OBJECTIVES

When you finish this section, you should be able to:

1. Define a line integral in the plane (p. 977)
2. Find the value of a line integral along a smooth curve (p. 979)
3. Find line integrals of the form \(\boldsymbol\int_{\!\scriptsize\boldsymbol C}~\boldsymbol{f(x}, \boldsymbol{y)}\, \boldsymbol{dx}\) and \(\boldsymbol\int_{\!\scriptsize\boldsymbol C}~\boldsymbol{f(x}, \boldsymbol{y)}\, \boldsymbol{dy}\) (p. 981)
4. Find line integrals along a piecewise-smooth curve (p. 984)
5. Find the value of a line integral in space (p. 985)

NEED TO REVIEW?

Smooth curves are discussed in Section 9.2, pp. 647-648.

spanDEFINITIONspan Line Integral of \(f\) Along \(C\)

Let \(C\) be a smooth curve defined by the parametric equations \(x=x(t)\) and \(y=y(t)\), \(a\leq t\leq b\). Let \(z=f(x,y)\) be a function defined on some region containing \(C\). The line integral of \(f\) along \(C\) from \(t=a\) to \(t=b\) is \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}f(x,y)\,ds=\lim\limits_{\Vert \Delta \Vert \rightarrow 0}\sum\limits_{i=1}^{n}f(x_{i}^{\ast },y_{i}^{\ast })\Delta s_{i} }}} \]

provided this limit exists.

THEOREM The Value of a Line Integral

Let \(C\) be a smooth curve defined by the parametric equations \[ x=x(t)\quad y=y(t)\qquad a\leq t\leq b \]

Let \(z=f(x,y)\) be a function that is continuous on some region that contains the curve \(C\). Then the line integral of \(f\) along \(C\) from \(t=a\) to \(t=b\) exists and \[ \begin{equation*} {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}f(x,y,z)\,ds=\int_{a}^{b}f(x(t),y(t))\sqrt{\left( \dfrac{dx}{dt}\right) ^{\!\!2}+\left( \dfrac{dy}{dt} \right) ^{\!\!2}}dt }}} \tag{1} \end{equation*} \]

NOTE

If \(f(x,y) =1\), the line integral of \(f\) along \(C\) is the arc length of \(C\) from \(t=a\) to \(t=b\).

NEED TO REVIEW?

Arc length is discussed in Section 9.2, pp. 651-653.

Finding the Value of a Line Integral Along a Smooth Curve

Find \(\int_{C}y\,ds\) if \(C\) is the curve defined by the parametric equations \(x(t)=t\) and \(y(t)=\sqrt{t}\), \(2\leq t\leq 6\)

Figure 13 \(x(t)=t, y(t)=\sqrt{t}\), \(2\leq t\leq 6\)

Solution The curve \(C\) is smooth and is part of the graph of \(y=\sqrt{x}\), from \((2,\sqrt{2})\) to \((6,\sqrt{6})\), as shown in Figure 13.

The differential \(ds\) of arc length along \(C\) is given by \[ ds=\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}\,dt \]

Since \(\dfrac{dx}{dt}=1\) and \(\dfrac{dy}{dt}=\dfrac{1}{2\sqrt{t}},\) we have \[ ds=\sqrt{1^{2}+\left( \dfrac{1}{2\sqrt{t}}\right) ^{2}}\,dt=\sqrt{1+\dfrac{1}{4t}}\,dt=\sqrt{\dfrac{4t+1}{4t}}\,dt=\dfrac{\sqrt{4t+1}}{2\sqrt{t}}\,dt \]

Now we use formula (1). \[ \int_{C}y\,ds\underset{\underset{\color{#0066A7}{\hbox{\(y=\sqrt{t}\)}}}{\color{#0066A7}{\uparrow}}}{=}\int_{2}^{6}\sqrt{t}\dfrac{\sqrt{4t+1}}{2\sqrt{t}}\,dt=\dfrac{1}{2}\int_{2}^{6}\sqrt{4t+1}\,dt=\dfrac{1}{2}\left[ \dfrac{(4t+1)^{3/2}}{4\,{\cdot}\, \dfrac{3}{2}}\right] _{2}^{6}=\dfrac{49}{6} \]

NOW WORK

Problem 9.

Finding the Value of a Line Integral

Find \[ \int_{C}(x^{2}+y)\,ds \]

if \(C\) is the line segment from \((0,0)\) to \((1,2)\) and \(C\) is expressed using the parametric equations:

1. \(x(t) =t\) and \(y(t) =2t,\quad 0\leq t\leq 1\)
2. \(x(t) =\sin t\) and \(y(t) =2\sin t,\quad 0\leq t\leq \dfrac{\pi }{2}\)

Figure 14 \(y = 2x\, \hbox{from}\, (0,0)\, \hbox{to}\,(1,2) \)

Solution The two sets of parametric equations in (a) and (b) are just different ways of representing a part of the line \(y=2x\) from \((0,0)\) to \((1,2)\), as shown in Figure 14.

