Line Integrals

Contents

Line Integrals

Surface Integrals

Volume Integrals

Greens Theorem

Stokes Theorem

Gauss’s Divergence Theorem

1M Questions

The concept of a line integral is a simple and natural generalization of the definite integral \[ \int\limits_{a}^{b} f(x) \, dx \] from elementary calculus.

While \(\int_{a}^{b} f(x) \, dx\) integrates \(f(x)\) along the x-axis from \(x = a\) to \(x = b\), a line integral integrates a given function along a curve in a plane or in space.

Given a vector field \(\bar{F} = F_{1} \hat{i} + F_{2} \hat{j} + F_{3} \hat{k}\) and an infinitesimal displacement vector \(d\bar{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}\), the line integral of \(\bar{F}\) along a curve \(C\) is: \[ \int\limits_{C} \bar{F} \cdot d\bar{r} = \int\limits_{C} F_{1} dx + F_{2} dy + F_{3} dz \]

If the path of integration \(C\) is a closed curve, the integral is denoted with a circle: \[ \oint\limits_{C} \bar{F} \cdot d\bar{r} = \oint\limits_{C} F_{1} dx + F_{2} dy + F_{3} dz \]

Applications of Line Integrals

  • Work done by a force: If \(\bar{F}\) represents the force acting on a particle moving along the arc \(P_1P_2\), then the work done by \(\bar{F}\) during the displacement from \(P_1\) to \(P_2\) is: \[ W = \int_{P_1}^{P_2} \bar{F} \cdot d\bar{r} \]
  • Circulation: If \(\bar{F}\) denotes the velocity of a fluid, then the circulation of \(\bar{F}\) around a simple closed curve \(C\) is defined as: \[ \text{Circulation} = \oint\limits_{C} \bar{F} \cdot d\bar{r} \]

Important Notes:

  1. If \(\oint\limits_{C} \bar{F} \cdot d\bar{r} = 0\), then \(\bar{F}\) is said to be irrotational.
  2. If the force \(\bar{F}\) is conservative, then there exists a scalar potential \(\phi\) such that \(\bar{F} = \nabla \phi\).
  3. For a conservative vector field \(\bar{F}\), the work done from \(P_1\) to \(P_2\) can be found as \(\phi(P_2) - \phi(P_1)\) without evaluating the line integral.

Example 1: Evaluate the line integral \(\int_C y^{2}dx - x^{2}dy\) about the triangle with vertices \((1,0)\), \((0,1)\), and \((-1,0)\).

Solution: Let \(A(1,0)\), \(B(0,1)\), and \(C(-1,0)\) be the vertices of \(\Delta ABC\). The path of integration \(C\) consists of three line segments: \(AB\), \(BC\), and \(CA\).

\[ \int\limits_{C} y^{2}dx - x^{2}dy = \int\limits_{AB} y^{2}dx - x^{2}dy + \int\limits_{BC} y^{2}dx - x^{2}dy + \int\limits_{CA} y^{2}dx - x^{2}dy \]

Equation of line AB:

\[ \frac{y-0}{x-1} = \frac{1-0}{0-1} \Rightarrow y = 1 - x \]

Thus, \(dy = -dx\) and \(x\) varies from 1 to 0:

\[ \int\limits_{AB} y^{2}dx - x^{2}dy = \int\limits_{1}^{0} (1-x)^{2}dx - x^{2}(-dx) = \int\limits_{1}^{0} [(1-2x+x^{2}) + x^{2}]dx \] \[ = \int\limits_{1}^{0} (1 - 2x + 2x^{2})dx = \left[ x - x^{2} + \frac{2}{3}x^{3} \right]_{1}^{0} = -\frac{2}{3} \]

Equation of line BC:

\[ \frac{y-1}{x-0} = \frac{0-1}{-1-0} \Rightarrow y = 1 + x \]

Thus, \(dy = dx\) and \(x\) varies from 0 to -1:

\[ \int\limits_{BC} y^{2}dx - x^{2}dy = \int\limits_{0}^{-1} (1+x)^{2}dx - x^{2}dx = \int\limits_{0}^{-1} (1 + 2x + x^{2} - x^{2})dx \] \[ = \int\limits_{0}^{-1} (1 + 2x)dx = \left[ x + x^{2} \right]_{0}^{-1} = 0 \]

Equation of line CA is \(y = 0\), so \(dy = 0\) and \(x\) varies from -1 to 1:

\[ \int\limits_{CA} y^{2}dx - x^{2}dy = \int\limits_{-1}^{1} 0 - x^{2}(0) = \int\limits_{-1}^{1} 0 = 0 \]
\[ \int\limits_{C} y^{2}dx - x^{2}dy = -\frac{2}{3} + 0 + 0 = \boxed{-\frac{2}{3}} \]

Example 2: Find the total work done by the force \(\bar{F} = 3xy\hat{i} - y\hat{j} + 2xz\hat{k}\) in moving a particle around the circle \(x^{2} + y^{2} = 4\).

