# Line Integrals (Theory and Examples) • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Definition of a Line Integral

Let $C$ be a smooth curve, with parametric equations $x=x(t),$ $y=y(t),$ and $z=z(t)$ for $a\leq t\leq b,$ that lies within the domain of a function $f(x,y,z).$ We say that $C$ is orientable if it is possible to describe direction along the curve for increasing $t.$ Partition $C$ into $n$ sub-arcs, the $k \text{th}$ of which has length $\triangle s_k.$ Let $\left(x_k^*,y_k^*,z_k^*\right)$ be a point chosen (arbitrarily) from the $k \text{th}$ sub-arc. Form the Riemann sum \begin{equation} \sum_{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle s_k\end{equation} and let $|\triangle s|$ denote the largest sub-arc length in the partition. The following limit \begin{equation} \lim_{|\text{$\triangle$s}|\to 0}\sum_{k=1}^n f \left(x_k^*,y_k^*,z_k^*\right)\triangle s_k\end{equation} is called the line integral of $f$ over $C$ and is denoted by $$\int_Cf(x,y,z)\, dS.$$ Additionally, if $C$ is a closed curve, then we denote the line integral by $$\oint_C f \, dS.$$

Theorem. Let $f,$ $f_1,$ and $f_2$ be continuous scalar functions defined on a piecewise smooth orientable curve $C.$ Then for any constants $k_1$and $k_2,$

$(1) \quad \int_C \left(k_1 f_1+k_2f_2\right) \, dS=k_1\int_Cf_1 \, dS+k_2\int_Cf_2 \, dS$

$(2) \quad \int_C f \, dS=\int_{C_1}f \,\, dS+\cdots+\int_{C_n}f \,\, dS$ where $C$ is the union of smooth orientable sub-arcs $C=C_1\cup C_2\cup \cdots \cup C_n$ with only endpoints in common.

$(3) \quad \int_{-C} f \, dS=-\int_C f \, dS.$

## Evaluating Line Integrals Using Parametrization

Theorem. Suppose that the function $f$ is continuous at each point on a smooth curve $C$, with parametric equations $x=x(t),$ $y=y(t),$ and $z=z(t)$ for $a\leq t\leq b,$ that lies within the domain of $f.$ Then $$\int_Cf(x,y,z)\, dS=\int_a^bf( x(t),y(t),z(t) )\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} \, dt .$$ where $\, dS=\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} \, dt.$

The definition of a line integral can be extended to curves that are piecewise smooth in the sense that they are the union of a finite number of smooth curves with only endpoints in common. In particular, if $C$ is comprised of a number of smooth sub-arcs $C_1,C_2,\ldots,C_n,$ then \begin{equation} \int_C f(x,y,z)\, dS=\int_{C_1} f(x,y,z)\, dS+\cdots +\int_{C_n} f(x,y,z)\, dS. \end{equation}

Example. Evaluate the line integral $$\int_Cy \sin z \, dS,$$ where $C$ is the circular helix given by the equations $x=\cos t, y=\sin t, z=t,$ and $0\leq t\leq 2\pi.$

Solution. We determine \begin{align} \int_C y \sin z \, dS & =\int_0^{2\pi }(\sin t) \sin t \sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t} \right)^2 + \left(\frac{d z}{d t}\right)^2}dt \\ & =\int_0^{2\pi }\sin ^2 t \sqrt{\sin ^2t+\cos ^2t+1}dt \\ & =\sqrt{2}\int_0^{2\pi }\frac{1}{2}(1-\cos 2t)dt \\ & = \frac{\sqrt{2}}{2}\left[t-\frac{1}{2}\sin 2t\right]_0^{2\pi } =\sqrt{2}\pi. \end{align}

Theorem. Let $C$ be a smooth curve and let $f(x,y,z)$ be a continuous function with domain containing the trace of $C.$ Then the value of the integrals \begin{equation} \int_C f \, dS, \quad \int_C f dx, \quad \int_C f dy, \quad \int_C f dz\end{equation} depen\, dS only on the initial point $A,$ terminal point $B,$ and the trace of $C.$ That is, two different parameterizations having the same trace from $A$ to $B$ yield the same values for these integrals.

