In single-variable calculus, you integrated a function $f(x)$ over an interval $[a,b]$ on the $x$-axis. $$\int_a^b f(x)\,dx$$ Now, we want to generalize this. What if we integrate a multivariable function $f(x,y)$ over a curve $C$ in the $xy$-plane?
To do this, we first need to understand how to "integrate" along a curve. Recall from Section 13.3 the definition of the derivative of a vector function: $$\vec{r}'(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}$$
This means for a very small $\Delta t$, the vector $\Delta \vec{r} = \vec{r}(t+\Delta t) - \vec{r}(t)$ is well-approximated by the tangent vector $\vec{r}'(t)\Delta t$. $$\Delta \vec{r} \approx \vec{r}'(t)\Delta t$$ The graph below shows this approximation. $\Delta \vec{r}$ is the secant vector (in blue) and $\vec{r}'(t)$ is the tangent vector (in black). Move the slider $h$ (which represents $\Delta t$) to 1 to see how $ds$ (arc length differential) is approximated by $|\vec{r}'(t)|h$.
The length of this small segment of the curve, $\Delta s$, is the length of the secant vector $\Delta \vec{r}$: $$\Delta s = |\Delta \vec{r}| \approx |\vec{r}'(t)|\Delta t$$
By summing up all these small pieces of arc length and taking the limit (a Riemann sum!), we get the arc length formula from Section 13.3: $$L = \int_a^b |\vec{r}'(t)|\,dt$$
From our approximation, we define the **differential of arc length**, $ds$, as this infinitesimal piece of length: $$ds = |\vec{r}'(t)|\,dt$$ Using this new notation, we can rewrite the arc length formula in a very compact way: $$L = \int_C 1\,ds$$ This makes intuitive sense: integrating the function '1' along a curve $C$ just means "sum up all the tiny pieces of length ($ds$)," which gives the total length of $C$.
Today, we ask: what if we integrate a function $f(x,y)$ instead of just $1$? What is the meaning of $\displaystyle \int_C f(x,y)\,ds$?
Geometrically, the integral $\int_C f(x,y)\,ds$ represents the area of a "curtain" whose base is the curve $C$ in the $xy$-plane and whose height at any point $(x,y)$ on $C$ is given by the surface $z=f(x,y)$. The graph below visualizes this exact concept.
This lecture (Part A) will focus entirely on this first type of line integral, the **scalar line integral** $\int_C f\,ds$. In Part B, we will look at a second type, the **vector line integral** $\int_C \vec{F} \cdot d\vec{r}$.
Summing them up and taking the limit gives the formal definition:
$$\int_C f(x,y)\,ds=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*,y_i^*)\,\Delta s_i.$$Bounds:
When $t=0$, $u=1+4(0)^2=1$.
When $t=1$, $u=1+4(1)^2=5$.
Integrate:
$$\int_0^1 t\sqrt{1+4t^2}\,dt = \int_1^5 \sqrt{u}\,\left(\frac{1}{8}du\right) = \frac{1}{8} \int_1^5 u^{1/2}\,du$$
$$=\frac{1}{8} \left[ \frac{u^{3/2}}{3/2} \right]_1^5 = \frac{1}{8} \cdot \frac{2}{3} \left[ u^{3/2} \right]_1^5$$
$$=\frac{1}{12} \left( 5^{3/2} - 1^{3/2} \right) = \frac{1}{12} (5\sqrt{5} - 1).$$
Problem. $C$ is the line from $(0,0)$ to $(3,4)$ and $f(x,y)=2$. Compute $\displaystyle \int_C f\,ds$.
$\displaystyle \int_C 36y\,ds=\int_0^1 \underbrace{(36t^3)}_{f(x(t),y(t))}\cdot \underbrace{\sqrt{1+9t^4}\,dt}_{ds}$
This is a u-substitution. Let $u=1+9t^4$. Then $du=36t^3\,dt$.
Bounds: When $t=0$, $u=1+9(0)^4=1$. When $t=1$, $u=1+9(1)^4=10$.
