Line integral x dy − y dx along the parabola y = x³

This problem asks to evaluate a line integral of the second kind along the parabola y = x³ using parametrization by the variable t.

Problem

Evaluate the line integral:

\int_{L} x d y - y d x, L : y = x^{3}, O (0; 0) \to A (1; 1)

Solution

Step 1

The curve L is given by the explicit equation y = x³. Introduce the parameter t = x, then y = t³. The point O(0;0) corresponds to t = 0, and the point A(1;1) to t = 1.

Why choose the parameter t = x?

When the curve is given as the graph of a function y = f(x), the simplest parametrization is x = t. Then y = f(t), and the parameter limits coincide with the limits of x along the curve.

x = t, y = t^{3}, t \in [0; 1]

Step 2

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

d x = d t, d y = 3 t^{2} d t

Step 3

Substitute into the integrand x dy − y dx:

x d y - y d x = t \cdot 3 t^{2} d t - t^{3} d t = 2 t^{3} d t

Step 4

Pass to a definite integral over the parameter t:

Why are the limits from 0 to 1?

The limits correspond to the values of the parameter t at the start and end points of the curve: t = 0 at O(0;0) and t = 1 at A(1;1). The direction of traversal is preserved — from O to A.

\int_{L} x d y - y d x = \int_{0}^{1} 2 t^{3} d t

Step 5

Evaluate the integral using the Newton–Leibniz formula:

2 \cdot \frac{t^{4}}{4} |_{0}^{1} = \frac{1}{2}

Answer

1/2.

Frequently asked questions

How do I evaluate a line integral of the second kind?

A line integral of the second kind ∫_L P dx + Q dy reduces to a definite integral by parametrizing the curve L: x = x(t), y = y(t), t ∈ [a; b]. Then dx = x′(t) dt, dy = y′(t) dt, and the integral becomes a definite integral in t.

How do I parametrize a curve given by y = f(x)?

If the curve is given as the graph y = f(x) for x from a to b, the natural parametrization is x = t, y = f(t), t ∈ [a; b]. This reduces the line integral to an ordinary definite integral.

What affects the sign of a line integral of the second kind?

The sign depends on the direction of traversal: reversing the direction changes the sign of the integral. In our problem the direction is given from O(0;0) to A(1;1), so the limits go from 0 to 1.

What is the difference between line integrals of the first and second kind?

A line integral of the first kind (over arc length) does not depend on the direction of traversal and its integrand contains dl. A line integral of the second kind depends on direction and contains differentials dx, dy (or their combinations).