Example. Suppose the smooth curves $C_1$ and $C_2$ are given by \begin{align} & C_1: \quad x=t, \, y=t^2 \quad \text{ for } 0\leq t\leq 1, \text{ and} \\ & C_2: \quad x=\sin t, \, y=\sin ^2 t\quad \text{ for } 0\leq t\leq \frac{\pi }{2}. \end{align} Evaluate $$\int_{C_1} x \, dS \quad \text{ and } \quad \int_{C_2} x \, dS.$$

Solution. Both $C_1$ and $C_2$ are smooth curves from $(0,0)$ to $(1,1)$ with the same trace which is the portion of the parabola $y=x^2$ for $0\leq x\leq 1.$ For $C_1,$ we have $x=t$ and $\, dS=\sqrt{1+4t^2}dt,$ therefore \begin{equation}
\int_{C_1} x \, dS=\int_0^1t\sqrt{1+4t^2}dt = \frac{1}{12} \left( 17^{3/2}-1 \right). \end{equation} For $C_2,$ we have $x=\sin t$ and $\, dS=\sqrt{\cos ^2t+4\sin ^2t\text{ }\cos ^2t}dt,$ therefore \begin{equation} \int_{C_2} x \, dS=\int_0^{\pi /2} \sin t\text{ }\cos t\sqrt{1+4\sin ^2t}dt=\frac{1}{12}\left(17^{3/2}-1\right). \end{equation}

## Line Integrals with Respect to Coordinate Variables

Other line integrals are obtained by replacing $\triangle s_k$ by $\triangle x_k=x_k-x_{k-1}.$ This is called the line integral of $f$ along $C$ with respect to $x$: \begin{equation} \int_C f(x,y,z) \,dx=\lim_{| \triangle x|\to 0}\sum_{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle x_k.\end{equation} Similarly with respect to $y$ and $z$, we define \begin{equation} \int_C f(x,y,z) \,dy=\lim_{| \triangle y |\to 0}\sum_{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle y_k\end{equation} \begin{equation} \int_C f(x,y,z) \, dz = \lim_{|\triangle z|\to 0}\sum_{k=1}^n f\left(x_k^*,y_k^*,z_k^*\right)\triangle z_k.\end{equation} When we want to distinguish the original line integral $\int_Cf \,\, dS$ from these, we call it the line integral with respect to arc length.

The following formulas say that the line integrals with respect to $x,$ $y,$ or $z$ can also be evaluated by expressing everything in terms of $t$: $x=x(t),$ $y=y(t),$ $z=z(t),$ $dx=x'(t),$ $dy=y'(t),$ and $dz=z'(t)$ yielding: \begin{equation}\int_C f(x,y,z)\,d x=\int_a^b f( x(t),y(t),z(t) )x'(t)\,dt\end{equation} \begin{equation}\int_C f(x,y,z)\,d y=\int_a^b f( x(t),y(t),z(t)) y'(t) \, dt\end{equation} and \begin{equation} \int_C f(x,y,z)\,d z = \int_a^b f(x(t), y(t), z(t)) z'(t)\,dt \end{equation} assuming $f$ is continuous and $C$ lies within the domain of $f.$

Example. Evaluate the line integral $$\int_C y \,dx+z \,dy+x \,dz,$$ where $C$ consists of the line segment $C_1$ from $(2,0,0)$ to $(3,4,5)$ followed by the vertical line segment $C_2$ from $(3,4,5)$ to $(3,4,0)$

Solution. The curve $C$ is the union of the curves \begin{align} & C_1: {R}1(t)=(1-t)\langle 2,0,0\rangle+t \langle 3,4,5\rangle = \langle 2+t,4t,5t\rangle \text{ for } 0\leq t\leq 1 \\ & C_2: {R}_2(t)=(1-t) \langle 3,4,5 \rangle+t\langle3,4,0\rangle=\langle3,4,5-5t\rangle \text{ for } 0\leq t\leq 1. \end{align} Thus \begin{align} \int_C y \, dx+z\, dy+x\, dz & =\int_{C_1} y \, dx+z\, dy+x\, dz+\int_{C_2} y \, dx+z \, dy+x \, dz \\ & =\int_0^1 ( (4t)+5t(4)+(2+t)5) \, dt +\int_0^1( 0+0+3(-5) )\, dt \\ & =\int_0^1 (10+29t) \, dt +\int_0^1 (-15)\, dt \\ & =\left[10t+29\frac{t^2}{2}\right]_0^1 +\left[-15t\right]_0^1 \\ & = 24.5 -15 =9.5 \end{align}