Solution:

Since the curve \(x^{2} + y^{2} = 4\) lies in the \(xy\)-plane:

\[ z = 0 \quad \text{and} \quad dz = 0 \]

The dot product \(\bar{F} \cdot d\bar{r}\) becomes:

\[ \bar{F} \cdot d\bar{r} = (3xy\hat{i} - y\hat{j} + 2xz\hat{k}) \cdot (dx\hat{i} + dy\hat{j}) = 3xy\,dx - y\,dy \]

Using polar coordinates for the circle of radius 2:

\[ x = 2\cos\theta, \quad y = 2\sin\theta \] \[ dx = -2\sin\theta\,d\theta, \quad dy = 2\cos\theta\,d\theta \]

The total work done is:

\[ W = \oint\limits_{C} \bar{F} \cdot d\bar{r} = \oint\limits_{C} (3xy\,dx - y\,dy) \]

Substituting the parametric equations:

\[ W = \int_{0}^{2\pi} \left[3(2\cos\theta)(2\sin\theta)(-2\sin\theta) - (2\sin\theta)(2\cos\theta)\right] d\theta \] \[ = \int_{0}^{2\pi} \left[-24\sin^{2}\theta\cos\theta - 4\sin\theta\cos\theta\right] d\theta \]

Breaking into two parts:

\[ \int -24\sin^{2}\theta\cos\theta\,d\theta = -24\left(\frac{\sin^{3}\theta}{3}\right) + C_1 \] \[ \int -4\sin\theta\cos\theta\,d\theta = -2\sin^{2}\theta + C_2 \]

Evaluating from 0 to \(2\pi\):

\[ W = \left[-8\sin^{3}\theta - 2\sin^{2}\theta\right]_{0}^{2\pi} = 0 - 0 = 0 \]
\[ W = \boxed{0} \]

Model Problems on Line Integrals

  1. If \(\bar{F} = 3xy\hat{i} - y^{2}\hat{j}\), evaluate \(\int\limits_{C} \bar{F} \cdot d\bar{r}\), where \(C\) is the curve \(y = 2x^{2}\) in the \(xy\)-plane from \((0,0)\) to \((1,2)\).
    Answer: \(-\dfrac{7}{6}\)
  2. Find the work done in moving a particle in the force field \(\bar{F} = 3x^{2}\hat{i} + \hat{j} + 2z\hat{k}\) along the straight line from \((0,0,0)\) to \((2,1,3)\).
    Answer: \(18\)
  3. Find the work done in moving a particle in the force field \(\bar{F} = 3x^{2}\hat{i} + \hat{j} + 2z\hat{k}\) along the curve defined by \(x^{2} = 4y\), \(3x^{3} = 8z\) from \(x = 0\) to \(x = 2\).
    Answer: \(16\)
  4. If \(\bar{F} = xy\hat{i} - z\hat{j} + x^{2}\hat{k}\) and \(C\) is the curve \(x = t^{2}\), \(y = 2t\), \(z = t^{3}\) from \(t = 0\) to \(t = 1\), evaluate \(\int\limits_{C} \bar{F} \cdot d\bar{r}\).
    Answer: \(\dfrac{51}{70}\)
  5. Find the total work done by the force \(\bar{F} = (2x-y-z)\hat{i} + (x+y-z)\hat{j} + (3x-2y-5z)\hat{k}\) along the curve \(C\) in the \(xy\)-plane given by the circle \(x^{2} + y^{2} = 9\) and \(z = 0\).
    Answer: \(18\pi\)
  6. If \(\bar{F} = 3xy\hat{i} - y^{2}\hat{j}\), evaluate \(\int\limits_{C} \bar{F} \cdot d\bar{r}\) along the curve \(C\) in the \(xy\)-plane \(y = x^{3}\) from the point \((1,1)\) to \((2,8)\).
    Answer: \(35\)
  7. Using the line integral calculate the work done by the force \(\bar{F} = (3x^{2}+6y)\hat{i} - 14yz\hat{j} + 20xz^{2}\hat{k}\) along the lines from \((0,0,0)\) to \((1,0,0)\), then to \((1,1,0)\) and then to \((1,1,1)\).
    Answer: \(\dfrac{23}{3}\)
  8. Find the total work done in moving a particle in a force field given by \(\bar{F} = 3xy\hat{i} - 5z\hat{j} + 10x\hat{k}\) along the curve \(x = t^{2}+1\), \(y = 2t^{2}\), \(z = t^{3}\) from \(t = 1\) to \(t = 2\).
    Answer: \(303\)

Back to Top