Assume that C is a (smooth) curve in a plane or space that is parameterized by the vector function \(\overline{l}\left(t\right)=\left\langle x(t),y(t),z(t)\right\rangle,\) where \(t\in[a,b].\)(Note that if is a curve in a plane, then we drop the third component z(t).)
Remember that line integrals and path integrals are equivalent.
What does \(\underset{\mathrm{c}}{\operatorname*{\int}}dl\) mean?
The integral \(\underset{\mathrm{c}}{\operatorname*{\int}}dl\) is a line integral of a scalar function f(x,y)=1 or f(x, y, z)=1, and represents the length of C.
To compute \(\underset{\mathrm{c}}{\operatorname*{\int}}dl\)
- In R2 (a plane),
\(\int\limits_{\mathrm{C}}dl=\int\limits_{a}^{b}\sqrt{\left(x'\left(t\right)\right)^2+\left(y'\left(t\right)\right)^2}dt=\int\limits_{a}^{b}\left|\overline{l'}\left(t\right)\right|dt.\)
- In R3 (space),
\(\int\limits_{\mathrm{C}}dl=\int\limits_{a}^{b}\sqrt{\left(x'\left(t\right)\right)^2+\left(y'\left(t\right)\right)^2+\left(z'\left(t\right)\right)^2} dt=\int\limits_{a}^{b}\left|\overline{l'}\left(t\right)\right|dt.\)
What does \(\underset{\mathrm{c}}{\operatorname*{\int}}d\overline{l}\) mean?
The line integral \(\underset{\mathrm{c}}{\operatorname*{\int}}d\overline{l}\) is the displacement vector of the curve / path \(\overline{l}\left(t\right).\) By definition
\(\int\limits_{\mathbb{C}}d\overline{l}=\int\limits_{a}^{b}\overline{l'}(t)dt,\)
where \(\overline{l^{\prime}}(t)\) is the tangent vector to the trajectory of a particle (or object) represented as a vector function \(\overline{l}\left(t\right).\)
The total distance along the curve \(\overline{l}\left(t\right)\) is calculated by \(\underset{\mathrm{C}}{\operatorname*{\int}}\left|\overline{l^{\prime}}(t)\right|dt.\)
Example. If \(\overline{l}\left(t\right)=\left\langle\cos t,\sin t\right\rangle \) for \(0\leq t\leq\pi,\) find the displacement of \(\overline{l}\left(t\right)\) and the total distance along the curve.
Solution. For the displacement of \(\overline{l}\left(t\right)\), we find that \(\overline{l^{\prime}}(t)=\langle-\sin t,\cos t\rangle \) and calculate
\(\int_0^\pi d\overline{l}=\int_0^\pi\overline{l^{\prime}}(t)dt=\int_0^\pi\left\langle-\sin t,\cos t\right\rangle dt=\left\langle\cos t\mid_0^\pi,\sin t\mid_0^\pi\right\rangle=\left\langle-2,0\right\rangle.\)
For the total distance along the curve, we calculate
\(\int_0^\pi|\overline{l^{\prime}}(t)|dt=\int_0^\pi\left|\langle-\sin t,\cos t\rangle\right|dt=\int_0^\pi\sqrt{\left(-\sin t\right)^2+\left(\cos t\right)^2}dt=\int_0^\pi1dt=\pi.\)
What does \(\underset{\mathrm{c}}{\operatorname*{\int}}fds\) mean?
The integral \(\underset{\mathrm{c}}{\operatorname*{\int}}fds\) is the line integral of a continuous real-valued function f(x, y) or f(x, y, z), i.e., it represents the area of the region under the surface z= f(x, y) along the curve \(\overline{l}\left(t\right).\) We calculate as follows:
\(\underset{\mathrm{C}}{\operatorname*{\int}}f ds=\int_{a}^{b}f\left(\overline{l}\left(t\right)\right)\left|\overline{l}\left(t\right)\right|dt.\)
What does \(\int dA\) mean where A is a region in the plane.
The integral \(\int dA\) is equal to the area of A.
What does \(\iint_R f(x, y)\, dA\) mean?
Suppose that f(x, y) is continuous on some region in the xy-plane. The integral \(\iint_R f(x, y)\, dA\) represents the volume over the region under the surface which is defined by z= f(x, y). Note that \(\iint_R f(x, y)\, dA\) is evaluated as an iterated integral and thus dA=dxdy or dA=dydx.
