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Line Integrals

# Geometric Applications of Line Integrals

Page 1
Problems 1-2
Page 2
Problems 3-10

Line integrals have many applications in mathematics, physics and engineering. In particular, they are used for computations of

• length of a curve;
• area of a region bounded by a closed curve;
• volume of a solid formed by rotating a closed curve about a line.

### Length of a Curve

Let $$C$$ be a piecewise smooth curve described by the position vector $$\mathbf{r}\left( t \right),\,\alpha \le t \le \beta .$$ Then the length of the curve is given by the line integral
${L = \int\limits_C {ds} } = {\int\limits_\alpha ^\beta {\left| {\frac{{d\mathbf{r}}}{{dt}}\left( t \right)} \right|dt} \text{ = }}\kern0pt {\int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt}}$ where $$\large\frac{{d\mathbf{r}}}{{dt}}\normalsize$$ is the derivative, and $$x\left( t \right),$$ $$y\left( t \right),$$ $$z\left( t \right)$$ are the components of the position vector $$\mathbf{r}\left( t \right).$$

If the curve $$C$$ is two-dimensional, the latter formula can be written in the form
${L = \int\limits_C {ds} } = {\int\limits_\alpha ^\beta {\left| {\frac{{d\mathbf{r}}}{{dt}}\left( t \right)} \right|dt} } = {\int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} .}$ If the curve $$C$$ is the graph of a continuous and differentiable function $$y = f\left( x \right)$$ in the $$xy$$-plane, the length of the curve is given by
$L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} .$ Finally, if the curve $$C$$ is given by the equation $$r = r\left( \theta \right),\,\alpha \le \theta \le \beta$$ in polar coordinates, and the function $$r\left( \theta \right)$$ is continuous and differentiable in the interval $$\left[ {\alpha ,\beta } \right],$$ the length of the curve is defined by the formula
$L = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dr}}{{d\theta }}} \right)}^2} + {r^2}} d\theta } .$

### Area of a Region Bounded by a Closed Curve

If $$C$$ is a closed smooth piecewise curve in the $$xy$$-plane (Figure $$1$$), the area of the region $$R$$ bounded by the curve is given by
${S = \oint\limits_C {xdy} }={ – \oint\limits_C {ydx} } = {\frac{1}{2}\int\limits_C {xdy – ydx} .}$ It is supposed here that the contour $$C$$ is traversed in the counterclockwise direction.

Figure 1.

If the closed curve $$C$$ is given in parametric form $$\mathbf{r}\left( t \right) =$$ $$\left( {x\left( t \right),y\left( t \right)} \right),$$ the area of the corresponding region can be calculated by the formula
${S = \int\limits_\alpha ^\beta {x\left( t \right)\frac{{dy}}{{dt}}dt} } = { – \int\limits_\alpha ^\beta {y\left( t \right)\frac{{dx}}{{dt}}dt} } = {\frac{1}{2}\int\limits_\alpha ^\beta {\left[ {x\left( t \right)\frac{{dy}}{{dt}} }\right.}-{\left.{ y\left( t \right)\frac{{dx}}{{dt}}} \right]dt} }$

Figure 2.

### Volume of a Solid Formed by Rotating a Closed Curve about the $$X$$-axis

Let $$R$$ be a region in the half-plane $$y \ge 0$$ bounded by a closed smooth piecewise curve $$C$$ traversed in the counterclockwise direction. Suppose that the solid $$\Omega$$ is formed by rotating the region $$R$$ about the $$x$$-axis (Figure $$2$$). Then the volume of the solid is given by
${V = – \pi \oint\limits_C {{y^2}dx} } = { – 2\pi \oint\limits_C {xydy} } = { – \frac{\pi }{2}\oint\limits_C {2xydy + {y^2}dx} .}$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Find the arc length of the plane curve $$a{y^2} = {x^3}$$ for $$0 \le x \le 5a,$$ $$y \ge 0.$$

### ✓Example 2

Find the length of the astroid $${x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {a^{\large\frac{2}{3}\normalsize}}.$$

### ✓Example 3

Find the length of the space curve parameterized by $$\mathbf{r}\left( t \right) =$$ $$\left( {3t,3{t^2},2{t^3}} \right),$$ where $$0 \le t \le 1.$$

### ✓Example 4

Find the arc length of the cycloid parameterized by $$\mathbf{r}\left( t \right) =$$ $$\big( {a\left( {t – \sin t} \right),}$$ $${a\left( {1 – \cos t} \right)} \big)$$ for $$0 \le t \le 2\pi$$ (Figure $$5\text{).}$$

### ✓Example 5

Calculate the length of the parabola $$y = {x^2}$$ for $$0 \le x \le 1.$$

### ✓Example 6

Find the length of the cardioid given in polar coordinates by the equation $$r = 5\left( {1 + \cos \theta } \right)$$ (Figure $$6$$).

### ✓Example 7

Find the area of the region bounded by the hyperbola $$y = {\large\frac{1}{x}\normalsize},$$ the $$x$$-axis, and the vertical lines $$x = 1,$$ $$x = 2$$ (Figure $$7$$).

### ✓Example 8

Find the area of the region bounded by the ellipse $$x = a\cos t,$$ $$y = b\sin t,$$ $$0 \le t \le 2\pi$$ (Figure $$8$$).

### ✓Example 9

Find the volume of the solid formed by rotating the region $$R$$ bounded by the curve $$y = 2 – \sin x$$ and the lines $$x = 0,$$ $$x = 2\pi,$$ $$y = 0$$ about the $$x$$-axis.

