In Part A, we introduced the **scalar line integral** $\int_C f\,ds$, which we used to find "curtain area" and the mass of a wire.
In Part B, we will first extend this scalar integral to 3D, and then introduce the second major type of line integral: the **vector line integral** (or **work integral**) $\int_C \vec{F} \cdot d\vec{r}$.
The concept of a scalar line integral extends naturally to 3D. If $C$ is a space curve parametrized by $\vec r(t)=\langle x(t),y(t),z(t)\rangle$ for $a\le t\le b$, and $f(x,y,z)$ is a scalar function defined on $C$, then:
This could represent the mass of a 3D wire with density $f(x,y,z)$, or the integral of a scalar field (like temperature) along a path.
Note: It is difficult to visualize this integral in the same way we did in 2D. The "curtain" for a 2D integral $\int_C f(x,y)\,ds$ lives in 3D (as $z=f(x,y)$). To visualize the "curtain" for a 3D integral $\int_C f(x,y,z)\,ds$, we would need a 4th spatial dimension to graph $w=f(x,y,z)$. Instead, we often visualize this by color-mapping the curve $C$ according to the value of $f$ at each point.
Start: $\vec{r}_0 = \langle 0,0,0 \rangle$.
End: $\vec{r}_1 = \langle 1,1,1 \rangle$.
Vector: $\vec{v} = \vec{r}_1 - \vec{r}_0 = \langle 1,1,1 \rangle$.
So, $\vec r(t)= \langle 0,0,0 \rangle + t\langle 1,1,1 \rangle = \langle t,t,t\rangle$, for $0\le t\le 1$.
Problem. Compute $\displaystyle \int_C z\,ds$ where $C$ is one turn of the helix $\vec{r}(t) = \langle \cos t, \sin t, t \rangle$ from $t=0$ to $t=2\pi$.
We now introduce the second major type of line integral. Instead of integrating a scalar field $f$ (like density or temperature), we will integrate a vector field $\vec{F}$ (like a force field or a fluid velocity field).
The main application is **Work**. Let's build this concept up from what you already know.
We can't just multiply. We must add up the small pieces of work. On a tiny piece of the path $\Delta \vec{r}$, the force $\vec{F}$ is *almost* constant. The small amount of work done is $\Delta W \approx \vec{F} \cdot \Delta \vec{r}$.
Summing these up and taking the limit gives the **Work Integral**: $$W = \int_C \vec{F} \cdot d\vec{r}$$
This new notation $\int_C \vec{F} \cdot d\vec{r}$ is what we must learn to compute.
Let $\vec{F} = \langle P(x,y), Q(x,y) \rangle$ be our force field.
Let $C$ be parametrized by $\vec{r}(t) = \langle x(t), y(t) \rangle$ for $a \le t \le b$.
The differential $d\vec{r}$ represents the infinitesimal displacement vector $\langle dx, dy \rangle$. From $\vec{r}'(t) = \langle x'(t), y'(t) \rangle$, we can write this as: $$d\vec{r} = \vec{r}'(t)\,dt = \langle x'(t), y'(t) \rangle\,dt = \langle x'(t)dt, y'(t)dt \rangle = \langle dx, dy \rangle$$
Now, we can write out the dot product $\vec{F} \cdot d\vec{r}$: $$\vec{F} \cdot d\vec{r} = \langle P, Q \rangle \cdot \langle dx, dy \rangle = P\,dx + Q\,dy$$
This gives us two new, equivalent notations for the work integral, and a way to compute it:
We've derived $W = \int P\,dx + Q\,dy$, but what do these new integrals mean on their own?
This also explains why direction matters: if the path moves to the left (so $dx$ is negative), the $xz$-projection accumulates negative area.
$x(t)=t^2 \implies x'(t)=2t \implies dx = 2t\,dt$.
$y(t)=t^3 \implies y'(t)=3t^2 \implies dy = 3t^2\,dt$.