## Line Integral of Vector Field Along a Curve

Theorem. Let $${V}(x,y,z) = u(x,y,z) {i} + v(x,y,z) {j} + w(x,y,z) {k}$$ be a continuous vector field, and let $C$ be a piecewise smooth orientable curve with parametric representation $${R}(t)=x(t){i}+y(t){j}+z(t) {k}$$ for $a\leq t\leq b.$ Using $d{R}=dx {i}+dy{j}+dz{k}$ we define the line integral of ${V}$ along $C$ by \begin{align*} & \int_C{V}\cdot d {R} \\ & = \int_C ( u \, dx +v \, dy +w \, dz) \\ & =\int_C \left({V}[{R}(t)]\cdot{R}'(t) \right)dt \\ & =\int_a^b \left[u(x,y,z)\frac{dx}{dt} + v(x,y,z)\frac{dy}{dt} + w(x,y,z)\frac{dz}{dt}\right]d t.\end{align*}

Theorem. Let ${F}$ be a continuous force field over a domain $D.$ Then the work $W$ performed as an object moves along a smooth curve $C$ in $D$ is given by the integral \begin{equation} W=\int_C {F}\cdot {T} \, \, dS=\int_C {F}\cdot d{R}\end{equation} where ${T}$ is the unit tangent at each point on $C$ and ${R}$ is the position vector of the object moving on $C.$

Example. Evaluate the line integral $$\oint_{F}\cdot \, d{R}$$ where ${F}=y^2{i} +x^2 {j}-(x+z){k}$ and $C$ is the boundary of the triangle with vertices $(0,0,0),$ $(1,0,0),$ and $(1,1,0)$, traversed once clockwise, as viewed from above.

Solution. First let $C_1$ be the line segment from $(1,0,0)$ to $(0,0,0).$ Then $C_1$ can be represented by the vector function $${R}_1(t)=(1-t)\langle 1,0,0\rangle +t\langle 0,0,0\rangle =\langle 1-t,0,0\rangle$$ with $0\leq t\leq 1.$ Let $C_2$ be the line segment from $(0,0,0)$ to $(1,1,0)$ and so can berepresented by the vector function, $${R}_2(t)=(1-t)\langle 0,0,0\rangle + t \langle 1,1,0 \rangle =\langle t,t,0\rangle$$ with $0\leq t\leq 1.$ Let $C_3$ be the line segment from $(1,1,0)$ to $(1,0,0)$ and so can be represented by the vector function, $${R}_3(t)=(1-t)\langle1,1,0\rangle+t\langle1,0,0\rangle =\langle 1,1-t,0\rangle$$ with $0\leq t\leq 1.$ Then \begin{align} & {F}_1=\left(1-t^2\right){j}-(1-t){k} & & d{R}_1=\langle -dt,0,0\rangle \\ & {F}_2 = t^2 {i} + t^2 {j} – t {k} & & d{R}_2=\langle dt,dt,0\rangle \\ &{F}_3=(1-t)^2{i}+{j}-{k} & & d{R}_3=\langle 0,-dt,0\rangle\end{align} We find \begin{align} \int_C{F}\cdot d{R} & =\underset{C_1}{\int }{F}_1\cdot d{R}_1+\underset{C_2}{\int }{F}_2\cdot d{R}_2+\underset{C_3}{\int }{F}_3\cdot d{R}_3 \\ & =\int_0^1 0 \, dt+\int_0^1 2t^2 \, dt+\int_0^1 (-1) \, dt=-\frac{1}{3}. \end{align}

## Exercises on Line Integrals

Exercise. Evaluate the line integral $$\int_C\frac{1}{3+y}\, dS$$ where $C$ is the curve with parametric equations $x=2t^{3/2}$ and $y=3t$ with $0\leq t\leq 1.$

Exercise. Evaluate the line integral $$\int_C\left(x^2+y^2\right)\, dS$$ where $C$ is the curve with parametric equations $x=e^{-t}\cos t$ and $y=e^{-t}\sin t$ with $0\leq t\leq \frac{\pi }{2}.$

Exercise. Evaluate the line integral $$\int_C xdy-ydx$$ where $C$ is the curve defined by $2x-4y=1$ with $4\leq x\leq 8.$