### ✓Example 10

Find the volume of the ellipsoid formed by revolving ellipse with semi-axis $$a$$ and $$b$$ about the $$x$$-axis (Figure $$10\text{).}$$

### Example 1.

Find the arc length of the plane curve $$a{y^2} = {x^3}$$ for $$0 \le x \le 5a,$$ $$y \ge 0.$$

#### Solution.

We can write the function as $${y^2} = {\large\frac{{{x^3}}}{a}\normalsize}$$ or $$y = \pm \sqrt {\large\frac{{{x^3}}}{a}\normalsize}.$$ As $$y \ge 0,$$ we take only the positive root in the equation of the curve (Figure $$3$$). The length of the arc is
${L = \int\limits_0^{5a} {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} } = {\int\limits_0^{5a} {\sqrt {1 + {{\left[ {\frac{d}{{dx}}\left( {\sqrt {\frac{{{x^3}}}{a}} } \right)} \right]}^2}} dx} } = {\int\limits_0^{5a} {\sqrt {1 + {{\left[ {\frac{{\frac{{3{x^2}}}{a}}}{{2\sqrt {\frac{{{x^3}}}{a}} }}} \right]}^2}} dx} }$

Figure 3.

$= {\int\limits_0^{5a} {\sqrt {1 + {{\left[ {\frac{{3\sqrt x }}{{2\sqrt a }}} \right]}^2}} dx} } = {\int\limits_0^{5a} {\sqrt {1 + \frac{{9x}}{{4a}}} dx} } = {\frac{1}{{2\sqrt a }}\int\limits_0^{5a} {\sqrt {4a + 9x} dx} } = {\frac{1}{{18\sqrt a }}\left[ {\left. {\left( {\frac{{{{\left( {9x + 4a} \right)}^{\large\frac{3}{2}\normalsize}}}}{{\frac{3}{2}}}} \right)} \right|_{x = 0}^{5a}} \right] } = {\frac{1}{{27\sqrt a }}\left[ {{{\left( {45a + 4a} \right)}^{\large\frac{3}{2}\normalsize}} – {{\left( {4a} \right)}^{\large\frac{3}{2}\normalsize}}} \right] } = {\frac{1}{{27\sqrt a }}\left[ {\sqrt {{{\left( {49a} \right)}^3}} – \sqrt {{{\left( {4a} \right)}^3}} } \right] } = {\frac{a}{{27}}\left( {\sqrt {{{49}^3}} – \sqrt {{4^3}} } \right) } = {\frac{a}{{27}}\left( {{7^3} – {2^3}} \right) } = {\frac{{335}}{{27}}.}$

### Example 2.

Find the length of the astroid $${x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {a^{\large\frac{2}{3}\normalsize}}.$$

Figure 4.

#### Solution.

The astroid is shown in Figure $$4.$$ By symmetry, we can calculate the length of the arc lying in the first quadrant and then multiply the result by $$4.$$ The equation of the astroid in the first quadrant is
${y = {\left( {{a^{\large\frac{2}{3}\normalsize}} – {x^{\large\frac{2}{3}\normalsize}}} \right)^{\large\frac{3}{2}\normalsize}},\;\;}\kern-0.3pt {\text{where}\;\;}\kern-0.3pt{x \in \left[ {0,a} \right].\;}$

Then
${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{ \frac{3}{2}{\left( {{a^{\large\frac{2}{3}\normalsize}} – {x^{\large\frac{2}{3}\normalsize}}} \right)^{\large\frac{1}{2}\normalsize}}\left( { – \frac{2}{3}{x^{ – \large\frac{1}{3}\normalsize}}} \right) } = { – \frac{{{{\left( {{a^{\large\frac{2}{3}\normalsize}} – {x^{\large\frac{2}{3}\normalsize}}} \right)}^{\large\frac{1}{2}\normalsize}}}}{{{x^{\large\frac{1}{3}\normalsize}}}},}$ so that
${{\left( {\frac{{dy}}{{dx}}} \right)^2} \text{ = }}\kern0pt {{\left[ { – \frac{{{{\left( {{a^{\large\frac{2}{3}\normalsize}} – {x^{\large\frac{2}{3}\normalsize}}} \right)}^{\large\frac{1}{2}\normalsize}}}}{{{x^{\large\frac{1}{3}\normalsize}}}}} \right]^2} } = {\frac{{{a^{\large\frac{2}{3}\normalsize}} – {x^{\large\frac{2}{3}\normalsize}}}}{{{x^{\large\frac{2}{3}\normalsize}}}} } = {\frac{{{a^{\large\frac{2}{3}\normalsize}}}}{{{x^{\large\frac{2}{3}\normalsize}}}} – 1.}$ Thus, the length of the astroid is
${L = 4\int\limits_0^a {\sqrt {1 + \frac{{{a^{\large\frac{2}{3}\normalsize}}}}{{{x^{\large\frac{2}{3}\normalsize}}}} – 1}\,dx} } = {4\int\limits_0^a {\frac{{{a^{\large\frac{1}{3}\normalsize}}}}{{{x^{\large\frac{1}{3}\normalsize}}}}dx} } = {4{a^{\large\frac{1}{3}\normalsize}}\int\limits_0^a {{x^{ – \large\frac{1}{3}\normalsize}}dx} } = {4{a^{\large\frac{1}{3}\normalsize}}\left[ {\left. {\left( {\frac{{{x^{\large\frac{2}{3}\normalsize}}}}{{\frac{2}{3}}}} \right)} \right|_0^a} \right] } = {4{a^{\large\frac{1}{3}\normalsize}} \cdot \frac{3}{2}{a^{\large\frac{2}{3}\normalsize}} } = {6a.}$

Page 1
Problems 1-2
Page 2
Problems 3-10