What does \(\int\bar{F}\cdot d\bar{A}\) mean?
To calculate \(\int\bar{F}\cdot d\bar{A}\), we write the integral as
\(\int\overline{F}\cdot d\overline{A}=\int\overline{F}\cdot\overline{n}dA,\)
where \(\bar{n}\) is the unit normal of the planar region A and dA is a small patch of the region. Note that \(d\bar{A}=\bar{n}dA.\)
If \(\bar{F}\) is a constant vector field, then we can evaluate the integral as follows.
\(\begin{aligned}
&\int\bar{F}\cdot d\bar{A} \\
&=\int\bar{F}\cdot\overline{n}dA,\quad d\overline{A}=\overline{n}dA \\
&=\int\left|\overline{F}\right|\left|\overline{n}\right|\cos\left(\theta\right)dA,\quad\text{Theorem }\overline{A}\cdot\overline{B}=AB\cos\left(\theta\right) \\
&=\int(F)(1)\cos(\theta)dA \\
&=F\cos(\theta)\int dA,\quad F\cos(\theta)\text{is a constant and}\int dA\text{ is the area of }A \\
&=FA\cos(\theta)
\end{aligned}\)
What does \(\int_Bf\left(x,y,z\right)dV\) mean where B is a region in space?
In general, there is no interpretation.
If f(x, y, z)= 1, then this triple integral represents the volume of B.
If f(x, y, z) represents density, then this triple integral gives the total mass of B.
This triple integral is computed as an iterated integral where dV= dxdydz (or any other order which is convenient to compute).
Line Integrals of Vector Fields (Theory)
Suppose that \(\overline{F}=P\overline{i}+Q\overline{j}+R\overline{k}\) is a continuous force field on R3 (i.e., in 3-dimensions; thus P, Q, and R are functions of x, y, and z). Note that a force field in R2 (i.e. in 2- dimesions) can be thought of as a special case where R=0 and P and Q depend only on x and y.
Now, a formula can be derived for a force \(\bar{F}\) acting upon a particle that moves along the curve C (from some initial point to some terminal point).
The force \(\bar{F}\) in moving the particle from Pi-1 to Pi is approximately
\(\bar{F}\left(x_i^*,y_i^*,z_i^*\right)\cdot\left[\Delta s_i\bar{T}\left(x_i^*,y_i^*,z_i^*\right)\right]=\left[\bar{F}\left(x_i^*,y_i^*,z_i^*\right)\cdot\bar{T}\left(x_i^*,y_i^*,z_i^*\right)\right]\Delta s_i\)
where \(\bar{T}\)(xi*, yi*, zi*) is the unit tangent vector at a point (xi*, yi*, zi*) between Pi-1 and Pi on C (details of this will be provided in a vector calculus course; we do not have all concepts available, such as Mean Value Theorem, for a precise derivation). Here, the actual displacement \(\Delta s_{i}\) along C is approximated by \(\Delta s_{i}\)\(\bar{T}\)(xi*, yi*, zi*) .
Moving the particle along C from the initial point to the terminal point is approximately
\(\sum_{i=1}^n\biggl[\bar{F}\left(x_i^*,y_i^*,z_i^*\right)\cdot\bar{T}\left(x_i^*,y_i^*,z_i^*\right)\biggr]\Delta s_i\)
Recall that the definition of a definite integral is:
\(\lim_{n\to\infty}\sum_{i=1}^n\Delta xf\left(x_i^*\right)=\int_a^bf\left(x\right)dx.\)
As n, the number of subintervals, increases, the approximation improves. Thus, the force field \(\bar{F}\) is defined as the limit of the Riemann sums, namely,
\(\int\limits_{\mathrm{C}}\bar{F}\left(x,y,z\right)\cdot\bar{T}\left(x,y,z\right)ds=\int\limits_{\mathrm{C}}\bar{F}\cdot\bar{T}ds\)
If the curve C is given by the vector equation \(\overline{l}\left(t\right)=x\left(t\right)\overline{i}+y\left(t\right)\overline{j}+z\left(t\right)\overline{k} ,\) where \(t\in[a,b],\) then the tangent unit vector is
\(\bar{T}\left(t\right)=\frac{\overline{l'}\left(t\right)}{\left|\overline{l'}\left(t\right)\right|}.\)
Then, using the definition of a path integral and the fact that \(ds=\begin{vmatrix}l^{\prime}(t)\end{vmatrix}dt,\) which follows from the fact that \(\frac{ds}{dt}\) is the rate of change of the arc length over time; i.e., the speed of the particle, and since the motion of the particle is described by a position function, we obtain
\(\int_a^b\biggl[\overline{F}\left(\overline{l}\left(t\right)\right)\cdot\frac{\overline{l'}\left(t\right)}{\left|\overline{l'}\left(t\right)\right|}\biggr]\biggl|\overline{l'}\left(t\right)\biggr|dt=\int_a^b\overline{F}\left(\overline{l}\left(t\right)\right)\cdot\overline{l'}\left(t\right)dt=\int_a^b\overline{F}\cdot d\overline{l}\)
Example. If \(\bar{F}=\langle2xy,y\rangle,\) evaluate \(\int\bar{F}\cdot d\bar{L}\) around L as shown in the figure below.
