(a^2-x^2)^1/2 Integral Formula

4 min read Jun 16, 2024
(a^2-x^2)^1/2 Integral Formula

Understanding the Integral of (a^2 - x^2)^1/2

The integral of (a^2 - x^2)^1/2 is a common integral that appears in various fields, including physics, engineering, and mathematics. This integral represents the area under the curve of the function (a^2 - x^2)^1/2, and it has a closed-form solution involving trigonometric functions.

The Formula

The integral of (a^2 - x^2)^1/2 is given by:

∫(a^2 - x^2)^1/2 dx = (1/2) * [x * (a^2 - x^2)^1/2 + a^2 * arcsin(x/a)] + C

where:

  • a is a constant
  • x is the variable of integration
  • C is the constant of integration

Derivation of the Formula

The derivation of this formula involves using trigonometric substitution and the following steps:

  1. Substitution: Let x = a * sin(θ), then dx = a * cos(θ) dθ
  2. Simplify: Substitute x and dx in the integral, and simplify using trigonometric identities.
  3. Integration: Integrate the simplified expression with respect to θ.
  4. Substitute back: Substitute θ back in terms of x using the original substitution.

Applications

The integral of (a^2 - x^2)^1/2 has numerous applications, including:

  • Finding the area of a semicircle: The integral represents the area of a semicircle with radius 'a'.
  • Calculating the volume of a sphere: The integral is used in calculating the volume of a sphere.
  • Solving problems in physics and engineering: The integral appears in various physics and engineering problems related to motion, oscillations, and waves.

Example

Let's calculate the definite integral of (a^2 - x^2)^1/2 from 0 to a:

∫(a^2 - x^2)^1/2 dx = (1/2) * [x * (a^2 - x^2)^1/2 + a^2 * arcsin(x/a)] + C

Evaluating at the limits of integration:

[(1/2) * [a * (a^2 - a^2)^1/2 + a^2 * arcsin(a/a)]] - [(1/2) * [0 * (a^2 - 0^2)^1/2 + a^2 * arcsin(0/a)]]

Simplifying:

[(1/2) * (a^2 * π/2)] - 0 = (1/4) * π * a^2

This result represents the area of a quarter circle with radius 'a'.

Conclusion

The integral of (a^2 - x^2)^1/2 is a fundamental integral with various applications in different fields. Understanding its derivation and applications allows for solving complex problems involving areas, volumes, and physical phenomena.

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