MULTIVARIABLE CALCULUS 10

Line Integrals

A line integral accumulates a quantity along a curve rather than across an interval, area or volume.

The main idea

An ordinary integral accumulates along a straight interval:

\[ \int_a^b f(x)\,dx. \]

A line integral allows the path itself to be curved.

\[ \boxed{ \int_C f\,ds } \]

Here \(C\) represents the curve and \(ds\) represents a small distance along that curve.

1. From an interval to a curve

In one-variable integration, a small contribution has the form

\[ f(x)\,dx. \]

Along a curve, the corresponding contribution is

\[ f\,ds. \]

So the total accumulation is obtained by adding contributions along the entire path.

\[ \boxed{ \text{small contribution} = f\times\text{small arc length} } \]

2. Parameterising a curve

A curve can be described using a parameter \(t\):

\[ \mathbf{r}(t) = \begin{pmatrix} x(t)\\ y(t) \end{pmatrix}, \qquad a\le t\le b. \]

As \(t\) changes, the point

\[ (x(t),y(t)) \]

moves along the curve.

3. Arc-length element

A small displacement along the curve is

\[ d\mathbf{r} = \begin{pmatrix} dx\\ dy \end{pmatrix}. \]

Its length is

\[ ds = \sqrt{dx^2+dy^2}. \]

If the curve is parameterised by \(t\), then

\[ \boxed{ ds = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 }\,dt } \]

4. Scalar line integral

For a scalar field

\[ f(x,y), \]

the line integral along \(C\) is

\[ \boxed{ \int_C f\,ds = \int_a^b f(x(t),y(t)) \|\mathbf{r}'(t)\| \,dt } \]

5. Simple example

Let \(C\) be the straight line

\[ \mathbf{r}(t) = \begin{pmatrix} t\\ t \end{pmatrix}, \qquad 0\le t\le1. \]

Suppose

\[ f(x,y)=x+y. \]

Along the curve,

\[ f(t,t)=2t. \]

6. Find \(ds\)

We have

\[ \mathbf{r}'(t) = \begin{pmatrix} 1\\ 1 \end{pmatrix}. \]

Therefore,

\[ \|\mathbf{r}'(t)\| = \sqrt{1^2+1^2} = \sqrt2. \]

\[ ds=\sqrt2\,dt. \]

7. Evaluate the integral

Therefore,

\[ \int_C f\,ds = \int_0^1 2t\sqrt2\,dt. \]

\[ = \sqrt2[t^2]_0^1. \]

Hence,

\[ \boxed{ \int_C f\,ds = \sqrt2 } \]

8. Physical interpretation: mass of a wire

Suppose a thin wire follows a curve \(C\) and has linear density

\[ \rho(x,y). \]

A small piece of length \(ds\) has approximate mass

\[ dm=\rho\,ds. \]

Therefore, the total mass is

\[ \boxed{ M = \int_C \rho\,ds } \]

9. Vector line integrals

Line integrals can also involve vector fields.

Let

\[ \mathbf{F}(x,y) = \begin{pmatrix} P(x,y)\\ Q(x,y) \end{pmatrix}. \]

Then the line integral

\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} } \]

measures how strongly the vector field acts along the direction of the path.

10. Parameter form

Since

\[ d\mathbf{r} = \mathbf{r}'(t)\,dt, \]

we have

\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \,dt } \]

11. Why the dot product appears

The vector field may not point in exactly the same direction as the path.

The dot product extracts the component of the field acting along the direction of movement.

\[ \boxed{ \mathbf{F}\cdot d\mathbf{r} = \text{field component along path} \times \text{small displacement} } \]

12. Work done by a force field

Suppose \(\mathbf{F}\) represents a force and an object moves along a path \(C\).

Then the work done is

\[ \boxed{ W = \int_C \mathbf{F}\cdot d\mathbf{r} } \]

Only the component of the force in the direction of movement contributes to work.

13. Simple vector-field example

Let

\[ \mathbf{F}(x,y) = \begin{pmatrix} x\\ y \end{pmatrix} \]

and let

\[ \mathbf{r}(t) = \begin{pmatrix} t\\ t \end{pmatrix}, \qquad 0\le t\le1. \]

Along the path,

\[ \mathbf{F}(\mathbf{r}(t)) = \begin{pmatrix} t\\ t \end{pmatrix}. \]

Also,

\[ \mathbf{r}'(t) = \begin{pmatrix} 1\\ 1 \end{pmatrix}. \]

14. Calculate the dot product

Then

\[ \mathbf{F}\cdot\mathbf{r}' = t+t = 2t. \]

Therefore,

\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 2t\,dt. \]

\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = 1 } \]

15. Diagram

16. Direction matters for vector line integrals

For

\[ \int_C \mathbf{F}\cdot d\mathbf{r}, \]

reversing the direction of the path changes the sign:

\[ \boxed{ \int_{-C} \mathbf{F}\cdot d\mathbf{r} = - \int_C \mathbf{F}\cdot d\mathbf{r} } \]

This is because the displacement vector \(d\mathbf{r}\) reverses direction.

17. Scalar line integrals behave differently

For a scalar line integral

\[ \int_C f\,ds, \]

the element \(ds\) represents distance and is nonnegative.

So reversing the path does not change the value:

\[ \boxed{ \int_{-C}f\,ds = \int_Cf\,ds } \]

18. Conservative vector fields

Suppose

\[ \mathbf{F}=\nabla\phi. \]

Then \(\mathbf{F}\) is a conservative vector field.

For such a field,

\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \phi(B)-\phi(A) } \]

where \(A\) and \(B\) are the starting and ending points of the path.

19. Why this is important

For a conservative field, the line integral depends only on the endpoints, not on the particular path taken between them.

\[ \boxed{ \text{conservative field} \Rightarrow \text{path independence} } \]

20. Closed curves

If the path begins and ends at the same point, it is called a closed curve.

The notation

\[ \oint_C \]

is often used for an integral around a closed curve.

For a conservative field,

\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 } \]

21. The fundamental theorem for line integrals

\[ \mathbf{F}=\nabla\phi, \]

then

\[ \boxed{ \int_C \nabla\phi\cdot d\mathbf{r} = \phi(B)-\phi(A) } \]

This is the multivariable analogue of

\[ \int_a^b F'(x)\,dx = F(b)-F(a). \]

So the Fundamental Theorem of Calculus extends naturally to curves and gradient fields.

22. Scalar and vector line integrals

\[ \boxed{ \int_C f\,ds = \text{accumulation of a scalar quantity along a curve} } \] \[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \text{accumulation of a vector field along the direction of a curve} } \]

23. The central idea

Line integrals extend integration from straight intervals to curved paths.

\[ \boxed{ \text{interval} \rightarrow \text{curve} } \] \[ \boxed{ dx \rightarrow ds \text{ or }d\mathbf{r} } \]

They connect geometry, vector fields, motion and accumulation in a single calculus framework.