Solution. We split L into three pieces as shown and consider each integral separately
We first consider L1. We will parametrize this line by x=2, y=t, 0≤t≤2. Then L1 is given by \(\overline{l}\left(t\right)=\left\langle2,t\right\rangle,0\leq t\leq2,\) and \(\overline{l^{\prime}}(t)=\langle0,1\rangle.\) Evaluating the integral, we obtain
\(\begin{aligned}\int_{L_1}\overline{F}\cdot d\overline{l}=\int_0^2\overline{F}\left(\overline{l}\left(t\right)\right)\cdot\overline{l^{\prime}}\left(t\right)dt=\int_0^2\left\langle4t,t\right\rangle\cdot\left\langle0,1\right\rangle dt=\int_0^2tdt=\frac{t^2}{2}|_0^2=2.\end{aligned}\)
Now we consider L2. We will parametrize this line by x=2-t, y=2, 0≤t≤2. Then L2 is given by \(\overline{l}\left(t\right)=\left\langle2-t,2\right\rangle,0\leq t\leq2,\) and \(\overline{l^{\prime}}(t)=\langle-1,0\rangle.\) Evaluating the integral, we obtain
\(\begin{aligned}
\int_{L_{2}}\bar{F}\cdot d\bar{l}& =\int_0^2\overline{F}\left(\overline{l}\left(t\right)\right)\cdot\overline{l'}\left(t\right)dt=\int_0^2\left\langle4\left(2-t\right),2\right\rangle\cdot\left\langle-1,0\right\rangle dt \\
&=\int_0^2-4\left(2-t\right)dt=-4\int_0^2\left(2-t\right)dt=-4\left[2t-\frac{t^2}2\right]_0^2=-8.
\end{aligned}\)
Finally, we consider L3. We will parameterize this line by x=2 sin t, y= 2cos t, \(0\leq t\leq\frac\pi2\text{.}\) Then L3 is given by \(\overline{l}\left(t\right)=\left\langle2\sin t,2\cos t\right\rangle,\) \(0\leq t\leq\frac\pi2\text{,}\) and \(\overline{l^{\prime}}(t)=\langle2\cos t,-2\sin t\rangle.\) Evaluating the integral, we obtain
\(\begin{aligned}
\int_{L_{3}}\bar{F}\cdot d\bar{l}& =\int_0^{\frac{\pi}{2}}\bar{F}\left(\overline{l}\left(t\right)\right)\cdot\overline{l^{\prime}}(t)dt \\
&=\int_0^{\frac\pi2}\left\langle8\sin t\cos t,2\cos t\right\rangle\cdot\left\langle2\cos t,-2\sin t\right\rangle dt \\
&=\int_0^{\frac\pi2}(16\sin t\cos^2t-4\sin t\cos t)dt.
\end{aligned}\)
We can evaluate this integral using substitution. Let u = cos t, then du = -sin t dt, and we evaluate from cos 0 = 1 to \(\cos\frac\pi2=0\) and obtain
\(\begin{aligned}&=\int_1^0\left(-16u^2+4u\right)du\\&=\left[-\frac{16}3u^3+2u^2\right]_1^0\\&=\frac{16}3-2=\frac{10}3\end{aligned}\)
Putting it all together, we obtain
\(\int_L\overline{F}\cdot d\overline{l}=\int_{L_1}\overline{F}\cdot d\overline{l}+\int_{L_2}\overline{F}\cdot d\overline{l}+\int_{L_3}\overline{F}\cdot d\overline{l}=2+\left(-8\right)+\frac{10}3=-\frac83.\)