Line Integrals with Respect to Arc Length
Integrate a scalar function along a curve-the "curtain area" interpretation - Kaplan §5.3
Prereq: §5.2 Line Integrals in the Plane
A wire bent into a circle has varying density-thicker in some spots, thinner in others. How do we add up all that mass?
What does $\mathrm{d}s$ actually mean?
Take a curve. Any curve. Now chop it into tiny straight pieces. Each little piece has some length $\Delta s$-you can measure it with a ruler.
How long is each piece? Just Pythagoras:
Use the slider below to chop a quarter-circle into more and more segments. Watch how the red highlighted segment gets shorter and the straight-line approximation hugs the curve more tightly.
In the limit, as $n\to\infty$, each $\Delta s$ becomes the infinitesimal $\mathrm{d}s$. That's all $\mathrm{d}s$ is: a tiny bit of arc length.
Now here's the key question. If the curve is parametrized by $t$-meaning $\mathbf{r}(t) = (x(t),\,y(t))$-what is $\mathrm{d}s$ in terms of $\mathrm{d}t$? Think about it for a second before reading on.
We just divide $\Delta s$ by $\Delta t$ and take the limit:
$$\frac{\mathrm{d}s}{\mathrm{d}t} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = |\mathbf{r}'(t)|$$
So:
That factor $|\mathbf{r}'(t)|$ is just the speed-how fast the point moves through space as $t$ changes. Fast parametrization means each $\mathrm{d}t$ covers more arc length. Makes sense.
Trap: $\mathrm{d}s$ is always positive. Even if you reverse direction along the curve, $\mathrm{d}s$ doesn't flip sign. This is fundamentally different from $\mathrm{d}x$ or $\mathrm{d}y$, which do change sign when you reverse. Keep this in mind-it matters later.
The curtain interpretation
Suppose we have a curve $C$ lying flat in the $xy$-plane, and a function $f(x,y) \ge 0$ giving a "height" at each point. Imagine lifting each point on $C$ straight up to height $f(x,y)$. What you get is a surface that looks like a curtain hanging from a curved rod.
The line integral $\int_C f\,\mathrm{d}s$ is the area of that curtain.
The green curve on the ground is $C$ (quarter-circle). The red line on top traces height $f(x,y)=0.5+0.8xy$. The shaded surface between them is the "curtain" whose area equals $\int_C f\,\mathrm{d}s$.
Let's build intuition with a prediction. If $f = 1$ everywhere-the curtain has constant height 1-what should the integral be?
Just the arc length of $C$. That's a nice sanity check: $\int_C 1\,\mathrm{d}s = \text{length of } C$.
Now, how do we actually compute this? We already know $\mathrm{d}s = |\mathbf{r}'(t)|\,\mathrm{d}t$, so we just substitute:
That's the whole recipe: plug the curve into $f$, multiply by the speed, integrate over $t$.
Quick check. Take the quarter-circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le\pi/2$, and $f(x,y) = xy$. Then $|\mathbf{r}'|=1$ (unit-speed!), so we just need $\int_0^{\pi/2} \cos t\sin t\,\mathrm{d}t = \frac{1}{2}\sin^2 t\big|_0^{\pi/2} = \frac{1}{2}$. Clean.
Why parametrization doesn't matter
Here's something genuinely beautiful. Two people can parametrize the same curve in completely different ways-one racing through, the other going slowly-and they'll get the exact same integral. Every time.
Why? Because $\mathrm{d}s$ absorbs the speed factor. If you speed up the parametrization, $|\mathbf{r}'(t)|$ gets bigger, but $\mathrm{d}t$ gets correspondingly smaller (you cover the same arc in less parameter-time). The product $|\mathbf{r}'(t)|\,\mathrm{d}t = \mathrm{d}s$ stays the same.
Let's see this concretely. Below, the same quarter-circle is parametrized two ways with $f(x,y) = x + y$. The shaded areas look totally different-different shapes, different widths-but they have the same area.
The integral sees the geometry-the actual curve and the function on it-not the arbitrary choice of how fast you trace through. That's what makes $\int_C f\,\mathrm{d}s$ a geometric quantity, not a parametric one.
Direction and other properties
Since $\mathrm{d}s$ is always positive (remember the trap from Card 1?), reversing the direction of travel along $C$ doesn't change $\int_C f\,\mathrm{d}s$. Walk the wire left-to-right or right-to-left-the total mass is the same.
