## 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 $B(1,1).$