Vector calculus. Line integrals of the second kind. Work…

What it is

A line integral of the second kind is the integral of a vector field along a directed curve:

C \int F \cdot d r = C \int P d x + Q d y + R d z .

Physically it is the work done by the force $F$ in moving a point along the curve $C$ . Unlike the integral of the first kind, it depends on the direction: reversing the orientation changes the sign.

How to compute it

Parametrise $r (t)$ with $t$ running from the start of the curve to its end, and substitute $d x = x^{'} (t) d t$ and so on:

C \int F \cdot d r = a \int b [P x^{'} (t) + Q y^{'} (t) + R z^{'} (t)] d t .

The segment from $A$ to $B$ is parametrised as $r (t) = A + t (B - A)$ , $t \in [0, 1]$ — then $d r = (B - A) d t$ and there are no derivatives to compute.

A potential field — the short way

If there is a function $φ$ with $F = \nabla φ$ , then

C \int F \cdot d r = φ (B) - φ (A),

that is, the integral does not depend on the path, and around a closed contour it is zero. The test for a potential field in a simply connected domain: $rot F = 0$ ; in the plane this is the single equation $\partial Q / \partial x = \partial P / \partial y$ .

Check for a potential before you start computing: if the field is potential, the whole work reduces to substituting two points.

Common mistakes

Losing the sign when $t$ decreases along the curve. If it is more convenient to parametrise "backwards", integrate and change the sign.
Computing $∣ F ∣ d s$ instead of $F \cdot d r$ — these are different integrals, and they agree only where the field is everywhere tangent to the curve.
Testing for a potential in a domain that is not simply connected. There the condition $\partial Q / \partial x = \partial P / \partial y$ is necessary but not sufficient; the classic counterexample is the field $(\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ around the origin.

Problems

Problem 1 (An integral of the second kind along an arc of a parabola). Compute

Γ \int (x^{2} - y^{2}) d x,

where $Γ$ is the arc of the parabola $y = x^{2}$ from the point $O (0; 0)$ to the point $A (2; 4)$ .

Solution.

Parametrisation. The curve is given as a graph and the integration is in $d x$ , so $x$ itself is the convenient parameter. It runs from $0$ (the point $O$ ) to $2$ (the point $A$ ); the order of the limits matters, because an integral of the second kind depends on the direction.

Substitution. On the curve $y = x^{2}$ , so $y^{2} = x^{4}$ :

Γ \int (x^{2} - y^{2}) d x = \int_{0}^{2} (x^{2} - x^{4}) d x .

There is no $d y$ term in the statement, so the derivative $y^{'} (x)$ is not needed here — it would only be needed for a term of the form $Q d y$ .

Integration.

\int_{0}^{2} x^{2} d x = [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3}, \int_{0}^{2} x^{4} d x = [\frac{x ^{5}}{5}]_{0}^{2} = \frac{32}{5} .

Γ \int (x^{2} - y^{2}) d x = \frac{8}{3} - \frac{32}{5} = \frac{40 - 96}{15} = - \frac{56}{15} .

A negative answer is to be expected: over most of the interval $x^{4} > x^{2}$ .

Answer: $- \frac{56}{15}$ .

Problem 2 (An integral of the second kind along a segment in space). Compute

A B \int (x d x + y d y + (x + y - 1) d z),

where $A B$ is the straight segment from the point $A (1; 1; 1)$ to the point $B (2; 3; 4)$ .

Solution.

Parametrising the segment. The segment from $A$ to $B$ is always parametrised the same way: $r (t) = A + t (B - A)$ , $t \in [0; 1]$ .

Here $B - A = (1; 2; 3)$ , so $x = 1 + t$ , $y = 1 + 2 t$ , $z = 1 + 3 t$ , $d x = d t$ , $d y = 2 d t$ , $d z = 3 d t$ .

The differentials are just the components of the vector $B - A$ ; there is nothing separate to differentiate.

Substitution. Compute the third factor: $x + y - 1 = (1 + t) + (1 + 2 t) - 1 = 1 + 3 t$ .

The integrand is then

(1 + t) \cdot 1 + (1 + 2 t) \cdot 2 + (1 + 3 t) \cdot 3 = (1 + t) + (2 + 4 t) + (3 + 9 t) = 6 + 14 t .

Integration.

\int_{0}^{1} (6 + 14 t) d t = [6 t + 7 t^{2}]_{0}^{1} = 6 + 7 = 13.

Answer: $13$ .

Problem 3 (The work of a force along a quarter circle). A particle moves along an arc of the unit circle from the point $A (1; 0)$ to the point $B (0; 1)$ (anticlockwise) under the action of the force $F = x^{2} i + y^{2} j$ .

