Geometric Applications of Line Integrals

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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 $r (t), α \leq t \leq β .$ Then the length of the curve is given by the line integral

L = \int_{C} d s = \int_{α}^{β} | \frac{d r}{d t} (t) | d t = \int_{α}^{β} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}} d t,

where $\frac{d r}{d t}$ is the derivative, and $x (t),$ $y (t),$ $z (t)$ are the components of the position vector $r (t) .$

If the curve $C$ is two-dimensional, the latter formula can be written in the form

L = \int_{C} d s = \int_{α}^{β} | \frac{d r}{d t} (t) | d t = \int_{α}^{β} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t .

If the curve $C$ is the graph of a continuous and differentiable function $y = f (x)$ in the $x y$ -plane, the length of the curve is given by

L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x .

Finally, if the curve $C$ is given by the equation $r = r (θ), α \leq θ \leq β$ in polar coordinates, and the function $r (θ)$ is continuous and differentiable in the interval $[α, β],$ the length of the curve is defined by the formula

L = \int_{α}^{β} \sqrt{{(\frac{d r}{d θ})}^{2} + r^{2}} d θ .

Area of a Region Bounded by a Closed Curve

If $C$ is a closed smooth piecewise curve in the $x y$ -plane (Figure $1$ ), the area of the region $R$ bounded by the curve is given by

S = \oint_{C} x d y = - \oint_{C} y d x = \frac{1}{2} \int_{C} x d y - y d x .

It is supposed here that the contour $C$ is traversed in the counterclockwise direction.

If the closed curve $C$ is given in parametric form $r (t) =$ $(x (t), y (t)),$ the area of the corresponding region can be calculated by the formula

S = \int_{α}^{β} x (t) \frac{d y}{d t} d t = - \int_{α}^{β} y (t) \frac{d x}{d t} d t = \frac{1}{2} \int_{α}^{β} [x (t) \frac{d y}{d t} - y (t) \frac{d x}{d t}] d t .

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

Let $R$ be a region in the half-plane $y \geq 0$ bounded by a closed smooth piecewise curve $C$ traversed in the counterclockwise direction. Suppose that the solid $Ω$ is formed by rotating the region $R$ about the $x$ -axis (Figure $2$ ).

Then the volume of the solid is given by

V = - π \oint_{C} y^{2} d x = - 2 π \oint_{C} x y d y = - \frac{π}{2} \oint_{C} 2 x y d y + y^{2} d x .

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 $0 \leq x \leq 5 a,$ $y \geq 0.$

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 $0 \leq x \leq 5 a,$ $y \geq 0.$

Solution.

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

The length of the arc is

L = \int_{0}^{5 a} \sqrt{1 + {[f^{'} (x)]}^{2}} d x = \int_{0}^{5 a} \sqrt{1 + {[\frac{d}{d x} (\sqrt{\frac{x^{3}}{a}})]}^{2}} d x = \int_{0}^{5 a} \sqrt{1 + {[\frac{\frac{3 x^{2}}{a}}{2 \sqrt{\frac{x^{3}}{a}}}]}^{2}} d x = \int_{0}^{5 a} \sqrt{1 + {[\frac{3 \sqrt{x}}{2 \sqrt{a}}]}^{2}} d x = \int_{0}^{5 a} \sqrt{1 + \frac{9 x}{4 a}} d x = \frac{1}{2 \sqrt{a}} \int_{0}^{5 a} \sqrt{4 a + 9 x} d x = \frac{1}{18 \sqrt{a}} [{(\frac{{(9 x + 4 a)}^{\frac{3}{2}}}{\frac{3}{2}}) |}_{x = 0}^{5 a}] = \frac{1}{27 \sqrt{a}} [{(45 a + 4 a)}^{\frac{3}{2}} - {(4 a)}^{\frac{3}{2}}] = \frac{1}{27 \sqrt{a}} [\sqrt{{(49 a)}^{3}} - \sqrt{{(4 a)}^{3}}] = \frac{a}{27} (\sqrt{49^{3}} - \sqrt{4^{3}}) = \frac{a}{27} (7^{3} - 2^{3}) = \frac{335 a}{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 = {(a^{\frac{2}{3}} - x^{\frac{2}{3}})}^{\frac{3}{2}}, where x \in [0, a] .

Then

\frac{d y}{d x} = \frac{3}{2} {(a^{\frac{2}{3}} - x^{\frac{2}{3}})}^{\frac{1}{2}} (- \frac{2}{3} x^{- \frac{1}{3}}) = - \frac{{(a^{\frac{2}{3}} - x^{\frac{2}{3}})}^{\frac{1}{2}}}{x^{\frac{1}{3}}},

so that

{(\frac{d y}{d x})}^{2} = {[- \frac{{(a^{\frac{2}{3}} - x^{\frac{2}{3}})}^{\frac{1}{2}}}{x^{\frac{1}{3}}}]}^{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_{0}^{a} \sqrt{1 + \frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}} - 1} d x = 4 \int_{0}^{a} \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}} d x = 4 a^{\frac{1}{3}} \int_{0}^{a} x^{- \frac{1}{3}} d x = 4 a^{\frac{1}{3}} [{(\frac{x^{\frac{2}{3}}}{\frac{2}{3}}) |}_{0}^{a}] = 4 a^{\frac{1}{3}} \cdot \frac{3}{2} a^{\frac{2}{3}} = 6 a .