Calculus

Line Integrals

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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

    \[
    {L = \int\limits_C {ds} }
    = {\int\limits_\alpha ^\beta {\left| {\frac{{d\mathbf{r}}}{{dt}}\left( t \right)} \right|dt} \text{ = }}\kern0pt
    {\int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt}}
    \]

    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

    \[
    {S = \oint\limits_C {xdy} }={ – \oint\limits_C {ydx} }
    = {\frac{1}{2}\int\limits_C {xdy – ydx} .}
    \]

    Area of a Region Bounded by a Closed Curve
    Figure 1.

    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

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

    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 \(2\)).

    Volume of a Solid Formed by Rotating a Closed Curve 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 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 \({x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {a^{\large\frac{2}{3}\normalsize}}.\)

    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 \(3\)).

    Arc of the plane curve ay^2=x^3
    Figure 3.

    The length of the arc is

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

    Example 2.

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

    Solution.

    The astroid is shown in Figure \(4.\)

    Astroid
    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