(a) For \(x(t) =t\) and \(y(t) =2t,\) we have \(\dfrac{dx}{dt}=1\) and \(\dfrac{dy}{dt}=2\). The differential \(ds\) of the arc length \(s\) is \(ds=\sqrt{1^{2}+2^{2}}dt=\sqrt{5}\,dt\). Then \[ \int_{C}(x^{2}+y)\,ds \underset{ \underset{ \underset{\color{#0066A7} {\hbox{\(y=2t\)}}} {\color{#0066A7}{\hbox{\(x=t\)}}}} {\color{#0066A7}{\uparrow }}} {=} \int_{0}^{1}(t^{2}+2t)\sqrt{5}\,dt=\left[\sqrt{5}\left( \dfrac{t^{3}}{3} +t^{2}\right) \right] _{0}^{1}=\dfrac{4\sqrt{5}}{3} \]

(b) For \(x(t) =\sin t\) and \(y(t) =2\sin t,\) we have \(\dfrac{dx}{dt}=\cos t\) and \(\dfrac{dy}{dt}=2\cos t\). Since \(\cos t\geq 0\) on \(0\leq t\leq \dfrac{\pi }{2},\) the differential \(ds\) of the arc length \(s\) is \[ ds=\sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}} \,dt=\sqrt{\cos ^{2}t+4\cos ^{2}t}\,dt=\sqrt{5\cos ^{2}t}\,dt=\sqrt{5}\cos t\,dt \]

Then \[ \begin{eqnarray*} \int_{C}(x^{2}+y)\,ds \underset{\underset{\underset{\color{#0066A7}{\hbox{\(y=2\sin t\)}}}{\color{#0066A7}{\hbox{\(x=\sin t\)}}}} {\color{#0066A7}{\uparrow }}}{=}\int_{0}^{\pi /2}(\sin ^{2}t+2\sin t)\sqrt{5}\cos t\,dt= \sqrt{5}\int_{0}^{\pi /2}(\sin ^{2}t+2\sin t)\cos t\,dt \\[-10pt] =\sqrt{5}\left[ \dfrac{\sin ^{3}t}{3}+\sin ^{2}t\right] _{0}^{\pi /2}=\dfrac{4\sqrt{5}}{3} \end{eqnarray*} \]

the same value found in (a).

NOW WORK

Problem 21.

Finding the Mass of a Wire with Variable Density

Find the mass \(M\) of a thin piece of wire in the shape of a semicircle \(x=2\cos t\), \(y=2\sin t\), \(0\leq t\leq \pi \), if the variable density of the wire is \(\rho (x,y)=y+2\).

Figure 16 Lateral surface area

Solution Using (2), we find that the mass \(M\) of the wire is \[ \begin{eqnarray*} M&=&\int_{C}\rho (x,y)\,ds=\int_{C}(y+2)\,ds=\int_{0}^{\pi }(y+2) \sqrt{\left( \dfrac{dx}{dt}\right) ^{2}+\left( \dfrac{dy}{dt}\right) ^{2}}dt\\[4pt] &=& \int_{0}^{\pi }(2\sin t+2)\sqrt{(-2\sin t)^{2}+(2\cos t)^{2}}\,dt \hspace{1.5pc}\color{#0066A7}{\dfrac{dx}{dt} =-2\sin t; \dfrac{dy}{dt} =2 \cos t}\\ &=&\int_{0}^{\pi }4(\sin t+1) \,dt=4\big[-\cos t+t\big]_{0}^{\pi }=4[\pi +2] \end{eqnarray*} \]

NOW WORK

Problem 67.

Finding the Lateral Surface Area of a Cylinder

Figure 17

Find the lateral surface area \(A\) of the cylinder that lies above the \(xy\)-plane, below the surface \(z = f(x,y)=x,\) and formed by lines parallel to the \(z\) -axis that intersect the curve \(y=x^{2}\), \(0\leq x\leq 2\).