Find the work done by the force $F$ over this displacement.

Solution.

Work is a line integral of the second kind:

A = Γ \int F \cdot d r = Γ \int x^{2} d x + y^{2} d y .

Test for a potential — before computing anything. Here $P = x^{2}$ , $Q = y^{2}$ , and

\frac{\partial Q}{\partial x} = 0 = \frac{\partial P}{\partial y} .

The condition holds on the whole plane, and the plane is simply connected, so the field is potential. The potential is easy to guess:

φ (x, y) = \frac{x ^{3} + y ^{3}}{3}, \nabla φ = (x^{2}; y^{2}) = F .

The work therefore does not depend on the path and equals the difference of the potentials:

A = φ (B) - φ (A) = \frac{0 + 1}{3} - \frac{1 + 0}{3} = 0.

A check by direct computation. Parametrise the arc: $x = cos t$ , $y = sin t$ , with $t$ from $0$ to $π /2$ . Then $d x = - sin t d t$ , $d y = cos t d t$ and

A = \int_{0}^{π /2} (- cos^{2} t sin t + sin^{2} t cos t) d t = [\frac{cos ^{3} t}{3} + \frac{sin ^{3} t}{3}]_{0}^{π /2} = \frac{1}{3} - \frac{1}{3} = 0.

Both ways give zero. Physically this means that the positive work on one part of the arc is exactly cancelled by the negative work on another — and the same will happen on any path from $A$ to $B$ , because the field is potential.

Answer: $A = 0$ .

Problem 4 (One integral along two paths). Compute $\int_{O A} x y d x + (y - x) d y$ from the point $O (0; 0)$ to the point $A (1; 1)$ :

along the straight segment $O A$ ;
along the arc of the parabola $y = x^{2}$ .

Explain why the answers differ.

Solution.

Along the segment. On the line $y = x$ , $d y = d x$ , with the parameter $x$ from $0$ to $1$ :

\int_{0}^{1} (x \cdot x + (x - x)) d x = \int_{0}^{1} x^{2} d x = \frac{1}{3} .

Along the parabola. On the curve $y = x^{2}$ , $d y = 2 x d x$ :

\int_{0}^{1} (x \cdot x^{2} + (x^{2} - x) \cdot 2 x) d x = \int_{0}^{1} (3 x^{3} - 2 x^{2}) d x = \frac{3}{4} - \frac{2}{3} = \frac{1}{12} .

Why they differ. An integral of the second kind is independent of the path only for a potential field. The test in the plane is $\partial Q / \partial x = \partial P / \partial y$ . Here $P = x y$ , $Q = y - x$ , and $\partial Q / \partial x = - 1$ while $\partial P / \partial y = x$ : they are equal nowhere except on the line $x = - 1$ . The field is not potential, and the work depends on the path taken.

Green's theorem also explains the difference itself. The path "out along the segment, back along the parabola" is a closed contour that goes round the region between the curves clockwise. Therefore

\frac{1}{3} - \frac{1}{12} = - D \iint (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y = D \iint (1 + x) d x d y,

where $D$ is the region $x^{2} ⩽ y ⩽ x$ , $0 ⩽ x ⩽ 1$ . And indeed: $\int_{0}^{1} (1 + x) (x - x^{2}) d x = \frac{1}{4}$ , that is $\frac{1}{3} - \frac{1}{12}$ .

Answer: along the segment $\frac{1}{3}$ , along the parabola $\frac{1}{12}$ ; the field is not potential.

Problem 5 (An integral that need not be computed along the curve). Compute

\int_{Γ} (e^{x} sin y + y) d x + (e^{x} cos y + x) d y,

where $Γ$ is the arc of the curve $y = \frac{π}{2} x^{5}$ from the point $O (0; 0)$ to the point $A (1; \frac{π}{2})$ .

Solution.

Substituting $y = \frac{π}{2} x^{5}$ leads to integrals like $\int e^{x} sin (\frac{π}{2} x^{5}) d x$ , which have no elementary antiderivative. The curve is chosen so that direct computation fails — the signal to test the field for a potential.

The test. $P = e^{x} sin y + y$ , $Q = e^{x} cos y + x$ :

\frac{\partial P}{\partial y} = e^{x} cos y + 1 = \frac{\partial Q}{\partial x} .

The equality holds on the whole plane, and the plane is simply connected — the field is potential, and the integral does not depend on the path.