$(2x)\,dx+(3y)\,dy = (2(t^2))\,(2t\,dt)+(3(t^3))\,(3t^2\,dt)$
$= 4t^3\,dt+9t^5\,dt = (4t^3+9t^5)\,dt.$
$\displaystyle \int_0^1 (4t^3+9t^5)\,dt = \left[t^4 + \frac{9t^6}{6}\right]_0^1 = \left[t^4 + \frac{3}{2}t^6\right]_0^1$
$ = \left(1^4 + \frac{3}{2}(1)^6\right) - (0) = 1 + \frac{3}{2} = \boxed{\frac{5}{2}}.$
(This is much faster than parametrizing. The full method would be: $\vec{r}_0 = \langle 0,0,0 \rangle$ and $\vec{v} = \langle 2,1,0 \rangle - \langle 0,0,0 \rangle = \langle 2,1,0 \rangle$. So, $\vec{r}(t) = \langle 2t, t, 0 \rangle$ for $0 \le t \le 1$. Then $d\vec{r} = \langle 2, 1, 0 \rangle dt$. The integral becomes $\int_0^1 \langle 1,2,3 \rangle \cdot \langle 2,1,0 \rangle dt = \int_0^1 4\,dt = 4$.)
Problem. Let $C$ be the quarter-circle $\vec r(t)=\langle \cos t,\ \sin t\rangle$, from $t=0$ to $t=\pi/2$. Compute $\displaystyle \int_C x\,dy - y\,dx$.
$x(t)=\cos t \implies dx = -\sin t\,dt$.
$y(t)=\sin t \implies dy = \cos t\,dt$.
(Note: The integral $\int_C x\,dy - y\,dx$ is famous. As we'll see, $\frac{1}{2}\int_C x\,dy - y\,dx$ computes the area enclosed by the curve $C$. The area of this quarter-circle is $\frac{1}{4}\pi r^2 = \frac{\pi}{4}$. Our integral gave $\frac{\pi}{2}$, which is $2 \cdot (\text{Area})$. This is a preview of Green's Theorem!)
Problem. Compute $\displaystyle \int_C x\,ds$ where $C$ is the helix $\vec{r}(t) = \langle \cos t, \sin t, t \rangle$ from $t=0$ to $t=\pi$.
$\displaystyle \int_C x\,ds = \int_0^\pi (\cos t) \cdot (\sqrt{2}\,dt) = \sqrt{2} \int_0^\pi \cos t\,dt$
$\displaystyle = \sqrt{2} [\sin t]_0^\pi = \sqrt{2}(\sin \pi - \sin 0) = \sqrt{2}(0-0) = \boxed{0}.$
Problem. Compute $\displaystyle \int_C (y\,dx - x\,dy)$ along $\vec r(t)=\langle t, t^3\rangle$, $0\le t\le 1$.
Problem. Compute $W = \displaystyle \int_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = \langle z, y, x \rangle$ and $C$ is the helix $\vec{r}(t) = \langle \cos t, \sin t, t \rangle$ from $t=0$ to $t=\pi/2$.
$x(t)=\cos t, \quad y(t)=\sin t, \quad z(t)=t$
$d\vec{r} = \langle -\sin t, \cos t, 1 \rangle\,dt$.
$\vec{F}(\vec{r}(t)) = \langle z(t), y(t), x(t) \rangle = \langle t, \sin t, \cos t \rangle$.
$\langle t, \sin t, \cos t \rangle \cdot \langle -\sin t, \cos t, 1 \rangle\,dt$
$= (-t\sin t + \sin t\cos t + \cos t)\,dt$
Split into three parts:
1. $\int_0^{\pi/2} \cos t\,dt = [\sin t]_0^{\pi/2} = 1-0 = 1$.
2. $\int_0^{\pi/2} \sin t\cos t\,dt = \int_0^{\pi/2} \frac{1}{2}\sin(2t)\,dt = \left[-\frac{1}{4}\cos(2t)\right]_0^{\pi/2} = -\frac{1}{4}(\cos\pi - \cos 0) = -\frac{1}{4}(-1 - 1) = \frac{1}{2}$.
3. $\int_0^{\pi/2} -t\sin t\,dt$: Use Integration by Parts. Let $u=-t, dv=\sin t\,dt$. Then $du=-dt, v=-\cos t$.
$\int u\,dv = uv - \int v\,du = [-t(-\cos t)]_0^{\pi/2} - \int_0^{\pi/2} (-\cos t)(-dt)$
$= [t\cos t]_0^{\pi/2} - \int_0^{\pi/2} \cos t\,dt = (\frac{\pi}{2}\cos\frac{\pi}{2} - 0) - [\sin t]_0^{\pi/2} = (0) - (1-0) = -1$.
Total: $1 + \frac{1}{2} + (-1) = \boxed{\frac{1}{2}}$.