The other key properties are what you'd expect from any sensible integral:
Linearity. Constants and sums pull through:
$$\int_C (a\,f + b\,g)\,\mathrm{d}s = a\!\int_C f\,\mathrm{d}s \;+\; b\!\int_C g\,\mathrm{d}s$$
Additivity. If you split $C$ into two pieces $C_1$ and $C_2$ joined at a point, the integrals add:
$$\int_C f\,\mathrm{d}s = \int_{C_1} f\,\mathrm{d}s + \int_{C_2} f\,\mathrm{d}s$$
Connection to $\int P\,dx + Q\,dy$. In §5.2 we worked with integrals like $\int_C P\,dx + Q\,dy$. How does that relate to $\int_C f\,\mathrm{d}s$?
If $\alpha$ is the angle the tangent makes with the $x$-axis, then $dx = \cos\alpha\,\mathrm{d}s$ and $dy = \sin\alpha\,\mathrm{d}s$, so:
$$\int_C P\,dx + Q\,dy = \int_C (P\cos\alpha + Q\sin\alpha)\,\mathrm{d}s$$
In other words, $\int P\,dx + Q\,dy$ is just a $\mathrm{d}s$-integral where $f = P\cos\alpha + Q\sin\alpha$. The difference: this one does depend on direction (because $\cos\alpha$ and $\sin\alpha$ flip sign when you reverse), while $\int f\,\mathrm{d}s$ does not.
Practice Problems - §5.3
From Kaplan, problems after §5.3
Evaluate $\displaystyle\int_C x\,dy$ along the straight line from $(0,0)$ to $(2,1)$.
Step 1: Parametrize the path.
The straight line from $(0,0)$ to $(2,1)$ can be parametrized as $x = 2t$, $y = t$, $0 \le t \le 1$. Then $dy = dt$.
Step 2: Substitute and evaluate.
$$\int_C x\,dy = \int_0^1 2t\,dt = \bigl[t^2\bigr]_0^1 = 1$$
Result: $\displaystyle\int_C x\,dy = 1$.
On a straight line the parametrization is linear, so line integrals reduce to simple polynomial integrals.Evaluate $\displaystyle\int_C \frac{y\,dx - x\,dy}{x^2+y^2}$ along the curve $x = \cos^3 t$, $y = \sin^3 t$, $0 \le t \le \pi/2$. (Hint: try $u = \tan^3 t$.)
Step 1: Compute the differentials.
$dx = -3\cos^2 t\sin t\,dt$, $dy = 3\sin^2 t\cos t\,dt$.
Step 2: Build the numerator.
$$y\,dx - x\,dy = \sin^3 t(-3\cos^2 t\sin t)\,dt - \cos^3 t(3\sin^2 t\cos t)\,dt$$ $$= -3\cos^2 t\sin^4 t\,dt - 3\sin^2 t\cos^4 t\,dt = -3\sin^2 t\cos^2 t\bigl(\sin^2 t + \cos^2 t\bigr)\,dt$$ $$= -3\sin^2 t\cos^2 t\,dt$$
Step 3: Simplify the denominator.
$x^2 + y^2 = \cos^6 t + \sin^6 t$. Using $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ with $a = \cos^2 t$, $b = \sin^2 t$:
$$\cos^6 t + \sin^6 t = 1 \cdot \bigl((\cos^2 t + \sin^2 t)^2 - 3\sin^2 t\cos^2 t\bigr) = 1 - 3\sin^2 t\cos^2 t$$
Step 4: Set up the integral and substitute.
The integral becomes $\displaystyle\int_0^{\pi/2}\frac{-3\sin^2 t\cos^2 t}{1 - 3\sin^2 t\cos^2 t}\,dt$. Dividing numerator and denominator by $\cos^6 t$ and setting $u = \tan^3 t$ (so $du = 3\tan^2 t\sec^2 t\,dt$), after simplification the integral transforms to
$$-\int_0^{\infty}\frac{du}{1 + u^2} = -\bigl[\arctan u\bigr]_0^{\infty} = -\frac{\pi}{2}$$
Result: $\displaystyle\int_C \frac{y\,dx - x\,dy}{x^2+y^2} = -\frac{\pi}{2}$.
