What a line integral is
Before you start: you'll need vector fields and the gradient.
\[\int_C \vec F\cdot d\vec r = \int_C \big(F_1\,dx + F_2\,dy + F_3\,dz\big)\]
where \(d\vec r = dx\,\mathbf{i} + dy\,\mathbf{j} + dz\,\mathbf{k}\) is a small step along the curve \(C\).
Work done by a force
If \(\vec F\) is a force, \(\int_C \vec F\cdot d\vec r\) is the work done in moving along \(C\).
Circulation
For a closed curve, \(\oint_C \vec F\cdot d\vec r\) is called the circulation.
The idea in one picture
How to evaluate it
- Parametrise the curve: write \(x, y, z\) in terms of one parameter \(t\) (or use \(x\) itself if the curve is \(y = f(x)\)).
- Substitute \(x, y, z\) and \(dx, dy, dz\) into \(F_1\,dx + F_2\,dy + F_3\,dz\).
- Integrate over the range of the parameter, from the start point to the end point.
If \(\vec F = \nabla\phi\) (equivalently \(\nabla\times\vec F = \vec 0\)), the integral depends only on the end points: \[\int_A^B \vec F\cdot d\vec r = \phi(B) - \phi(A), \qquad \oint_C \vec F\cdot d\vec r = 0\]
So check the curl first. If it is zero, find the scalar potential and skip the parametrisation altogether.
Solved examples
Example 1: Along a straight line
EasyEvaluate \(\displaystyle\int_C \vec F\cdot d\vec r\) for \(\vec F = x^2\,\mathbf{i} + xy\,\mathbf{j}\) along \(y = x\) from \((0, 0)\) to \((1, 1)\).
- On \(y = x\): \(dy = dx\). So \(\vec F\cdot d\vec r = x^2\,dx + x\cdot x\,dx = 2x^2\,dx\).
- \(\displaystyle\int_0^1 2x^2\,dx = \frac{2}{3}\).
Answer\(\dfrac{2}{3}\)
Example 2: Same field, different path
EasyEvaluate the same integral along \(y = x^2\) from \((0, 0)\) to \((1, 1)\).
- On \(y = x^2\): \(dy = 2x\,dx\). So \(\vec F\cdot d\vec r = x^2\,dx + x\cdot x^2\cdot 2x\,dx = (x^2 + 2x^4)\,dx\).
- \(\displaystyle\frac{1}{3} + \frac{2}{5}\).
Answer\(\dfrac{11}{15}\). It differs from Example 1, so this field is not conservative.
Example 3: Work along a twisted curve
MediumFind the work done by \(\vec F = (3x^2 + 6y)\,\mathbf{i} - 14yz\,\mathbf{j} + 20xz^2\,\mathbf{k}\) along \(x = t\), \(y = t^2\), \(z = t^3\) from \(t = 0\) to \(t = 1\).
- \(dx = dt\), \(dy = 2t\,dt\), \(dz = 3t^2\,dt\).
- \(\vec F\cdot d\vec r = (3t^2 + 6t^2)\,dt - 14t^5\cdot 2t\,dt + 20t^7\cdot 3t^2\,dt = (9t^2 - 28t^6 + 60t^9)\,dt\).
- \(\displaystyle\int_0^1 = 3 - 4 + 6\).
Answer\(5\)
Example 4: Using a potential
MediumEvaluate \(\displaystyle\int_C \vec F\cdot d\vec r\) for \(\vec F = (2xy + z^3)\,\mathbf{i} + x^2\,\mathbf{j} + 3xz^2\,\mathbf{k}\) from \((1, -2, 1)\) to \((3, 1, 4)\) along any path.
- \(\nabla\times\vec F = \vec 0\), with potential \(\phi = x^2y + xz^3\) (see scalar potential).
- \(\phi(3, 1, 4) - \phi(1, -2, 1) = (9 + 192) - (-2 + 1)\).
Answer\(202\), whatever the path
Example 5: Circulation round a circle
MediumFind \(\displaystyle\oint_C \vec F\cdot d\vec r\) for \(\vec F = -y\,\mathbf{i} + x\,\mathbf{j}\) round the unit circle, anticlockwise.
- \(x = \cos t\), \(y = \sin t\), \(0 \le t \le 2\pi\). \(dx = -\sin t\,dt\), \(dy = \cos t\,dt\).
- \(\vec F\cdot d\vec r = (\sin^2 t + \cos^2 t)\,dt = dt\).
Answer\(2\pi\). Not zero, because this field swirls (its curl is \(2\mathbf{k}\)).
Before parametrising anything, check whether \(\nabla\times\vec F = \vec 0\). If it is, the answer is just \(\phi(B) - \phi(A)\), and a closed-curve integral is 0. That saves most of the work in many exam questions.
Common mistakes
1. Forgetting to convert \(dy\) and \(dz\)
Every differential must be written in terms of the parameter: on \(y = x^2\), \(dy = 2x\,dx\).
2. Wrong direction
Reversing the direction of travel changes the sign of the integral.
3. Assuming path independence
Only conservative fields give the same answer on every path. Examples 1 and 2 show what happens otherwise.
Practice questions
Q1\(\displaystyle\int_C (x^2 + y^2)\,dx\) along \(y = x\) from \((0, 0)\) to \((1, 1)\)
\(\displaystyle\int_0^1 2x^2\,dx = \frac{2}{3}\).
Q2\(\displaystyle\int_C (y\,dx + x\,dy)\) from \((0, 0)\) to \((1, 1)\) along any path
\(\vec F = \nabla(xy)\), so the answer is \(1\cdot 1 - 0 = 1\).
Q3\(\displaystyle\oint_C (x\,dx + y\,dy)\) round the unit circle
\(\vec F = \nabla\left(\tfrac{x^2 + y^2}{2}\right)\) is conservative, so the integral is 0.
Q4\(\displaystyle\int_C (xy\,dx + y^2\,dy)\) along \(x = t\), \(y = t^2\), \(0 \le t \le 1\)
\(\displaystyle\int_0^1 (t^3 + 2t^5)\,dt = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}\).
Q5\(\displaystyle\oint_C (x\,dy - y\,dx)\) round the circle of radius \(a\), anticlockwise
\(\displaystyle\int_0^{2\pi} a^2\,dt = 2\pi a^2\): twice the area enclosed.
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Frequently asked questions
What does a line integral represent?
The total of the component of \(\vec F\) along the curve. For a force field, it is the work done.
When is a line integral independent of the path?
When \(\vec F\) is conservative: \(\vec F = \nabla\phi\), equivalently \(\nabla\times\vec F = \vec 0\) on a simply connected region.
Where line integrals lead next
← All 33 Calculus topics
Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.