Line integral of the first kind along a helix

This problem asks to evaluate a line integral of the first kind along half a turn of a helix.

Problem

Evaluate the line integral of the first kind:

\int_{L} \frac{z d l}{\sqrt{x^{2} + y^{2}}}, L : x = cos t, y = sin t, z = t, t \in [0; π]

Solution

Step 1

The curve L is given parametrically: x = cos t, y = sin t, z = t. As the parameter t varies from 0 to π, the point traces half a turn of the helix around the Oz axis, simultaneously rising to height π.

x = cos t, y = sin t, z = t, t \in [0; π]

Step 2

Find the derivatives of the coordinates with respect to t:

x^{'} (t) = - sin t, y^{'} (t) = cos t, z^{'} (t) = 1

Step 3

Evaluate the arc-length element dl using the formula for a parametrically given curve:

Why is dl = √2 dt?

The arc-length element for a parametrically given space curve is dl = √((x′)² + (y′)² + (z′)²) dt. Here (x′)² = sin²t, (y′)² = cos²t, (z′)² = 1. The sum sin²t + cos²t + 1 = 2 by the basic trigonometric identity, hence dl = √2 dt — a constant along the entire curve.

d l = \sqrt{{(- sin t)}^{2} + {(cos t)}^{2} + 1^{2}} d t = \sqrt{2} d t

Step 4

Simplify the denominator of the integrand using the basic trigonometric identity:

\sqrt{x^{2} + y^{2}} = \sqrt{{cos}^{2} t + {sin}^{2} t} = 1

Step 5

Substitute the found expressions into the integrand:

\frac{z d l}{\sqrt{x^{2} + y^{2}}} = \frac{t \cdot \sqrt{2} d t}{1} = t \sqrt{2} d t

Step 6

Pass to a definite integral over the parameter t:

\int_{L} \frac{z d l}{\sqrt{x^{2} + y^{2}}} = \sqrt{2} \int_{0}^{π} t d t

Step 7

Evaluate the integral using the Newton–Leibniz formula:

\sqrt{2} \cdot \frac{t^{2}}{2} |_{0}^{π} = \sqrt{2} \cdot \frac{π^{2}}{2} = \frac{π^{2} \sqrt{2}}{2}

Answer

π²√2 / 2.

Frequently asked questions

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

A line integral of the first kind ∫_L f(x, y, z) dl reduces to a definite integral by parametrizing the curve: x = x(t), y = y(t), z = z(t), t ∈ [a; b]. Then dl = √((x′)² + (y′)² + (z′)²) dt, and the integral becomes ∫_a^b f(x(t), y(t), z(t))·√((x′)² + (y′)² + (z′)²) dt.

What is a helix and what are its properties?

A helix is a space curve given by x = R cos t, y = R sin t, z = ct. It winds around a circular cylinder of radius R with a step of 2πc per full turn. In our problem R = 1, c = 1, so the projection onto the Oxy plane is the unit circle, and the height grows linearly with t.

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 contains dl. A line integral of the second kind depends on direction and contains dx, dy, dz. In our problem the integral is of the first kind, so the answer does not change if the direction of traversal is reversed.

Why does the answer contain π² and √2?

The factor √2 comes from the arc-length element dl = √2 dt (the sum of squares of derivatives equals 2). The factor π² arises when integrating t dt from 0 to π: ∫₀^π t dt = π²/2. Together we obtain π²√2/2.