Solution Figure 17 illustrates the cylinder. Along \(C\), \(y=x^{2},\) the differential \(ds\) of arc length is \[ ds=\sqrt{1+\left( \dfrac{dy}{dx}\right) ^{2}}\,dx\underset{ \underset{\color{#0066A7} {\hbox{\(\dfrac{dy}{dx}=2x\)}}} {\color{#0066A7}{\uparrow }}}{=} \sqrt{1+4x^{2}}\,dx\\[-11pt] \]

The lateral surface area \(A\) of the cylinder is \[ \begin{eqnarray*} A& =&\int_{C}f(x,y) ds=\int_{C}x\,ds=\int_{0}^{2}x\sqrt{1+4x^{2}} \,dx \qquad \color{#0066A7}{{\rm Let}\ u=1+4x^{2};\ {\rm then}\ du=8x\,dx\ {\rm or} \ x\,dx= \dfrac{1}{8}\, du.}\\ && \hspace{14.5pc}\color{#0066A7}{\hbox{When \(x=0\), \(u=1\); when \(x=2\), \(u=17\).}} \\ & =&\int_{1}^{17}\dfrac{1}{8}\sqrt{u}\,du=\left[ \dfrac{1}{4}\dfrac{u^{3/2}}{3}\right] _{1}^{17}=\dfrac{17\sqrt{17}-1}{12} \end{eqnarray*} \]

NOW WORK

Problem 55.

THEOREM The Value of a Line Integral Along \(C\) with Respect to \(x\) or \(y\)

Let \(C\) be a smooth curve defined by the parametric equations \[\begin{equation*} x=x(t)\quad y=y(t)\qquad a\leq t\leq b \end{equation*}\]

Let \(f=f(x,y)\) be a function that is continuous on some region that contains the curve \(C\). Then the line integral of \(f\) along \(C\) with respect to \(x\) is given by \[ \begin{equation*} {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}f(x,y)\,dx=\int_{a}^{b}f(x(t),y(t))x' (t)\,dt }}} \tag{3} \end{equation*} \]

and the line integral of \(f\) along \(C\) with respect to \(y\) is given by \[ \begin{equation*} {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}f(x,y)\,dy=\int_{a}^{b}f(x(t),y(t))y' (t)\,dt }}} \tag{4} \end{equation*} \]

Finding a Line Integral Along \(C\) with Respect to \(x\) and with Respect to \(y\)

Find \[ \int_{C}(x-3y)\,dx\qquad\hbox{and}\qquad \int_{C}(x-3y)\,dy \]

if \(C\) is the part of the parabola \(x=y^{2}\) that joins the points \((1,1)\) and \((4,2)\).

Solution The function \(f(x,y) =x-3y\) is continuous, and \(C\) is a smooth curve everywhere in the plane. So, we can use the formulas (3) and (4). Using the parametric equations of the curve \(C,\) \(x=t^{2}\) and \(y=t\), \(\ 1\leq t\leq 2\), we find \(dx=2t\,dt\) and \(dy=dt\). Then \[ \begin{eqnarray*} \int_{C}(x-3y)\,dx &=&\int_{1}^{2}(t^{2}-3t)2t\,dt=2\int_{1}^{2}(t^{3}-3t^{2}) dt=-\dfrac{13}{2} \notag \\[4pt] \int_{C}(x-3y)\,dy &=&\int_{1}^{2}(t^{2}-3t)\,dt=-\dfrac{13}{6} \end{eqnarray*} \]

NOW WORK

Problem 25.

spanDEFINITIONspan Line Integral \(\int_{C} (P\,dx+Q\,dy) \) Along a Smooth Curve \(C\)

Let \(C\) denote a smooth curve, and let \(P=P(x,y)\) and \(Q=Q(x,y)\) be functions of two variables that are continuous on some region containing \(C\) . The line integral of \(P\,dx+Q\,dy\) along \(C\) is defined as \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}(P\,dx+Q\,dy)=\int_{C}P(x,y)\,dx+\int_{C}Q(x,y)\,dy }}} \]

Finding a Line Integral of the Form \(\int_{C}( P\,dx+Q\,dy)\)

Find the line integral \(\int_{C}(y^{2}\,dx-x^{2}\,dy)\) along

\(C_{1}\): The parabola \(y=x^{2}\) joining the two points \((0,0)\) and \((2,4)\)

\(C_{2}\): The line \(y=2x\) joining the two points \((0,0)\) and \((2,4)\)

Solution The curves \(C_{1}\) and \(C_{2}\) are smooth, and the functions \(P(x,y) =y^{2}\) and \(Q(x,y) = -x^{2}\) are continuous in the \(xy\)-plane.