$\displaystyle \int_1^{10} \sqrt{u}\,du = \int_1^{10} u^{1/2}\,du = \left[\frac{u^{3/2}}{3/2}\right]_1^{10} = \frac{2}{3}\left[u^{3/2}\right]_1^{10} = \frac{2}{3}(10^{3/2} - 1^{3/2}) = \frac{2}{3}(10\sqrt{10} - 1).$
Bounds: When $\theta=0$, $u=\cos(0)=1$. When $\theta=\pi$, $u=\cos(\pi)=-1$.
$$\int_0^\pi \cos^2\theta\,\sin\theta\,d\theta=\int_{1}^{-1} u^2(-du)=\int_{-1}^{1} u^2\,du =\left[\frac{u^3}{3}\right]_{-1}^{1}=\frac{1}{3}-\left(-\frac{1}{3}\right)=\frac{2}{3}.$$
Problem. $C$ is $\vec r(t)=\langle t^2,\,t\rangle$, $0\le t\le 2$. Set up and compute $\displaystyle \int_C (x/y)\,ds$.
Let $u=4t^2+1$. Then $du=8t\,dt$, which means $t\,dt = \frac{1}{8}du$.
Bounds:
When $t=0$, $u=4(0)^2+1=1$.
When $t=2$, $u=4(2)^2+1=17$.
$$\int_1^{17} \sqrt{u} \left(\frac{1}{8}du\right) = \frac{1}{8} \int_1^{17} u^{1/2}\,du$$
$$= \frac{1}{8} \left[ \frac{u^{3/2}}{3/2} \right]_1^{17} = \frac{1}{8} \cdot \frac{2}{3} \left[ u^{3/2} \right]_1^{17}$$
$$= \frac{1}{12} \left( 17^{3/2} - 1^{3/2} \right) = \boxed{\frac{17\sqrt{17}-1}{12}}$$
A common question is: "What if I parametrize the curve differently? Will I get the same answer?"
Yes. For a **scalar line integral** ($\int_C f\,ds$), the value is **independent of the parametrization** you choose, as long as the curve $C$ is traced exactly once.
For example, in 2B, we could have used $\vec{r}(t) = \langle t, \sqrt{1-t^2} \rangle$ for $-1 \le t \le 1$. The calculation would be much harder, but the final answer would still be $\frac{2}{3}$. This is because $ds$, the arc length, is an intrinsic geometric property of the curve itself. It doesn't depend on how fast you "walk" along it.
This is in sharp contrast to the integrals we will see in Part B, which do depend on direction.
If a thin wire is shaped like a curve $C$ and has a linear density function $\rho(x,y)$ (units of mass per unit length), we can find its total mass and center of mass by integrating.
A small piece of the wire with length $ds$ has mass $dm = \rho(x,y)\,ds$. Integrating this gives the total mass: $$m=\int_C \rho(x,y)\,ds$$
The moments about the $x$ and $y$ axes are found by weighting the density by the distance ($y$ or $x$): $$M_x=\int_C y\,\rho(x,y)\,ds \qquad \text{and} \qquad M_y=\int_C x\,\rho(x,y)\,ds$$
The center of mass $(\bar x,\bar y)$ is the balance point, given by: $$(\bar x,\bar y)=\left(\frac{M_y}{m},\frac{M_x}{m}\right)$$
Answer: $\boxed{(\bar x,\bar y) = \left(0, \frac{2R}{\pi}\right)}$.
A common and excellent question is: "How can the center of mass not be on the wire?"
The center of mass $(\bar{x}, \bar{y})$ is the balance point of the entire mass distribution, not a point that must physically be on the object. Think of a donut or a ring: its center of mass is in the empty hole, as that's the only point where it would balance.
Our 'C'-shaped wire is the same. The center of mass is the point where the total rotational forces (moments) from all parts of the wire perfectly cancel out. The calculated point $(0, \frac{2R}{\pi}) \approx (0, 0.637 R)$ is the unique balance point below the top of the arc $(0,R)$.
