Line Integrals in the Form ∫P dx + Q dy

Line integrals of vector fields are often written without vectors at all: $\int_C P\,dx + Q\,dy$. This is the same thing as $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle P, Q \rangle$, written component by component. The notation is handy when the curve is a graph or is made of horizontal and vertical pieces. It builds on line integrals of vector fields.

Key idea

With $d\mathbf{r} = \langle dx, dy, dz \rangle$, $$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C P\,dx + Q\,dy + R\,dz = \int_a^b \left[P\,x'(t) + Q\,y'(t) + R\,z'(t)\right] dt.$$ On a horizontal segment $dy = 0$; on a vertical segment $dx = 0$. On a graph $y = g(x)$ you can use $x$ itself as the parameter: $dy = g'(x)\,dx$.

Why it works

$dx = x'(t)\,dt$ is the differential of $x = x(t)$, the same substitution as in u-substitution. Writing $\langle P, Q \rangle \cdot \langle x'(t), y'(t) \rangle\,dt$ out in components gives $P\,x'(t)\,dt + Q\,y'(t)\,dt$, which is $P\,dx + Q\,dy$. On a horizontal segment $y$ is constant, so $y'(t) = 0$ and the $Q$ term disappears.

The field $\langle y, x^2 \rangle$ along the parabola $y = x^2$ from $(0, 0)$ to $(1, 1)$: the running total ends at $\frac{5}{6}$, the value found in the first example.

Orientation still matters in this notation: reversing the curve changes the sign of $\int_C P\,dx + Q\,dy$, because every $dx$ and $dy$ changes sign. The pieces $\int_C P\,dx$ and $\int_C Q\,dy$ can also be computed separately and added, which helps when one of them is easy. For the path from $(0, 0)$ to $(1, 0)$ and then up to $(1, 1)$, $\int_C y\,dx$ is ${}0$ on the first piece (where $y = 0$) and ${}0$ on the second (where $dx = 0$), so only $\int_C x^2\,dy$ needs work. In space, $\int_C P\,dx + Q\,dy + R\,dz$ works the same way, with $dz = z'(t)\,dt$.

Worked examples

Worked example: Along a parabola

Evaluate $\int_C y\,dx + x^2\,dy$, where $C$ is $y = x^2$ from $(0, 0)$ to $(1, 1)$.

Use $x$ as the parameter: $y = x^2$, $dy = 2x\,dx$, $x$ from ${}0$ to ${}1$.
The integrand becomes $x^2\,dx + x^2 \cdot 2x\,dx = (x^2 + 2x^3)\,dx$.
$\int_0^1 (x^2 + 2x^3)\,dx = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$.

Answer: $\frac{5}{6}$

Worked example: Along two segments

Evaluate the same integral $\int_C y\,dx + x^2\,dy$ along $(0, 0) \to (1, 0) \to (1, 1)$.

First piece: $y = 0$, $dy = 0$, $x$ from ${}0$ to ${}1$: $\int_0^1 0\,dx = 0$.
Second piece: $x = 1$, $dx = 0$, $y$ from ${}0$ to ${}1$: $\int_0^1 1\,dy = 1$.
Total ${}1$, not $\frac{5}{6}$: for this field the path matters.

Answer: ${}1$

Worked example: Around a circle

Evaluate $\oint_C x\,dy - y\,dx$ around the unit circle, counterclockwise.

$x = \cos t$, $y = \sin t$: $dx = -\sin t\,dt$, $dy = \cos t\,dt$.
$$x\,dy - y\,dx = (\cos^2 t + \sin^2 t)\,dt = dt.$$
$\int_0^{2\pi} dt = 2\pi$, twice the area of the disk. (Green's theorem explains why.)

Answer: ${}2\pi$

Line Integrals in the Form ∫P dx + Q dy

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