $t = \pi/2$ the point is $(2, -4)$, below the center, so the motion is clockwise.
Answer: $\langle 2 + 3\cos t, -1 - 3\sin t \rangle$
Worked example: A segment and its reverse in space
- $B - A = \langle 3, -2, 0 \rangle$, so $\mathbf{r}(t) = \langle 3t, 1 - 2t, 2 \rangle$, ${}0 \le t \le 1$.
- From $B$ to $A$: $B + t(A - B) = \langle 3 - 3t, -1 + 2t, 2 \rangle$, which is also $\mathbf{r}(1 - t)$.
Answer: $\langle 3t, 1 - 2t, 2 \rangle$ and $\langle 3 - 3t, -1 + 2t, 2 \rangle$
Common mistake