Along \(C_{1},\) parametric equations for the parabola are \(x(t) =t,\) \(y(t) =t^{2}.\) Then \(dx=dt\) and \(dy=2t\,dt,\) so \[ \begin{eqnarray*} \int_{C_{1}}(y^{2}\,dx-x^{2}\,dy)&=&\int_{C_{1}}y^{2}\,dx-\int_{C_{1}}x^{2}\,dy=\int_{0}^{2}(t^2) ^{2}\,dt-\int_{0}^{2}t^{2}(2t\,dt)\\ &=& \int_{0}^{2}t^{4}\,dt-2\int_{0}^{2}t^{3}\,dt=\left[ \dfrac{t^{5}}{5}\right] _{0}^{2}-\left[ \dfrac{t^{4}}{2}\right] _{0}^{2}=-\dfrac{8}{5} \end{eqnarray*} \]

Figure 18

Along \(C_{2}\), parametric equations for the line segment are \(x(t) =t\), \(y(t) =2t.\) Then \(dx=dt\) and \(dy=2\,dt,\) so \[ \begin{eqnarray*} \int_{C_{2}}(y^{2}\,dx-x^{2}\,dy)&=&\int_{C_{2}}y^{2}\,dx-\int_{C_{2}}x^{2}\,dy=\int_{0}^{2}(2t) ^{2}\,dt-\int_{0}^{2}t^{2}(2\,dt)\\[4pt] &=& \left[ \dfrac{4t^{3}}{3}\right] _{0}^{2}-\left[ \dfrac{2t^{3}}{3}\right] _{0}^{2}=\dfrac{16}{3} \end{eqnarray*} \]

Figure 18 shows the curves \(C_{1}\) and \(C_{2}\) joining the points \(( 0,0) \) and \((2,4) .\)

NOW WORK

Problem 29.

NEED TO REVIEW?

The dot product is discussed in Section 10.4, pp. 715-716.

Finding a Line Integral of the Form \(\int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}\)

Find \[ \int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r} \]

if \(\mathbf{F}(x,y)=x\mathbf{i}+xy\mathbf{j}\) and the curve \(C\) is traced out by the vector function \(\mathbf{r}(t)=t\,\mathbf{i}\,+\,t^{2} \mathbf{j}\), \(0\leq t\leq 2\).

Solution Parametric equations of the curve \(C\) are \[ x(t) =t\qquad y(t) =t^{2}\qquad 0\leq t\leq 2 \]

So, \[ \mathbf{F}=x\mathbf{i}+xy\,\mathbf{j}=t\mathbf{i}+(t) (t^{2}) \mathbf{j}=t\mathbf{i}+t^{3}\mathbf{j} \]

and \[ d\mathbf{r}=\dfrac{d\mathbf{r}}{dt}\,dt=\dfrac{d}{dt}( t\,\mathbf{i} +t^{2}\mathbf{j})\, dt=(\mathbf{i}+2t\mathbf{j})\,dt \]

Then \[ \mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}=( t\mathbf{i}+t^{3}\,\mathbf{j}) \,{\boldsymbol\cdot}\, ( \mathbf{i}+2t\,\mathbf{j}) \,dt=( t+2t^{4}) \,dt \]

so that \[ \int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}=\int_{0}^{2}(t+2t^{4}) \,dt=\dfrac{74}{5} \]

NOW WORK

Problem 47.

spanDEFINITIONspan Line Integral \(\int_{C} (P\,dx+Q\,dy) \) Along a Piecewise-Smooth Curve \(C\)

Let \(C\) be a piecewise-smooth curve consisting of \(n\geq 2\) smooth curves \(C_{1},\) \(C_{2}, \ldots , C_{n}.\) Let \(P=P(x,y)\) and \(Q=Q(x,y)\) be functions of two variables that are each continuous on some region containing \(C\). The line integral \(\int_{C}( P\,dx+Q\,dy)\) along \(C\) is defined as \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}(P\,dx+Q\,dy)=\int_{C_{1}}(P\,dx+Q\,dy)+\int_{C_{2}}(P\,dx+Q\,dy)+ \cdots +\int_{C_{n}}(P\,dx+Q\,dy) }}} \]

If \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}\) and \(C\) is traced out by \(\mathbf{r} =\mathbf{r}(t) =x(t) \mathbf{i}+y(t) \mathbf{j}\), \(a\leq t\leq b,\) then \[ {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C_{1}}\mathbf{F}\,{\cdot}\, d\mathbf{r}+\int_{C_{2}}\mathbf{F}\,{\cdot}\, d\mathbf{r}+ \cdots +\int_{C_{n}}\mathbf{F} \,{\cdot}\, d\mathbf{r} }}} \]

Finding a Line Integral Along a Piecewise-Smooth Curve

Find \(\int_{C}(xy\,dx+x^{2}\,dy)\) along the piecewise-smooth curve \(C\) illustrated in Figure 21.

