Line integral x dy − y dx along a circle

This problem asks to evaluate a line integral of the second kind along the full traversal of the unit circle.

Problem

Evaluate the line integral:

\int_{L} x d y - y d x, L : x = cos t, y = sin t, t \in [0; 2 π]

Solution

Step 1

The curve L is given parametrically: x = cos t, y = sin t. As the parameter t varies from 0 to 2π, the point traverses the unit circle counterclockwise, returning to the starting point.

x = cos t, y = sin t, t \in [0; 2 π]

Step 2

Find the differentials dx and dy in terms of the parameter t:

d x = - sin t d t, d y = cos t d t

Step 3

Substitute into the integrand x dy − y dx:

x d y - y d x = cos t \cdot cos t d t - sin t \cdot (- sin t) d t

= {cos}^{2} t d t + {sin}^{2} t d t

Step 4

Apply the basic trigonometric identity:

Why does the integrand simplify to dt?

After substitution we obtain cos²t dt + sin²t dt. The sum cos²t + sin²t equals 1 by the basic trigonometric identity. Hence the entire integrand reduces to dt — a typical result for a full traversal of a circle.

{cos}^{2} t + {sin}^{2} t = 1 \Rightarrow x d y - y d x = d t

Step 5

Pass to a definite integral and evaluate it:

\int_{L} x d y - y d x = \int_{0}^{2 π} d t = 2 π

Answer

2π.

Geometric meaning

A line integral of the form ∫_L x dy − y dx equals twice the area of the region bounded by the closed curve L (when traversed counterclockwise). For the unit circle the area is π, so the integral is 2π. This confirms the obtained result.

Frequently asked questions

How do I evaluate a line integral along a circle?

If the circle is given parametrically (x = R cos t, y = R sin t, t ∈ [0; 2π]), substitute the parametrization into the integrand, find dx and dy, and pass to a definite integral over t. The integrand often simplifies via trigonometric identities.

Why is the integral equal to 2π for a full traversal?

With the parametrization x = cos t, y = sin t the integrand x dy − y dx reduces to dt. Integrating over t from 0 to 2π gives the interval length 2π. Geometrically this is twice the area of the unit disk.

What changes if we traverse the circle clockwise?

Traversing clockwise changes the sign of the integral. This is because a line integral of the second kind depends on the direction of traversal. Mathematically this corresponds to reversing the limits of integration in t to [2π; 0].

How are line integrals and areas related?

There is a formula for the area via a line integral of the second kind: S = ½ ∮_L (x dy − y dx). It is valid for a closed curve L traversed counterclockwise and is a special case of Green's formula.