# Calculus

## Line Integrals # Geometric Applications of Line Integrals

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

$L = \int\limits_C {ds} = \int\limits_\alpha ^\beta {\left| {\frac{{d\mathbf{r}}}{{dt}}\left( t \right)} \right|dt} = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt},$

where $$\frac{{d\mathbf{r}}}{{dt}}$$ 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.

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

$S = \int\limits_\alpha ^\beta {x\left( t \right)\frac{{dy}}{{dt}}dt} = - \int\limits_\alpha ^\beta {y\left( t \right)\frac{{dx}}{{dt}}dt} = \frac{1}{2}\int\limits_\alpha ^\beta {\left[ {x\left( t \right)\frac{{dy}}{{dt}} - y\left( t \right)\frac{{dx}}{{dt}}} \right]dt.}$

## 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 or tap a problem to see the 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 $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}.$

### 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 $${y^2} = {\frac{{{x^3}}}{a}}$$ or $$y = \pm \sqrt {\frac{{{x^3}}}{a}}.$$ As $$y \ge 0,$$ we take only the positive root in the equation of the curve (Figure $$3$$).

The length of the arc is

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

### Example 2.

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

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

$y = {\left( {{a^{\frac{2}{3}}} - {x^{\frac{2}{3}}}} \right)^{\frac{3}{2}}},\;\; \text{where}\;\; x \in \left[ {0,a} \right].$

Then

$\frac{{dy}}{{dx}} = \frac{3}{2}{\left( {{a^{\frac{2}{3}}} - {x^{\frac{2}{3}}}} \right)^{\frac{1}{2}}}\left( { - \frac{2}{3}{x^{ - \frac{1}{3}}}} \right) = - \frac{{{{\left( {{a^{\frac{2}{3}}} - {x^{\frac{2}{3}}}} \right)}^{\frac{1}{2}}}}}{{{x^{\frac{1}{3}}}}},$

so that

$\left( {\frac{{dy}}{{dx}}} \right)^2 = \left[ { - \frac{{{{\left( {{a^{\frac{2}{3}} - x^{\frac{2}{3}}}} \right)}^{\frac{1}{2}}}}}{{{x^{\frac{1}{3}}}}}} \right]^2 = \frac{{{a^{\frac{2}{3}} - x^{\frac{2}{3}}}}}{{{x^{\frac{2}{3}}}}} = \frac{{{a^{\frac{2}{3}}}}}{{{x^{\frac{2}{3}}}}} - 1.$

Thus, the length of the astroid is

$L = 4\int\limits_0^a {\sqrt {1 + \frac{{{a^{\frac{2}{3}}}}}{{{x^{\frac{2}{3}}}}} - 1}\,dx} = 4\int\limits_0^a {\frac{{{a^{\frac{1}{3}}}}}{{{x^{\frac{1}{3}}}}}dx} = 4{a^{\frac{1}{3}}}\int\limits_0^a {{x^{ - \frac{1}{3}}}dx} = 4{a^{\frac{1}{3}}}\left[ {\left. {\left( {\frac{{{x^{\frac{2}{3}}}}}{{\frac{2}{3}}}} \right)} \right|_0^a} \right] = 4{a^{\frac{1}{3}}} \cdot \frac{3}{2}{a^{\frac{2}{3}}} = 6a.$

See more problems on Page 2.