Multivariable calculus and vector analysis

Alternate notation for line integrals

We initially denoted line integrals using the notation

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_a^b\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt& & \hfill \text{ (1)}\end{array}$$

where $\mathbf{c}\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)$. We can derive a new notation by “multiplying out” the dot products on both sides of equation (1).

Let’s assume we are in three dimensions to $\mathbf{F}=\left(F_1,F_2,F_3\right)$. If we multiply out the left hand side, thinking of $d\mathbf{s}$ being the “vector” $d\mathbf{s}=\left(dx,dy,dz\right)$, it becomes

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_C{F}_{1}dx+F_{2}dy+F_{3}dz.& & \hfill \end{array}$$

Also, since

$$\begin{array}{lll}\hfill {\mathbf{c}}^{\text{'}}\left(t\right)=\left(\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t},\frac{\mathrm{d}z}{\mathrm{d}t}\right),& & \hfill \end{array}$$

we could “multiply out” the right hand side of of equation (1) to write it as

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_a^b\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt={\int \; <!--nolimits\; -->}_a^b\left(F_1\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}x}{\mathrm{d}t}+F_2\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}y}{\mathrm{d}t}+F_3\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}z}{\mathrm{d}t}\right)dt.& & \hfill \end{array}$$

Frequently, line integral problems are presented in this notation. In this case, one can compute them directly using the formula

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C{F}_{1}dx+F_{2}dy+F_{3}dz={\int \; <!--nolimits\; -->}_a^b\left(F_1\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}x}{\mathrm{d}t}+F_2\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}y}{\mathrm{d}t}+F_3\left(\mathbf{c}\left(t\right)\right)\frac{\mathrm{d}z}{\mathrm{d}t}\right)dt.& & \hfill \text{ (2) }\end{array}$$

Example

Evaluate

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C}ydx+\left(x+y\right)dy+dz& & \hfill \end{array}$$

where thee curve $C$ is parameterized by $\mathbf{c}\left(t\right)=\left(t,1-t,1\right)$, $0\le t\le 1$.

Solution: If we write $\mathbf{c}\left(t\right)=\left(t,1-t,1\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)$, Then $\frac{\mathrm{d}x}{\mathrm{d}t}=1$, $\frac{\mathrm{d}y}{\mathrm{d}t}=-1$, and $\frac{\mathrm{d}z}{\mathrm{d}t}=0$. Since $F_1\left(x,y,z\right)=y$, $F_2\left(x,y,z\right)=x+y$, and $F_3\left(x,y,z\right)=1$, we can use formula (2) to calculate the integral.

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_{C}ydx+\left(x+y\right)dy& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(y\left(t\right)\frac{\mathrm{d}x}{\mathrm{d}t}+\left(x\left(t\right)+y\left(t\right)\right)\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{\mathrm{d}z}{\mathrm{d}t}\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left[\left(1-t\right)\left(1\right)+\left(t+\left(1-t\right)\right)\left(-1\right)\right]dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left[\left(1-t\right)+\left(-1\right)\right]dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(-t\right)dt={\left.-\frac{t^2}{2}\right|}_0^1=-\frac{1}{2}& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

4d. Path integrals and line integrals

• Path integral of a scalar-valued function
• Path integral examples
• Line integral of a vector field
• Alternate notation for line integrals
• Line integrals as circulation
• Line integral examples

4e. Surface integrals

5. The fundamental theorems of vector calculus

6. Review material