Answer: $\boxed{(\bar x,\bar y) = \left(0, \frac{\pi R}{4}\right)}$. (Note: $\frac{\pi}{4} \approx 0.785$, which is higher than $\frac{2}{\pi} \approx 0.637$. This makes sense, as the wire is now heavier at the top).
Problem. A wire $C$ is given by the parabola $\vec{r}(t) = \langle t^2, t \rangle$ from $t=0$ to $t=2$. Its density is $\rho(x,y)=y$. Find the total mass $m = \int_C y\,ds$.
$\displaystyle m=\int_C \rho\,ds=\int_0^2 (t)\cdot (\sqrt{4t^2+1}\,dt)$
Let $u=4t^2+1$. Then $du=8t\,dt \implies t\,dt = \frac{1}{8}du$.
Bounds: When $t=0$, $u=4(0)^2+1=1$. When $t=2$, $u=4(2)^2+1=17$.
$\displaystyle m = \int_1^{17} \sqrt{u} \left(\frac{1}{8}du\right) = \frac{1}{8} \int_1^{17} u^{1/2}\,du = \frac{1}{8} \left[\frac{u^{3/2}}{3/2}\right]_1^{17} = \frac{1}{12}\left[u^{3/2}\right]_1^{17} = \frac{1}{12}(17^{3/2} - 1^{3/2}) = \boxed{\frac{17\sqrt{17}-1}{12}}.$
The first step in any line integral is to parametrize the curve $C$. Here is a guide to the most common types you will encounter.
To parametrize a line segment from a starting point $\vec{r}_0$ to an ending point $\vec{r}_1$, we use the formula: $$\vec{r}(t) = \vec{r}_0 + t\vec{v}, \quad 0 \le t \le 1$$ where $\vec{v} = \vec{r}_1 - \vec{r}_0$ is the direction vector from the start to the end.
Problem: Parametrize the line segment $C$ from $A=(1, 2)$ to $B=(4, -1)$.
This is the simplest case. If $C$ is a graph $y=f(x)$ from $x=a$ to $x=b$, just let $x$ be the parameter.
Parametrization: Let $x=t$. Then $y=f(t)$. $$\vec{r}(t) = \langle t, f(t) \rangle, \quad a \le t \le b$$
Problem: Parametrize $y=x^2$ from $x=0$ to $x=1$.
This is similar to the case above, but now we let $y$ be the parameter.
Parametrization: Let $y=t$. Then $x=g(t)$. $$\vec{r}(t) = \langle g(t), t \rangle, \quad c \le t \le d$$
Problem: Parametrize $x=y^2$ from $y=0$ to $y=2$.
We use polar/cylindrical coordinates. For a circle of radius $R$ centered at the origin, $x=R\cos t$ and $y=R\sin t$. The bounds for $t$ (which is $\theta$) determine how much of the circle is traced.
Problem 1: A full circle of radius $R$, counter-clockwise.
Solution: $\vec{r}(t) = \langle R\cos t, R\sin t \rangle$, for $0 \le t \le 2\pi$.
Problem 2: The top semicircle of radius $R$, from right to left.
Solution: This is $y \ge 0$. This corresponds to $t$ (or $\theta$) from $0$ to $\pi$.
$\vec{r}(t) = \langle R\cos t, R\sin t \rangle$, for $0 \le t \le \pi$. (This matches Examples 2B, 3A, 3B).
Problem 3: The quarter circle of radius $R$ in the first quadrant, from bottom to top.
Solution: This corresponds to $t$ from $\pi/2$ (at $(0,R)$) down to $0$ (at $(R,0)$). Oh, wait! From *bottom to top* means starting at $(R,0)$ and ending at $(0,R)$.
This corresponds to $t$ from $0$ to $\pi/2$.
$\vec{r}(t) = \langle R\cos t, R\sin t \rangle$, for $0 \le t \le \pi/2$.
In Part B, we will explore the second type of line integral: the **vector line integral** (or **work integral**), $\int_C \vec{F} \cdot d\vec{r}$.