The key insight is that the substitution $u = \tan^3 t$ converts a complicated trigonometric integrand into the standard $1/(1+u^2)$ form. The appearance of $\pi$ in the answer is a hint that an arctangent lurks inside the problem.Evaluate $\displaystyle\oint_C x^2 y^2\,dx - xy^3\,dy$ where $C$ is the triangle with vertices $(0,0)$, $(1,0)$, $(1,1)$, traversed counterclockwise.
Step 1: Side 1 - bottom: $(0,0) \to (1,0)$.
Here $y = 0$, $dy = 0$, so $x^2 y^2\,dx - xy^3\,dy = 0$. Integral $= 0$.
Step 2: Side 2 - right: $(1,0) \to (1,1)$.
Here $x = 1$, $dx = 0$, $y$ goes from $0$ to $1$: $$\int_0^1 -1 \cdot y^3\,dy = -\left[\frac{y^4}{4}\right]_0^1 = -\frac{1}{4}$$
Step 3: Side 3 - hypotenuse: $(1,1) \to (0,0)$.
Here $y = x$, $dy = dx$. With $x$ going from $1$ to $0$: $$\int_1^0 \bigl(x^2 \cdot x^2 - x \cdot x^3\bigr)\,dx = \int_1^0 (x^4 - x^4)\,dx = 0$$
Result: $\displaystyle\oint_C x^2 y^2\,dx - xy^3\,dy = 0 - \frac{1}{4} + 0 = -\frac{1}{4}$.
Two of the three sides contribute nothing - the bottom because $y = 0$ kills every term, and the hypotenuse because $y = x$ makes the two pieces cancel exactly. Only the vertical side survives.$x = 2\cos t$, $y = 2\sin t$, $0\le t\le 2\pi$.
$x'(t) = -2\sin t$, $y'(t) = 2\cos t$ $\Rightarrow$ $|\mathbf{r}'(t)| = \sqrt{4\sin^2 t + 4\cos^2 t} = 2$.
$f(\mathbf{r}(t)) = 4\cos^2 t - 4\sin^2 t = 4\cos 2t$.
$\displaystyle\int_0^{2\pi} 4\cos(2t)\cdot 2\,\mathrm{d}t = 8\int_0^{2\pi}\cos(2t)\,\mathrm{d}t = 8\cdot 0 = 0$.
Not surprising-$x^2 - y^2$ is positive in some quadrants and negative in others, and the circle's symmetry makes everything cancel perfectly.
$x = t$, $y = t$, $0\le t\le 1$.
$x'=1$, $y'=1$ $\Rightarrow$ $|\mathbf{r}'| = \sqrt{1+1} = \sqrt{2}$.
$\displaystyle\int_0^1 t\cdot\sqrt{2}\,\mathrm{d}t = \sqrt{2}\left[\frac{t^2}{2}\right]_0^1 = \frac{\sqrt{2}}{2}$.
$x = t$, $y = t^2$, $0\le t\le 1$. Here $f = 1$ (we're just measuring length).
$x'=1$, $y'=2t$ $\Rightarrow$ $|\mathbf{r}'| = \sqrt{1+4t^2}$.
We need $\displaystyle\int_0^1 \sqrt{1+4t^2}\,\mathrm{d}t$. This is an arc length integral - the classic $\sqrt{1+(\text{something})^2}$ pattern. The standard trick: let $2t = \sinh u$, so $\mathrm{d}t = \tfrac{1}{2}\cosh u\,\mathrm{d}u$ and the square root simplifies beautifully: $\sqrt{1+\sinh^2 u} = \cosh u$. When $t=0$, $u=0$; when $t=1$, $u = \operatorname{arcsinh}2 = \ln(2+\sqrt{5})$.
The integral becomes $\displaystyle\frac{1}{2}\int_0^{\ln(2+\sqrt{5})}\cosh^2 u\,\mathrm{d}u$. Using $\cosh^2 u = \tfrac{1}{2}(1 + \cosh 2u)$, we get $\displaystyle\frac{1}{2}\left[\frac{u}{2} + \frac{\sinh 2u}{4}\right]_0^{\ln(2+\sqrt{5})}$.
At the upper limit, $\sinh(2u) = 2\sinh u\cosh u = 2 \cdot 2 \cdot \sqrt{5} = 4\sqrt{5}$, so the answer works out to $\dfrac{\sqrt{5}}{2} + \dfrac{\ln(2+\sqrt{5})}{4}$.
This is a typical arc length answer - irrational, involves a logarithm. If you see $\sqrt{1+4t^2}$ on an exam, reach for the hyperbolic substitution.