Figure 21

Solution The values of the line integral along each of the smooth curves \(C_{1},\) \(C_{2},\) \(C_{3}\), and \(C_{4}\) are \[ \begin{eqnarray*} C_{1}{:}&& y =\dfrac{1}{3}x,\quad dy=\dfrac{1}{3}dx;\qquad 0\leq x\leq 3 \\ &&{\int_{C_{1}}(xy\,dx+x^{2}\,dy)=\int_{0}^{3}\left[ x\left( \dfrac{1}{3}x\right) dx+x^{2}\dfrac{1}{3}dx\right] =6} \\ C_{2}{:}&& x =3,\quad dx=0;\qquad 1\leq y\leq 2 \\ &&{\int_{C_{2}}(xy\,dx+x^{2}\,dy)=\int_{1}^{2}9\,dy=9} \\ C_{3}{:}&& y =2, \quad dy=0;\qquad \color{#0066A7}{\hbox{Watch the orientation here: \(x\) varies from 3 to 2.}} \\ &&{\int_{C_{3}}(xy\,dx+x^{2}\,dy)=\int_{3}^{2}2x\,dx=\left[x^{2}\right]^{2}_{3}=-5} \\ C_{4}{:}&& y =x, \quad dy=dx;\qquad \color{#0066A7}{\hbox{Watch the orientation here: \(x\) varies from 2 to 0.}} \\ &&{\int_{C_{4}}(xy\,dx+x^{2}\,dy)=\int_{2}^{0}(x^{2}\,dx+x^{2}\,dx)=2\int_{2}^{0}x^{2}dx=-\dfrac{16}{3}} \end{eqnarray*} \]

Then \[ \int_{C}(xy\,dx+x^{2}\,dy)=6+9-5-\dfrac{16}{3}=\dfrac{14}{3} \]

NOW WORK

Problem 39.

THEOREM The Value of a Line Integral in Space

Let \(C\) be a smooth curve in space defined by the parametric equations \[ x=x(t)\quad y=y(t)\quad z=z(t)\quad a\leq t\leq b \]

Let \(w=f(x,y,z)\) be a function that is continuous on some solid that contains the smooth curve \(C\). Then the line integral of \(f\) along \(C\) from \(t=a\) to \(t=b\) exists and \[ \begin{equation*} {\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C}f(x,y,z)\,ds=\int_{a}^{b}f(x(t),y(t),z(t) )\sqrt{\left( \dfrac{dx}{dt}\right) ^{\!\!2}+\left( \dfrac{dy}{dt} \right) ^{\!\!2}+\left( \dfrac{dz}{dt}\right) ^{\!\!2}}dt }}} \tag{5} \end{equation*} \]

Finding a Line Integral in Space

Find \(\int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}\) if \[ \begin{equation*} \mathbf{F}(x,y,z)=xy^{2}\,\mathbf{i}+x^{2}z\,\mathbf{j}-(y-x)\,\mathbf{k} \end{equation*} \]

and the curve \(C\) is traced out by \(\mathbf{r}(t)=t\,\mathbf{i}+t^{2}\mathbf{j}+t^{3}\,\mathbf{k}\), \(\ 0\leq t\leq 1\).

Solution Parametric equations of the curve \(C\) are \[ \begin{equation*} x=t\qquad y=t^{2}\qquad z=t^{3}\qquad 0 \le t \le 1 \end{equation*} \]

So, \[ \mathbf{F}=xy^{2}\,\mathbf{i}+x^{2}z\,\mathbf{j}-(y-x)\,\mathbf{k}=t^{5} \mathbf{i}+t^{5}\mathbf{j}-(t^{2}-t)\,\mathbf{k} \]

and \[ d\mathbf{r}=\dfrac{d\mathbf{r}}{dt}\,dt=(\mathbf{i}+2t\mathbf{j}+3t^{2} \mathbf{k})\,dt \]

Then \[ \begin{eqnarray*} \mathbf{F}\,{\cdot}\, d\mathbf{r}&=&[ t^{5}\mathbf{i}+t^{5}\,\mathbf{j} -(t^{2}-t)\,\mathbf{k}] \,{\cdot}\, ( \mathbf{i}+2t\,\mathbf{j}+3t^{2}\, \mathbf{k}) \,dt=[ t^{5}+2t^{6}-3t^{2}(t^{2}-t)] \,dt\\ &=&( 2t^{6}+t^{5}-3t^{4}+3t^{3}) \,dt \end{eqnarray*} \]

so that \[ \int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}=\int_{0}^{1}(2t^{6}+t^{5}-3t^{4}+3t^{3}) \,dt=\dfrac{253}{420} \]

NOW WORK

Problem 51.