The potential. We look for $φ$ with $φ_{x} = P$ , $φ_{y} = Q$ . Integrating the first equation in $x$ : $φ = e^{x} sin y + x y + C (y)$ . Substituting into the second: $φ_{y} = e^{x} cos y + x + C^{'} (y)$ , whence $C^{'} (y) = 0$ . So $φ = e^{x} sin y + x y$ .

The answer through the potential.

\int_{Γ} P d x + Q d y = φ (A) - φ (O) = e sin \frac{π}{2} + \frac{π}{2} - 0 = e + \frac{π}{2} .

The part $y d x + x d y$ is worth recognising at once: it is $d (x y)$ . Practice in spotting exact differentials saves half the work of finding a potential.

Answer: $e + \frac{π}{2}$ .

Problem 6 (For which value of the parameter is the field potential). Consider the field $F = (a x y^{2} + y) i + (2 x^{2} y + x) j$ .

For which value of $a$ is the field potential?
For that $a$ , find the work done by the field in moving from the point $A (1; 1)$ to the point $B (2; 3)$ .

Solution.

The condition for a potential. The field is defined on the whole plane, and the plane is simply connected, so being potential is equivalent to $\partial P / \partial y = \partial Q / \partial x$ :

\frac{\partial P}{\partial y} = 2 a x y + 1, \frac{\partial Q}{\partial x} = 4 x y + 1.

The equality must hold identically, for all $x$ and $y$ , not at isolated points. The coefficients of $x y$ give $2 a = 4$ , that is $a = 2$ .

The potential. With $a = 2$ we have $P = 2 x y^{2} + y$ . Integrating in $x$ : $φ = x^{2} y^{2} + x y + C (y)$ . Then $φ_{y} = 2 x^{2} y + x + C^{'} (y)$ , and comparing with $Q$ gives $C^{'} (y) = 0$ . So $φ = x^{2} y^{2} + x y$ .

Check: $φ_{x} = 2 x y^{2} + y = P$ , $φ_{y} = 2 x^{2} y + x = Q$ .

The work. It does not depend on the path:

A = φ (B) - φ (A) = (4 \cdot 9 + 6) - (1 + 1) = 42 - 2 = 40.

For any other $a$ the work from $A$ to $B$ would depend on the path, and the second question would have no single answer — which is why it is asked only for the $a$ found.

Answer: $a = 2$ ; the work is $40$ .

Problem 7* (A curl-free field with non-zero circulation). Consider the field $F = \frac{- y i + x j}{x ^{2} + y ^{2}}$ .

Check that $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ at every point where the field is defined.
Compute the circulation of the field around the circle $x^{2} + y^{2} = 4$ , traversed anticlockwise.
Compute the circulation around the circle $(x - 3)^{2} + y^{2} = 1$ .
Explain why the test in part 1 did not make the circulation in part 2 zero.

Solution.

1. The derivatives. $P = - \frac{y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ :

\frac{\partial Q}{\partial x} = \frac{( x ^{2} + y ^{2} ) - 2 x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}, \frac{\partial P}{\partial y} = \frac{- ( x ^{2} + y ^{2} ) + 2 y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} .

They are equal everywhere except the origin, where the field is not defined.

2. The circle of radius 2. $x = 2 cos t$ , $y = 2 sin t$ , $t \in [0; 2 π]$ . Then $x^{2} + y^{2} = 4$ and

- y d x + x d y = 4 sin^{2} t d t + 4 cos^{2} t d t = 4 d t, \oint \frac{- y d x + x d y}{x ^{2} + y ^{2}} = \int_{0}^{2 π} d t = 2 π .

The radius cancelled: around any circle centred at the origin the circulation is $2 π$ .

3. A circle away from the origin. The disc $(x - 3)^{2} + y^{2} ⩽ 1$ does not contain the origin, the field is smooth in it, and Green's theorem applies: the integrand $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is zero, so the circulation is $0$ as well.

4. Why the test did not work. The equality $\partial Q / \partial x = \partial P / \partial y$ guarantees a potential only in a simply connected domain. The plane without the origin is not simply connected: the circle from part 2 cannot be shrunk to a point without passing through the removed point. Green's theorem cannot be applied to the disc $x^{2} + y^{2} ⩽ 4$ — there is a singularity inside.

This field does have a potential, but only locally: it is the polar angle, $φ = arctan \frac{y}{x}$ up to a constant. Going once round the origin, the angle increases by $2 π$ and comes back to the starting point with a different value — that is the non-zero circulation. Near the circle from part 3 the angle is single-valued, hence zero.

Answer: $2 π$ and $0$ .

Vector calculus. Line integrals of the second kind. Work of a field

What it is

How to compute it

A potential field — the short way

Common mistakes

Problems

line-integral-second-kind.ipynb