Exercise. Evaluate the line integral $$\int_C -xdy+\left(y^2-x^2\right)dx$$ where $C$ is the quarter-circle $x^2+y^2=4$ from $(0,2)$ to $(2,0).$

Exercise. Evaluate the line integral $$\int_C (x+y)^2dx-(x-y)^2 dy$$ where $C$ is the curve defined by $y=|2x|$ from $(-1,2)$ to $(1,2).$

Exercise. Evaluate the line integral $$\oint_{F} \cdot d{R}$$ where ${F}=y^2 {i} +x^2 {j}-(x+z){k}$ and $C$ is the boundary of the triangle with vertices $(0,0,0),$ $(1,0,0),$ and $(1,1,0)$, transversed once clockwise, as viewed from above.

Exercise. Let ${F}=y \,{i}-x \,{j}$ and let $C_1$ and $C_2$ be the following two paths joining $(0,0)$ to $(1,1).$ $$C_1: y=x \text{ for } 0\leq x\leq 1 \quad \text{and} \quad C_2: y=x^2 \text{ for } 0\leq x\leq 1.$$ Show that $$\int_{C_1}{F}\cdot d {R}\neq \int_{C_2}F\cdot d{R}.$$ Explain what this means?

Exercise. Evaluate the closed line integral $$\oint_C {F}\cdot \, dS$$ where ${F}=-x \, {i}+2\,{j}$ and $C$ is the boundary of the trapezoid with vertices $(0,0),$ $(1,0),$ $(2,1)$ and $(0,1)$ transversed once clockwise, as viewed from above.

Exercise. Find the work done by the force field $${F}(x,y,z)=\left(y^2-z^2\right) {i}+2 y z {j}-x^2 {k}$$ on any object moving along the curve $C$ where $C$ is the path given by $x(t)=t,$ $y(t)=t^2,$ and $z(t)=t^3,$ for $0\leq t\leq 1.$

Exercise. Find the work done by the force field $${F}(x,y,z)=2 x y \, {i}+\left(x^2+2\right)\,{j}+y \,{k}$$ on any object moving along the curve $C$ where $C$ is the line segment from $(1,0,2)$ to $(3,4,1).$

Exercise. Show that the vector field $${F}(x,y)=(x+2y)\, {i}+(2x+y)\, {j}$$ is conservative and find a scalar potential function $f$ for $F.$ Then evaluate the line integral $\int_C{F}\cdot d{R}$ where $C$ is any smooth path connecting $A(0,0)$to $B(1,1).$

Exercise. Show that the vector field $${F}(x,y)=\left(y-x^2\right) \, {i}+\left(x+y^2\right) \,{j}$$ is conservative and find a scalar potential function $f$ for ${F}.$ Then evaluate the line integral $$\int_C {F}\cdot dR$$ where $C$ is any smooth path connecting $A(0,0)$to $B(1,1).$

Exercise. Show that the vector field $${F}(x,y)=e^{-y}\,{i}-x e^{-y}\,{j}$$ is conservative and find a scalar potential function $f$ for ${F}.$ Then evaluate the line integral $$\int_C{F}\cdot d{R}$$ where $C$ is any smooth path connecting $A(0,0)$to $B(1,1).$

Exercise. Show that the vector field $${F}(x,y,z)=e^{x y}y z \,{i}+e^{x y}x z \, {j}+e^{x y} \,{k}$$ is conservative and find a scalar potential function $f$ for ${F}.$

Exercise. Show that the vector field ${F}$ with component functions $f(x,y,z) = \left(2 x z^3-e^{-x y}y \sin z\right)$, $g(x,y,z)=-x e^{-x y}\sin z$, and $$h(x,y,z)=\left(3x^2z^2+e^{-x y}\cos z\right)$$ is conservative and find a scalar potential function $f$ for ${F}.$ Then evaluate the line integral $$\int_C{F}\cdot d{R}$$ where $C$ is any smooth path connecting $A(1,0,-1)$to $B(0,-1,1).$

Exercise. Show that the vector field $${F}(x,y)=\frac{(y+1){i}-x {j}}{(y+1)^2}$$ is conservative and find a scalar potential $f$ for ${F}.$ Then evaluate the line integral $$\int_C{F}\cdot d {R}$$ where $C$ is any smooth path connecting $A(0,0)$ to $B(1,1).$